THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Hard

1 Hard

Analyzing +4 | -1

Assertion :- At constant pressure of the change

\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})

work done is negative.

Reason :- During phase transition work done is always negative.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False

Option C is correct because the assertion is true as the work done is negative during expansion, but the reason is false as work done is not always negative during phase transition.

Solution approach

Correct Answer + Core Reason: If Assertion is True but the Reason is False. Work done is negative during expansion, but not always during phase transition.

Step-by-Step Solution:

Step 1: Analyze Assertion: $\mathrm{H}_2\mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{g})$ is a phase transition from solid to gas, which involves expansion.

Step 2: Work done during expansion: At constant pressure, work done is $w = -P\Delta V$ . Since $\Delta V$ is positive (expansion), $w$ is negative.

Step 3: Analyze Reason: Phase transitions can involve expansion (e.g., solid to gas) or contraction (e.g., gas to liquid).

Step 4: Work done during phase transition: If the volume increases during the phase transition, the work done is negative. If the volume decreases, the work done is positive.

Step 5: Conclusion: The assertion is true because the work done is negative. The reason is false because work done is not always negative during phase transition.

Why other options are wrong:
- Option A: Incorrect because the reason is false.
- Option B: Incorrect because the reason is false.
- Option D: Incorrect because the assertion is true.

Key Concepts:
- Work done by a gas
- Phase transition
- Expansion and contraction

Memory Aids:
- Expansion: Work done is negative.
- Compression: Work done is positive.

Quick Tips:
- Remember the sign convention for work: work done *by* the system is negative, work done *on* the system is positive.

2 Hard

2 of 111Question permalink

Application +4 | -1

For the reaction,

\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1}

at

300 \mathrm{~K}

. Hence,

\Delta G

is

View answer and explanation

Correct option: B

-2.7 \mathrm{~kcal}

Option B (-2.7 kcal) is correct because it is calculated using the formula

\Delta G = \Delta H - T\Delta S

, where

\Delta H

is derived from

\Delta U

and the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: -2.7 kcal. $\Delta G$ is calculated using the Gibbs-Helmholtz equation and the relationship between $\Delta H$ and $\Delta U$ .

Step-by-Step Solution:

Step 1: Calculate $\Delta H$ using the relation $\Delta H = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in the number of moles of gas.

Step 2: In the reaction $\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g})$ , $\Delta n_g = 2 - 0 = 2$ .

Step 3: Convert R to kcal/mol.K: $R = 2 \text{{ cal/mol.K}} = 0.002 \text{{ kcal/mol.K}}$

Step 4: Calculate $\Delta H$ : $\Delta H = 2.1 \text{{ kcal}} + (2 \text{{ mol}}) \times (0.002 \text{{ kcal/mol.K}}) \times (300 \text{{ K}}) = 2.1 + 1.2 = 3.3 \text{{ kcal}}$

Step 5: Calculate $\Delta G$ using the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$ .

Step 6: Substitute the values: $\Delta G = 3.3 \text{{ kcal}} - (300 \text{{ K}}) \times (20 \text{{ cal/K}}) \times (1 \text{{ kcal}}/1000 \text{{ cal}}) = 3.3 - 6 = -2.7 \text{{ kcal}}$

Why other options are wrong:
- Option A: 2.7 kcal is incorrect because it has the wrong sign. The correct calculation yields a negative value.
- Option C: 9.3 kcal is incorrect. This value likely arises from adding T $\Delta$ S to $\Delta$ U instead of subtracting it from $\Delta$ H.
- Option D: -9.3 kcal is incorrect. This value is not obtained from the correct application of the Gibbs-Helmholtz equation and the relationship between $\Delta$ H and $\Delta$ U.

Key Concepts:
- Gibbs Free Energy ( $\Delta G$ )
- Enthalpy ( $\Delta H$ )
- Internal Energy ( $\Delta U$ )
- Gibbs-Helmholtz Equation: $\Delta G = \Delta H - T\Delta S$
- Relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + \Delta n_g RT$

Memory Aids:
- G = H - TS (Gibbs = Hats - TimeS)

Quick Tips:
- Always convert units to be consistent (e.g., calories to kilocalories) before performing calculations.
- Pay attention to the sign conventions for $\Delta H$ , $\Delta S$ , and $\Delta G$ .

3 Hard

3 of 111Question permalink

Understanding +4 | -1

During complete combustion of one mole of butane,

2658 \mathrm{~kJ}

of heat is released. The thermochemical reaction for above change is

View answer and explanation

Correct option: C

\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option C (

\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is correct because it represents the complete combustion of one mole of butane with the correct stoichiometric coefficients and the given enthalpy change.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . This equation correctly represents the combustion of one mole of butane with the given enthalpy change.

Step-by-Step Solution:
Step 1: Write the balanced chemical equation for the complete combustion of butane:
$\mathrm{C}_4 \mathrm{H}_{10}(g) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(l)$
Step 2: Balance the equation:
$\mathrm{C}_4 \mathrm{H}_{10}(g) + \frac{13}{2} \mathrm{O}_2(g) \rightarrow 4 \mathrm{CO}_2(g) + 5 \mathrm{H}_2\mathrm{O}(l)$
Step 3: Include the enthalpy change, which is given as -2658 kJ/mol (since heat is released).
$\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Why other options are wrong:
- Option A: This equation represents the combustion of two moles of butane, not one, and the enthalpy change is for two moles.
- Option B: The equation is for one mole of butane, but the enthalpy change is half of the given value. This is incorrect because the problem states 2658 kJ is released per mole.
- Option D: The equation is correct, but the sign of the enthalpy change is incorrect. Combustion is an exothermic process, so the enthalpy change must be negative.

Key Concepts:
- Thermochemical equations
- Enthalpy of combustion
- Balancing chemical equations

Memory Aids:
- Remember that combustion reactions are exothermic, so $\Delta H$ is negative.

Quick Tips:
- Always balance the chemical equation before writing the thermochemical equation.
- Pay attention to the sign of the enthalpy change (positive for endothermic, negative for exothermic).

4 Hard

4 of 111Question permalink

Application +4 | -1

What is the internal energy

({kJ})

change occurs when

36 {~g}

of

{H}_2 {O}(l)

converted to

{H}_2 {O}({g})

?

\Delta {H}^{\circ}

(vapourisation)

=

40.79 {~kJ} / {mol}

View answer and explanation

Correct option: A

75.38

Option A (75.38) is correct because it is calculated by finding the change in internal energy using the enthalpy of vaporization and accounting for the work done during the phase change.

Solution approach

Correct Answer + Core Reason: 75.38. The internal energy change is calculated using the enthalpy of vaporization and accounting for the work done during the phase change.

Step-by-Step Solution:
Step 1: Calculate the number of moles of water: Molar mass of water = 18 g/mol. Moles = Mass / Molar mass = 36 g / 18 g/mol = 2 mol.
Step 2: Calculate the enthalpy change: $\Delta H = n \times \Delta H_{vap} = 2 \times 40.79 kJ/mol = 81.58 kJ$ .
Step 3: Calculate the change in the number of moles of gas: $\Delta n_g = n_{gas} - n_{liquid} = 2 - 0 = 2$ .
Step 4: Calculate the work done: $W = \Delta n_g RT = 2 \times 8.314 J/mol K \times 373 K = 6202.744 J = 6.203 kJ$ .
Step 5: Calculate the internal energy change: $\Delta U = \Delta H - W = 81.58 kJ - 6.203 kJ = 75.377 kJ ≈ 75.38 kJ$ .

Why other options are wrong:
- Option B: 80.98. This is incorrect; the calculation leads to 75.38 kJ.
- Option C: 70.98. This is incorrect; the calculation leads to 75.38 kJ.
- Option D: 45.89. This is incorrect; the calculation leads to 75.38 kJ.

Key Concepts:
- Enthalpy of Vaporization
- Internal Energy Change
- Work Done
- Relationship between $\Delta H$ and $\Delta U$

Memory Aids:
- Remember the formula: $\Delta H = \Delta U + \Delta n_g RT$

Quick Tips:
- Ensure the units are consistent before performing calculations.
- Pay attention to the phase transition occurring.

5 Hard

5 of 111Question permalink

Application +4 | -1

What is the change in internal energy when a gas contracts from

325 \mathrm{~ml}

to

125 \mathrm{~ml}

at a constant pressure of

2 \mathrm{~bar},

while at the same time being cooled by removing

124 \mathrm{~J}

heat?

View answer and explanation

Correct option: B

-84 \mathrm{~J}

Option B (-84 J) is correct because it is calculated using the first law of thermodynamics, considering both the heat removed and the work done due to contraction.

Solution approach

Correct Answer + Core Reason: -84 J. The change in internal energy is calculated using the first law of thermodynamics: $\Delta U = Q + W$ .

Step-by-Step Solution:

Step 1: Identify the given information: Volume change $\Delta V = 125 \text{ ml} - 325 \text{ ml} = -200 \text{ ml} = -200 \times 10^{-6} \text{ m}^3$ , Pressure $P = 2 \text{ bar} = 2 \times 10^5 \text{ Pa}$ , Heat $Q = -124 \text{ J}$ .

Step 2: Calculate the work done ( $W$ ) at constant pressure using the formula: $W = -P\Delta V$ .

Step 3: Substitute the values: $W = -(2 \times 10^5 \text{ Pa})(-200 \times 10^{-6} \text{ m}^3)$ .

Step 4: Calculate the work: $W = 40 \text{ J}$ .

Step 5: Apply the first law of thermodynamics: $\Delta U = Q + W = -124 \text{ J} + 40 \text{ J}$ .

Step 6: Calculate the change in internal energy: $\Delta U = -84 \text{ J}$ .

Why other options are wrong:
- Option A (-524 J): This is incorrect because it likely involves an error in calculating the work done or misapplying the sign convention for heat and work.
- Option C (-164 J): This might be obtained if the work is incorrectly calculated or if the sign of the work is not properly considered.
- Option D (+84 J): This is incorrect because the heat removed is negative, and the work done is positive, leading to a net negative change in internal energy.

Key Concepts:
- First law of thermodynamics: $\Delta U = Q + W$
- Work done at constant pressure: $W = -P\Delta V$
- Sign conventions for heat and work

Memory Aids:
- $\Delta U = Q + W$ (Internal energy change equals heat added plus work done on the system)
- Work done *by* the system is negative; work done *on* the system is positive.

Quick Tips:
- Always pay attention to the sign conventions for heat and work.
- Ensure consistent units (e.g., convert ml to m³ and bar to Pa).
- Remember that work done by the system is negative, while work done on the system is positive.

6 Hard

6 of 111Question permalink

Analyzing +4 | -1

An ideal gas was subjected to a reversible adiabatic expansion then its initial volume was restored by reversible, isothermal compression. If '

q

' denotes the heat added to the system and

w

the work done on the system for the overall process then

View answer and explanation

Correct option: A

w

< 0

, q

< 0

Option A (w < 0, q < 0) is correct because in a reversible adiabatic expansion, no heat is exchanged (q=0), and the gas does work, so w < 0. During isothermal compression, work is done on the system, and heat is released, so w < 0 and q < 0 overall.

Solution approach

Correct Answer + Core Reason: w $< 0$ , q $< 0$ . In a reversible adiabatic expansion, no heat is exchanged, and the gas does work. During isothermal compression, work is done on the system, and heat is released.

Step-by-Step Solution:
Step 1: Analyze the reversible adiabatic expansion: In an adiabatic process, $q = 0$ . The gas expands, doing work on the surroundings, so $w < 0$ . The internal energy decreases, and the temperature drops.
Step 2: Analyze the reversible isothermal compression: The volume is restored to its initial value. Work is done on the gas, so $w > 0$ . Since the temperature is constant, the internal energy remains constant. Heat is released to the surroundings, so $q < 0$ .
Step 3: Consider the overall process: The net work done is negative because the work done by the gas during expansion is greater than the work done on the gas during compression. The net heat exchange is negative because heat is released during compression. Therefore, $w < 0$ and $q < 0$ .

Why other options are wrong:
- Option B: $w>0$ , q $< 0$ . This is incorrect because the net work done is negative.
- Option C: $w < 0, q>0$ . This is incorrect because the net heat exchange is negative.
- Option D: $w>0$ , q $>0$ . This is incorrect because both the net work done and heat exchange are negative.

Key Concepts:
- Adiabatic Process
- Isothermal Process
- Work Done
- Heat Exchange
- Thermodynamics

Memory Aids:
- Remember that in an adiabatic process, $q = 0$ .
- Remember that in an isothermal process, the temperature is constant.

Quick Tips:
- Understand the sign conventions for work and heat.
- Know the characteristics of adiabatic and isothermal processes.

7 Hard

7 of 111Question permalink

Application +4 | -1

A

20 \mathrm{~g}

piece of iron (specific heat capacity

=0.5 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}

) at

200^{\circ} \mathrm{C}

is immersed in

100 \mathrm{~g}

water of

27^{\circ} \mathrm{C}

.
What will be the temperature

\left({ }^{\circ} \mathrm{C}\right)

of this system at thermal equilibrium?

View answer and explanation

Correct option: A

31

Option A (31) is correct because it is calculated using the principle of calorimetry, where heat lost by iron equals heat gained by water until thermal equilibrium is reached.

Solution approach

Correct Answer + Core Reason: 31. The final temperature is calculated by equating heat lost by iron to heat gained by water.

Step-by-Step Solution:

Step 1: Identify the principle: Heat lost by iron = Heat gained by water.

Step 2: Write the formula for heat transfer: $Q = mc\Delta T$ , where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.

Step 3: Heat lost by iron: $Q_{iron} = m_{iron} \times c_{iron} \times (T_{iron, initial} - T_{final})$ \nHeat gained by water: $Q_{water} = m_{water} \times c_{water} \times (T_{final} - T_{water, initial})$

Step 4: Substitute the given values:
$Q_{iron} = 20 \times 0.5 \times (200 - T_{final})$
$Q_{water} = 100 \times 4.2 \times (T_{final} - 27)$

Step 5: Equate heat lost and heat gained:
$20 \times 0.5 \times (200 - T_{final}) = 100 \times 4.2 \times (T_{final} - 27)$
$10 \times (200 - T_{final}) = 420 \times (T_{final} - 27)$
$2000 - 10T_{final} = 420T_{final} - 11340$

Step 6: Solve for $T_{final}$ :
$430T_{final} = 13340$
$T_{final} = \frac{13340}{430} = 31.02 \approx 31^{\circ}C$

Why other options are wrong:
- Option B (0): This is incorrect. It's an unreasonable temperature given the initial temperatures of the iron and water.
- Option C (0): Same as Option B.
- Option D (0): Same as Option B.

Key Concepts:
- Calorimetry
- Heat transfer
- Specific heat capacity

Memory Aids:
- $Q = mc\Delta T$

Quick Tips:
- Ensure consistent units are used throughout the calculation.
- Remember that heat lost by one substance is gained by another in a closed system.

8 Hard

8 of 111Question permalink

Application +4 | -1

The enthalpy of vaporization of a liquid is 30 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

and entropy of vaporization is 75 J

\mathrm{mo}{\mathrm{l}^{ - 1}}

. The boiling point of liquid at 1 atm is :

View answer and explanation

Correct option: B

400 K

Option B (400 K) is correct. It is calculated using the Clausius-Clapeyron equation, where boiling point is determined by the enthalpy and entropy of vaporization.

Solution approach

Correct Answer + Core Reason: 400 K. The boiling point is calculated using the relationship between enthalpy and entropy of vaporization.

Step-by-Step Solution:

Step 1: Given: $\Delta H_{vap} = 30 \text{{ kJ/mol}} = 30000 \text{{ J/mol}}$ and $\Delta S_{vap} = 75 \text{{ J/mol K}}$

Step 2: At boiling point, $\Delta G = 0$ . We know that $\Delta G = \Delta H - T\Delta S$ .

Step 3: Therefore, at boiling point, $0 = \Delta H_{vap} - T_b\Delta S_{vap}$ .

Step 4: Rearrange to solve for $T_b$ (boiling point): $T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}}$

Step 5: Substitute the given values: $T_b = \frac{30000 \text{{ J/mol}}}{75 \text{{ J/mol K}}} = 400 \text{{ K}}$

Why other options are wrong:
- Option A (250 K): This is incorrect. It likely arises from dividing entropy by enthalpy instead of enthalpy by entropy, or from unit conversion errors.
- Option C (450 K): This is incorrect. It likely arises from calculation errors or using incorrect formulas.
- Option D (600 K): This is incorrect. It likely arises from calculation errors or using incorrect formulas.

Key Concepts:
- Gibbs Free Energy
- Enthalpy of Vaporization
- Entropy of Vaporization
- Boiling Point

Memory Aids:
- $\Delta G = \Delta H - T\Delta S$
- At boiling point, $\Delta G = 0$

Quick Tips:
- Ensure units are consistent before calculations.
- Remember the relationship between Gibbs Free Energy, Enthalpy, and Entropy.

9 Hard

9 of 111Question permalink

Analyzing +4 | -1

Statement I: A reaction which is spontaneous and
accompanied by decrease of randomness must be
exothermic.

Statement II: All exothermic reactions are
accompanied by decrease of randomness.

View answer and explanation

Correct option: C

If Statement I is CORRECT, but Statement II is INCORRECT

Option C is correct. Statement I is correct because for a spontaneous reaction with decreasing randomness (negative ΔS), the reaction must be exothermic (negative ΔH) to ensure ΔG is negative. Statement II is incorrect because not all exothermic reactions decrease randomness.

Solution approach

Correct Answer + Core Reason: If Statement I is CORRECT, but Statement II is INCORRECT. Spontaneity depends on both enthalpy and entropy changes.

Step-by-Step Solution:

Step 1: Recall the Gibbs Free Energy equation: $\Delta G = \Delta H - T\Delta S$

Step 2: Analyze Statement I: A spontaneous reaction has $\Delta G < 0$ . If the reaction is accompanied by a decrease in randomness, then $\Delta S < 0$ . To ensure $\Delta G < 0$ , $\Delta H$ must be negative (exothermic) and large enough to overcome the $-T\Delta S$ term. Therefore, Statement I is correct.

Step 3: Analyze Statement II: All exothermic reactions have $\Delta H < 0$ . However, this does not necessarily mean that they are accompanied by a decrease in randomness. Some exothermic reactions can lead to an increase in randomness ( $\Delta S > 0$ ). For example, the combustion of methane: $\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(g)$ is exothermic and increases randomness. Therefore, Statement II is incorrect.

Step 4: Conclusion: Statement I is correct, and Statement II is incorrect.

Why other options are wrong:
- Option A: Both statements are not correct.
- Option B: Both statements are not correct.
- Option D: Statement II is not correct.

Key Concepts:
- Gibbs Free Energy
- Spontaneity
- Enthalpy
- Entropy

Memory Aids:
- $\Delta G = \Delta H - T\Delta S$ (Gives spontaneity)

Quick Tips:
- Remember that spontaneity depends on both enthalpy and entropy changes.

10 Hard

10 of 111Question permalink

Application +4 | -1

The heat evolved on combustion of

1 \, \mathrm{gm}

of starch,

\left({C}_6 {H}_{10}{O}_5\right)

, into

{CO}_2({g})

and

{H}_2 {O}({l})

is

4.18 \, \text{Kcal}.

What is the standard enthalpy of formation of

1 \, \mathrm{gm}

of starch? Heat of formation of

{CO}_2({g})

and

{H}_2 {O}({l})

are

-94.05

and

-68.32 \, \mathrm{Kcal} / \mathrm{mol}

.

View answer and explanation

Correct option: C

-1.41 \, \mathrm{Kcal}

Option C (

-1.41 \text{ Kcal}

) is correct because it is calculated by using the heat of combustion of starch and the heats of formation of CO2 and H2O to find the standard enthalpy of formation of starch.

Solution approach

Correct Answer + Core Reason: $-1.41 \text{ Kcal}$ . The standard enthalpy of formation of starch is calculated using the heat of combustion and the heats of formation of the products.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the combustion of starch:
$(C_6H_{10}O_5)_n(s) + 6nO_2(g) \rightarrow 6nCO_2(g) + 5nH_2O(l)$ \nFor 1 mole of $C_6H_{10}O_5$ : $C_6H_{10}O_5(s) + 6O_2(g) \rightarrow 6CO_2(g) + 5H_2O(l)$

Step 2: Calculate the molar mass of starch ( $C_6H_{10}O_5$ ):
$6(12) + 10(1) + 5(16) = 72 + 10 + 80 = 162 \text{ g/mol}$

Step 3: Calculate the heat of combustion per mole of starch:
$\Delta H_{combustion} = 4.18 \text{ Kcal/g} \times 162 \text{ g/mol} = 677.16 \text{ Kcal/mol}$ \nSince heat is evolved, $\Delta H_{combustion} = -677.16 \text{ Kcal/mol}$

Step 4: Use the equation for heat of combustion:
$\Delta H_{combustion} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
$-677.16 = [6 \times \Delta H_f(CO_2) + 5 \times \Delta H_f(H_2O)] - [\Delta H_f(C_6H_{10}O_5)]$

Step 5: Substitute the given heats of formation:
$-677.16 = [6(-94.05) + 5(-68.32)] - [\Delta H_f(C_6H_{10}O_5)]$
$-677.16 = [-564.3 - 341.6] - [\Delta H_f(C_6H_{10}O_5)]$
$-677.16 = -905.9 - \Delta H_f(C_6H_{10}O_5)]$

Step 6: Solve for $\Delta H_f(C_6H_{10}O_5)$ :
$\Delta H_f(C_6H_{10}O_5) = -905.9 + 677.16 = -228.74 \text{ Kcal/mol}$

Step 7: Calculate the heat of formation per gram of starch:
$\Delta H_f(\text{per gram}) = \frac{-228.74 \text{ Kcal/mol}}{162 \text{ g/mol}} = -1.41 \text{ Kcal/g}$

Why other options are wrong:
- **Option A ( $-2.82 \text{ Kcal}$ ): This is incorrect because it likely involves an error in the calculation of the molar mass or the heat of combustion per mole.
- Option B ( $-0.71 \text{ Kcal}$ ): This is incorrect because it likely involves dividing by the wrong factor or making a mistake in the arithmetic.
- Option D ( $-8.46 \text{ Kcal}$ ): This is incorrect because it likely involves a significant error in the calculation, possibly due to incorrect application of Hess's Law or incorrect unit conversions.

Key Concepts:
- Heat of combustion
- Heat of formation
- Hess's Law

Memory Aids:
- Products - Reactants (for enthalpy changes)

Quick Tips:**
- Always balance the chemical equation first.
- Pay attention to units and convert them if necessary.

11 Hard

11 of 111Question permalink

Application +4 | -1

An engine operating between 150℃ and 25℃ takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is

View answer and explanation

Correct option: D

147.7 J

Option D (147.7 J) is the correct work done by the engine, calculated using the Carnot efficiency formula.

Solution approach

Correct Answer + Core Reason: 147.7 J. The work done is calculated using Carnot efficiency.

Step-by-Step Solution:

Step 1: Convert temperatures to Kelvin:
- $T_H = 150^\circ C = 150 + 273 = 423 \text{{ K}}$
- $T_C = 25^\circ C = 25 + 273 = 298 \text{{ K}}$

Step 2: Calculate the Carnot efficiency ( $\eta$ ):
- $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{298}{423}$
- $\eta = 1 - 0.7045 = 0.2955$

Step 3: Calculate the work done (W) by the engine:
- $W = \eta \times Q_H$
- $W = 0.2955 \times 500 \text{{ J}} = 147.75 \text{{ J}}$

Step 4: Round to the nearest tenth: W = 147.7 J

Why other options are wrong:
- Option A: This option is incorrect, likely due to an error in calculating the efficiency or applying it to the heat input.
- Option B: This option is incorrect, likely due to an error in calculating the efficiency or applying it to the heat input.
- Option C: This option is incorrect, likely due to an error in calculating the efficiency or applying it to the heat input.

Key Concepts:
- Carnot efficiency: $\eta = 1 - \frac{T_C}{T_H}$
- Work done by an engine: $W = \eta \times Q_H$
- Temperature conversion from Celsius to Kelvin

Memory Aids:
- Remember the Carnot efficiency formula and its relation to work done.

Quick Tips:
- Always convert temperatures to Kelvin when dealing with thermodynamic calculations.
- Ensure you are using the correct heat input value ( $Q_H$ ) in the work calculation.

12 Hard

12 of 111Question permalink

Application +4 | -1

Heat of formation of two moles of

\mathrm{NH}_3(\mathrm{~g})

is

-90 \mathrm{KJ}

. Bond energies of

\mathrm{H}-\mathrm{H}

and

\mathrm{N}-\mathrm{H}

bonds are

435 \mathrm{KJ}

and

390 \mathrm{KJ} \mathrm{mol}^{-1}

respectively. The value of the bond energy of

N=N

will be

View answer and explanation

Correct option: D

945 \mathrm{KJ}

Option D (

945 \text{ KJ}

) is correct because it is calculated using the heat of formation and the given bond energies of H-H and N-H bonds, applying Hess's Law.

Solution approach

Correct Answer + Core Reason: $945 \text{ KJ}$ . The bond energy of N=N is calculated using the heat of formation of NH3 and the given bond energies.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation for the formation of ammonia:
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

Step 2: Define the heat of formation in terms of bond energies:
$\Delta H_f = \sum \text{Bond Energies (Reactants)} - \sum \text{Bond Energies (Products)}$

Step 3: Write the equation with given values:
$-90 \text{ KJ} = [1 \times BE(N=N) + 3 \times BE(H-H)] - [6 \times BE(N-H)]$ \nNote: 2 moles of $NH_3$ have 6 N-H bonds.

Step 4: Substitute the given bond energies:
$-90 = [BE(N=N) + 3(435)] - [6(390)]$
$-90 = BE(N=N) + 1305 - 2340$

Step 5: Solve for BE(N=N):
$BE(N=N) = -90 - 1305 + 2340$
$BE(N=N) = 945 \text{ KJ}$

Why other options are wrong:
- **Option A ( $-472.5 \text{ KJ}$ ): This is incorrect because it likely involves an incorrect sign or division by 2, failing to properly account for the stoichiometry and bond breaking/forming.
- Option B ( $-945 \text{ KJ}$ ): This is incorrect because bond energies are always positive values. This likely results from an incorrect sign convention in applying Hess's Law.
- Option C ( $472.5 \text{ KJ}$ ):** This is incorrect because it likely involves dividing the correct answer by 2, possibly confusing the bond energy per mole of $N_2$ with the value for 2 moles of $NH_3$ .

Key Concepts:
- Heat of formation
- Bond energy
- Hess's Law

Memory Aids:
- Reactants - Products (for enthalpy changes)

Quick Tips:
- Always balance the chemical equation first.
- Remember to multiply bond energies by the number of bonds in the molecule.

13 Hard

13 of 111Question permalink

Application +4 | -1

If,

\mathrm{S} + {\mathrm{O}_2} \to \mathrm{S}{\mathrm{O}_2};∆H = - 298.2\mathrm{kJ}

...(i)

\mathrm{S}{\mathrm{O}_2} + \frac{1}{2}{\mathrm{O}_2} \to \mathrm{S}{\mathrm{O}_3};∆H = - 98.7\mathrm{kJ}

...(ii)

\mathrm{S}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4};∆H = - 130.2\mathrm{kJ}

...(iii)

{\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{O};∆H = - 227.3\mathrm{kJ}

...(iv)The enthalpy of formation of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

at 298 K will be:

View answer and explanation

Correct option: A

- 754.4\mathrm{kJ}

Option A (-754.4 kJ) is correct because it represents the sum of the enthalpy changes of the reactions that form H2SO4 from its elements.

Solution approach

Correct Answer + Core Reason: $- 754.4\mathrm{kJ}$ . The enthalpy of formation is calculated by summing the enthalpy changes of the given reactions.

Step-by-Step Solution:

Step 1: Identify the target reaction: $\mathrm{H}_2 + \mathrm{S} + 2{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$

Step 2: Add the given reactions to obtain the target reaction:
(i) $\mathrm{S} + {\mathrm{O}_2} \to \mathrm{S}{\mathrm{O}_2};\Delta H = - 298.2\mathrm{kJ}$
(ii) $\mathrm{S}{\mathrm{O}_2} + \frac{1}{2}{\mathrm{O}_2} \to \mathrm{S}{\mathrm{O}_3};\Delta H = - 98.7\mathrm{kJ}$
(iii) $\mathrm{S}{\mathrm{O}_3} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4};\Delta H = - 130.2\mathrm{kJ}$
(iv) ${\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{O};\Delta H = - 227.3\mathrm{kJ}$

Step 3: Sum the enthalpy changes:
$\Delta H_\mathrm{formation} = (-298.2) + (-98.7) + (-130.2) + (-227.3) = -754.4\mathrm{kJ}$

Why other options are wrong:
- Option B: This value is incorrect; it likely involves adding instead of subtracting some of the enthalpy values or making a sign error.
- Option C: This value is incorrect; it likely involves adding only some of the enthalpy values, omitting one or more of the reactions.
- Option D: This value is incorrect; it likely involves incorrect arithmetic or omitting one or more of the reactions.

Key Concepts:
- Hess's Law
- Enthalpy of Formation

Memory Aids:
- Hess's Law: The enthalpy change of a reaction is independent of the path taken.

Quick Tips:
- Always check the signs of the enthalpy changes when adding reactions.
- Ensure the target reaction is obtained by correctly adding the given reactions.

14 Hard

14 of 111Question permalink

Application +4 | -1

The enthalpy of vapourisation of a certain liquid at its boiling point of

{35^0}C

is

24.64\,\,KJ/mol

. The value of change in entropy for the process is

View answer and explanation

Correct option: B

80\,\,J{K^{ - 1}}mo{l^{ - 1}}

Option B (

80\,J{K^{ - 1}}mo{l^{ - 1}}

) is correct because it is the result of dividing the enthalpy of vaporization by the boiling point in Kelvin.

Solution approach

Correct Answer + Core Reason: $80\,J{K^{ - 1}}mo{l^{ - 1}}$ . The entropy change is calculated by dividing the enthalpy of vaporization by the boiling point in Kelvin.

Step-by-Step Solution:

Step 1: Convert the boiling point from Celsius to Kelvin: $T(K) = T(^\circ C) + 273 = 35 + 273 = 308 \,K$

Step 2: Use the formula for entropy change during vaporization: $\Delta S = \frac{{\Delta H_{vap}}}{T}$

Step 3: Substitute the given values: $\Delta S = \frac{{24.64 \times 10^3 \,J/mol}}{{308 \,K}} = 80 \,J/K.mol$

Why other options are wrong:
- **Option A ( $704\,J{K^{ - 1}}mo{l^{ - 1}}$ ): This is incorrect due to an error in calculation, possibly multiplying instead of dividing or not converting units correctly.
- Option C ( $24.64\,J{K^{ - 1}}mo{l^{ - 1}}$ ): This is incorrect because it doesn't divide the enthalpy by the temperature.
- Option D ( $7.04\,J{K^{ - 1}}mo{l^{ - 1}}$ ): This is incorrect due to an error in calculation, possibly dividing by a wrong temperature or making a unit conversion mistake.

Key Concepts:**
- Entropy change during phase transition: $\Delta S = \frac{{\Delta H}}{T}$
- Conversion between Celsius and Kelvin: $T(K) = T(^\circ C) + 273$

Memory Aids:
- Remember the formula for entropy change during phase transition.

Quick Tips:
- Always convert the temperature to Kelvin before performing the calculation.
- Ensure the units are consistent.

15 Hard

15 of 111Question permalink

Understanding +4 | -1

For an isothermal free expansion of an ideal gas correct option is:

View answer and explanation

Correct option: D

q=0, w=0, \Delta H=0

Option D (

q=0, w=0, \Delta H=0

) is correct because in an isothermal free expansion of an ideal gas, there is no heat exchange (q=0), no work done (w=0), and no change in enthalpy (

\Delta H=0

).

Solution approach

Correct Answer + Core Reason: $q=0, w=0, \Delta H=0$ . For isothermal free expansion of an ideal gas, heat exchange, work done, and enthalpy change are all zero.

Step-by-Step Solution:

Step 1: Define isothermal process.\nAn isothermal process is a process that occurs at constant temperature ( $\Delta T = 0$ ).

Step 2: Define free expansion.\nA free expansion is an expansion against zero external pressure ( $P_{ext} = 0$ ).

Step 3: Apply the first law of thermodynamics.
$\Delta U = q + w$ , where $\Delta U$ is the change in internal energy, q is heat, and w is work.\nFor an ideal gas, internal energy depends only on temperature. Since the process is isothermal, $\Delta U = 0$ .\nTherefore, $q = -w$ .

Step 4: Calculate work done.
$w = -P_{ext} \Delta V$ . Since $P_{ext} = 0$ , $w = 0$ .

Step 5: Calculate heat.\nSince $q = -w$ and $w = 0$ , $q = 0$ .

Step 6: Calculate enthalpy change.
$\Delta H = \Delta U + \Delta (PV) = \Delta U + nR\Delta T$ . Since $\Delta U = 0$ and $\Delta T = 0$ , $\Delta H = 0$ .

Step 7: Conclusion: $q=0, w=0, \Delta H=0$ .

Why other options are wrong:
- Option A: $q=0, w=0, \Delta H \neq 0$ is incorrect because $\Delta H$ is also zero for an isothermal free expansion of an ideal gas.
- Option B: $q \neq 0, w=0, \Delta H=0$ is incorrect because q is also zero.
- Option C: $q \neq 0, w \neq 0, \Delta H \neq 0$ is incorrect because q, w, and $\Delta H$ are all zero.

Key Concepts:
- Isothermal process
- Free expansion
- First law of thermodynamics
- Enthalpy

Memory Aids:
- Isothermal: Constant temperature.
- Free expansion: Expansion against zero external pressure.

Quick Tips:
- Remember that for an ideal gas, internal energy depends only on temperature.

16 Hard

16 of 111Question permalink

Understanding +4 | -1

The temperature coefficient of e.m.f. of a cell can be given by:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because all the given expressions are valid representations of the temperature coefficient of e.m.f. of a cell.

Solution approach

Correct Answer + Core Reason: All of these. All options correctly represent the temperature coefficient of e.m.f. of a cell.

Step-by-Step Solution:

Step 1: Recall the definition of the temperature coefficient of e.m.f.

Step 2: Remember the following relationships:
- ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \frac{{\Delta S}}{{nF}}$ (relates temperature coefficient to entropy change)
- ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = {\left( {\frac{{{{E_2} - {E_1}}}}{{{{T_2} - {T_1}}}}} \right)_P}$ (experimental determination)
- ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \left[ {\frac{{\Delta H}}{{nF}} + E} \right]\frac{1}{T}$ (relates temperature coefficient to enthalpy change)

Step 3: Evaluate each option:
- Option A: ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \frac{{\Delta S}}{{nF}}$ Correct.
- Option B: ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = {\left( {\frac{{{{E_2} - {E_1}}}}{{{{T_2} - {T_1}}}}} \right)_P}$ Correct.
- Option C: ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \left[ {\frac{{\Delta H}}{{nF}} + E} \right]\frac{1}{T}$ Correct.

Step 4: Conclude that all options are correct.

Why other options are wrong:
- Option A: Incorrect because option D is the correct answer.
- Option B: Incorrect because option D is the correct answer.
- Option C: Incorrect because option D is the correct answer.
- Option D: Correct since A, B, and C are all correct.

Key Concepts:
- Temperature coefficient of e.m.f., Entropy change, Enthalpy change, Nernst equation

Memory Aids:
- Remember the relationships between the temperature coefficient of e.m.f. and thermodynamic quantities.

Quick Tips:
- Use the appropriate formula based on the given information (entropy change, enthalpy change, or experimental data).

17 Hard

17 of 111Question permalink

Analyzing +4 | -1

Given,

\mathrm{KOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{KCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(l); \Delta H=-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots (i)

\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}); \Delta \mathrm{H}=-286.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots.....(ii)

\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \xrightarrow{A q} \mathrm{HCl}(\mathrm{aq}); \Delta H=-164.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots(\mathrm{iii)}

\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{aq} \longrightarrow \mathrm{KOH}(\mathrm{aq}) ; \Delta H=-487.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \ldots(\mathrm{iv})

\mathrm{KCl}(\mathrm{s})+\mathrm{aq} \longrightarrow \mathrm{KCl}(\mathrm{aq}); \Delta H=+18.4 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}

.....(v)

Heat of formation of

\mathrm{KCl}

will be

View answer and explanation

Correct option: B

-441.3 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option B is correct because the heat of formation of KCl can be calculated by manipulating the given equations using Hess's Law.

Solution approach

Correct Answer + Core Reason: $-441.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . This is calculated using Hess's Law by manipulating the given equations.

Step-by-Step Solution:

Step 1: Target reaction: $\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{KCl}(\mathrm{s}) ; \Delta H = ?$

Step 2: Given equations:
(i) $\mathrm{KOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{KCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(l); \Delta H=-57.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}); \Delta \mathrm{H}=-286.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iii) $\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \xrightarrow{A q} \mathrm{HCl}(\mathrm{aq}); \Delta H=-164.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iv) $\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{aq} \longrightarrow \mathrm{KOH}(\mathrm{aq}) ; \Delta H=-487.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(v) $\mathrm{KCl}(\mathrm{s})+\mathrm{aq} \longrightarrow \mathrm{KCl}(\mathrm{aq}); \Delta H=+18.4 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$

Step 3: Manipulate equations to obtain the target reaction:\nWe want: $\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{KCl}(\mathrm{s}) + \mathrm{H}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{~g})$

Step 4: Add equations (iv), (iii), and (i) to get:
$\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})+\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) + aq \longrightarrow \mathrm{KCl}(\mathrm{aq}) + \mathrm{H}_2\mathrm{O}(l) + \mathrm{aq} ; \Delta H = -487.4 -164.4 -57.3 = -709.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Step 5: Subtract equation (ii) from the result of Step 4:
$\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) + aq \longrightarrow \mathrm{KCl}(\mathrm{aq}) + \mathrm{aq} ; \Delta H = -709.1 - (-286.2) = -422.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Step 6: Subtract equation (v) from the result of Step 5:
$\mathrm{K}(\mathrm{s})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{KCl}(\mathrm{s}) ; \Delta H = -422.9 - 18.4 = -441.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Why other options are wrong:
- Option A: Incorrect; this value is not obtained through correct manipulation of the given equations.
- Option C: Incorrect; the sign is wrong, and the value is not obtained through correct manipulation of the given equations.
- Option D: Incorrect; this value is too large and not obtained through correct manipulation of the given equations.

Key Concepts:
- Hess's Law
- Enthalpy of Formation
- Manipulation of Thermochemical Equations

Memory Aids:
- Hess's Law: The enthalpy change of a reaction is independent of the path taken.

Quick Tips:
- Carefully manipulate the given equations to match the target equation, paying attention to signs and coefficients.

18 Hard

18 of 111Question permalink

Application +4 | -1

A process has

\Delta H = 200J\mathrm{ }mo{l^{ - 1}}

and

\Delta S = 40J{K^{ - 1}}mo{l^{ - 1}}\,

. Out of the values given below, choose the minimum temperature above which the process will be spontaneous

View answer and explanation

Correct option: C

5 K

Option C (5 K) is correct because it represents the minimum temperature above which the process becomes spontaneous, calculated using the Gibbs free energy equation.

Solution approach

Correct Answer + Core Reason: 5 K. The process is spontaneous when $\Delta G < 0$ , which occurs above a certain temperature.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H = 200 \text{{ J/mol}}$ , $\Delta S = 40 \text{{ J/(K\u00b7mol)}}$ . We need to find the minimum temperature $T$ for which the process is spontaneous.

Step 2: Recall the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$ .

Step 3: For a spontaneous process, $\Delta G < 0$ . Therefore, we need to find the temperature $T$ such that $\Delta H - T\Delta S < 0$ .

Step 4: Rearrange the inequality: $T\Delta S > \Delta H$ , which means $T > \frac{{\Delta H}}{{\Delta S}}$ .

Step 5: Substitute the given values: $T > \frac{{200 \text{{ J/mol}}}}{{40 \text{{ J/(K\u00b7mol)}}}} = 5 \text{{ K}}$ .

Step 6: Therefore, the minimum temperature above which the process will be spontaneous is 5 K.

Why other options are wrong:
- Option A (20 K): This temperature is higher than the minimum required temperature for spontaneity. While the process would be spontaneous at 20 K, the question asks for the *minimum* temperature.
- Option B (4 K): At this temperature, $\Delta G$ would be positive, making the process non-spontaneous.
- Option D (12 K): This temperature is higher than the minimum required temperature for spontaneity. While the process would be spontaneous at 12 K, the question asks for the *minimum* temperature.

Key Concepts:
- Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$
- Spontaneity: $\Delta G < 0$ for a spontaneous process

Memory Aids:
- "Gotta Have This Situation" (G = H - TS) to remember the Gibbs Free Energy equation.

Quick Tips:
- Always check the units of $\Delta H$ and $\Delta S$ to ensure they are consistent before performing calculations.
- Remember that spontaneity is temperature-dependent when both $\Delta H$ and $\Delta S$ are non-zero.

19 Hard

19 of 111Question permalink

Application +4 | -1

In an isobaric process, the ratio of heat supplied to the system (dθ) and work done by the system (dW) for diatomic gas is

View answer and explanation

Correct option: B

7:2

Option B (

7:2

) is correct because for a diatomic gas in an isobaric process, the ratio of heat supplied to work done is 7/2.

Solution approach

Correct Answer + Core Reason: $7:2$ . The ratio of heat supplied to work done in an isobaric process for a diatomic gas is calculated using the degrees of freedom.

Step-by-Step Solution:

Step 1: Identify the process: Isobaric process (constant pressure).

Step 2: Write the formula for heat supplied at constant pressure: $d\theta = nC_p dT$

Step 3: Write the formula for work done at constant pressure: $dW = P dV = nR dT$

Step 4: Determine the value of $C_p$ for a diatomic gas. For a diatomic gas, the number of degrees of freedom is 5. Therefore, $C_v = \frac{5}{2}R$ and $C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$ .

Step 5: Calculate the ratio of heat supplied to work done:
$\frac{d\theta}{dW} = \frac{nC_p dT}{nR dT} = \frac{C_p}{R} = \frac{\frac{7}{2}R}{R} = \frac{7}{2}$

Step 6: Final Answer: $7:2$

Why other options are wrong:
- Option A (1:1): This is incorrect because it would only be true if $C_p = R$ , which is not the case for a diatomic gas.
- Option C (7:5): This is incorrect because it uses the ratio of $C_p$ to $C_v$ instead of $C_p$ to $R$ .
- Option D (5:7): This is incorrect because it uses the inverse ratio of $C_p$ to $R$ .

Key Concepts:
- Isobaric Process
- Heat Capacity
- Degrees of Freedom

Memory Aids:
- Remember that $C_p = C_v + R$
- For diatomic gas, $C_v = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$

Quick Tips:
- Identify the type of gas (monoatomic, diatomic, etc.) to determine the correct values of $C_v$ and $C_p$ .
- Remember the relationship between $C_p$ , $C_v$ , and $R$ .

20 Hard

20 of 111Question permalink

Application +4 | -1

The heat of combustion of yellow phosphorous is

-9.91 {~kJ}

and of red phosphorous is

-8.78 {~kJ}

. The heat of transition of yellow phosphorus to red phosphorus is :

View answer and explanation

Correct option: D

-1.13 {~kJ}

Option D (

-1.13 \text{ kJ}

) is correct because it is calculated by subtracting the heat of combustion of red phosphorus from the heat of combustion of yellow phosphorus.

Solution approach

Correct Answer + Core Reason: $-1.13 \text{ kJ}$ . The heat of transition is the difference between the heats of combustion of the two allotropes.

Step-by-Step Solution:

Step 1: Write the equation for the transition:
$P_{yellow} \rightarrow P_{red}$

Step 2: Use Hess's Law to find the heat of transition:
$\Delta H_{transition} = \Delta H_{combustion}(P_{yellow}) - \Delta H_{combustion}(P_{red})$

Step 3: Substitute the given values:
$\Delta H_{transition} = -9.91 \text{ kJ} - (-8.78 \text{ kJ})$

Step 4: Calculate the heat of transition:
$\Delta H_{transition} = -9.91 + 8.78 = -1.13 \text{ kJ}$

Why other options are wrong:
- **Option A ( $-9.91 \text{ kJ}$ ): This is incorrect because it only considers the heat of combustion of yellow phosphorus and does not account for the heat of combustion of red phosphorus.
- Option B ( $-8.78 \text{ kJ}$ ): This is incorrect because it only considers the heat of combustion of red phosphorus and does not account for the heat of combustion of yellow phosphorus.
- Option C ( $-9.34 \text{ kJ}$ ): This is incorrect because it likely involves an arithmetic error in the subtraction or an incorrect application of Hess's Law.

Key Concepts:
- Heat of combustion
- Heat of transition
- Hess's Law

Memory Aids:
- Hess's Law: The enthalpy change of a reaction is independent of the path taken.

Quick Tips:**
- Pay attention to the signs of the enthalpy changes.
- Remember to subtract the final state from the initial state.

21 Hard

21 of 111Question permalink

Application +4 | -1

The molar heat capacity of water at constant pressure

\mathrm{P}

is

75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}.

When

1.0 \mathrm{~kJ}

of heat is supplied to

1000 \mathrm{~g}

of water, which is free to expand, the increase in temperature of water is

View answer and explanation

Correct option: D

0.24 \mathrm{~K}

Option D (0.24 K) is correct because it is the increase in temperature calculated using the formula relating heat supplied, mass, specific heat capacity, and temperature change.

Solution approach

Correct Answer + Core Reason: $0.24 \mathrm{~K}$ . The increase in temperature is calculated using the formula relating heat supplied, mass, specific heat capacity, and temperature change.

Step-by-Step Solution:

Step 1: Identify the given information:\nHeat supplied (q) = 1.0 kJ = 1000 J\nMass of water (m) = 1000 g\nMolar heat capacity of water at constant pressure (Cp,m) = 75 J K-1 mol-1

Step 2: Calculate the number of moles of water (n):\nMolar mass of water (H2O) = 18 g/mol\nn = mass / molar mass = 1000 g / 18 g/mol = 55.56 mol

Step 3: Use the formula relating heat supplied, number of moles, molar heat capacity, and temperature change:\nq = n * Cp,m * Δ T

Step 4: Solve for the increase in temperature (Δ T):
Δ T = q / (n * Cp,m)

Step 5: Substitute the values:
Δ T = 1000 J / (55.56 mol * 75 J K-1 mol-1)
Δ T = 1000 / (55.56 * 75) K
Δ T = 1000 / 4167 K
Δ T = 0.24 K

Why other options are wrong:
- Option A (1.2 K): This is incorrect because it uses incorrect values or an incorrect formula.
- Option B (2.4 K): This is incorrect because it uses incorrect values or an incorrect formula.
- Option C (4.8 K): This is incorrect because it uses incorrect values or an incorrect formula.

Key Concepts:
- Heat Capacity
- Molar Heat Capacity
- Formula: q = n * Cp,m * Δ T

Memory Aids:
- Remember the formula relating heat supplied, number of moles, molar heat capacity, and temperature change.

Quick Tips:
- Ensure the units are consistent.
- Remember the definition of molar heat capacity.

22 Hard

22 of 111Question permalink

Application +4 | -1

If

1.0 \mathrm{~kcal}

of heat is added to

1.2 \mathrm{~L}

of

\mathrm{O}_2

in a cylinder at constant pressure of

1 \mathrm{~atm}

, the volume increases to

1.5 \mathrm{~L}

. Calculate

\Delta \mathrm{U}

and

\Delta \mathrm{H}

of the process.

(1 \mathrm{~L}-\mathrm{atm}=100 \mathrm{~J}, 1 \mathrm{~cal}=4.2 \mathrm{~J})

View answer and explanation

Correct option: A

\Delta \mathrm{U}=4170 \mathrm{~J}; \Delta \mathrm{H}=1 \mathrm{~kcal}

Option A (ΔU=4170 J; ΔH=1 kcal) is correct because it involves calculating the change in internal energy and enthalpy using the given heat added and volume change at constant pressure.

Solution approach

Correct Answer + Core Reason: $\Delta \mathrm{U}=4170 \mathrm{~J}; \Delta \mathrm{H}=1 \mathrm{~kcal}$ . The change in internal energy and enthalpy are calculated using the given heat added and volume change at constant pressure.

Step-by-Step Solution:

Step 1: Calculate the heat added in joules.
$q = 1.0 \text{{ kcal}} = 1.0 \times 4.2 \times 10^3 \text{{ J}} = 4200 \text{{ J}}$

Step 2: Calculate the work done at constant pressure.
$w = -P\Delta V = -1 \text{{ atm}} \times (1.5 - 1.2) \text{{ L}} = -1 \times 0.3 \text{{ L atm}}$

Step 3: Convert the work done to joules.
$w = -0.3 \text{{ L atm}} \times 100 \text{{ J/L atm}} = -30 \text{{ J}}$

Step 4: Calculate the change in internal energy.
$\Delta U = q + w = 4200 - 30 = 4170 \text{{ J}}$

Step 5: Calculate the change in enthalpy.
$\Delta H = q_p = 1 \text{{ kcal}}$ (since the process is at constant pressure, the heat added is equal to the change in enthalpy).

Why other options are wrong:
- Option B: $\Delta \mathrm{U}=4180 \mathrm{~J}; \Delta \mathrm{H}=2 \mathrm{~kcal}$ . This is incorrect because the change in enthalpy should be equal to the heat added, which is 1 kcal.
- Option C: $\Delta \mathrm{U}=4270 \mathrm{~J}; \Delta \mathrm{H}=0 \mathrm{~kcal}$ . This is incorrect because the change in enthalpy should be equal to the heat added, which is 1 kcal, and the change in internal energy is calculated incorrectly.
- Option D: $\Delta \mathrm{U}=5170 \mathrm{~J}; \Delta \mathrm{H}=4 \mathrm{~kcal}$ . This is incorrect because both the change in internal energy and enthalpy are calculated incorrectly.

Key Concepts:
- First Law of Thermodynamics
- Enthalpy Change at Constant Pressure
- Work Done at Constant Pressure

Memory Aids:
- $\Delta U = q + w$
- $\Delta H = q_p$

Quick Tips:
- Remember to convert all units to a consistent system (e.g., joules, liters, atm).
- At constant pressure, the heat added is equal to the change in enthalpy.

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Application +4 | -1

A sample of ideal gas

(\gamma=1.4)

is heated at constant pressure if an amount of

85 \mathrm{~J}

of heat is supplied to gas, find

{ }_{\Delta U}

.

View answer and explanation

Correct option: A

5

The question is incomplete and does not provide enough information to determine the change in internal energy (ΔU). It seems there is a typo and the correct answer cannot be determined with the given information.

Solution approach

Correct Answer + Core Reason: 5. This question is incomplete and cannot be solved. The value of Δ U cannot be determined with the given information.

Step-by-Step Solution:\nSince the question is incomplete, a step-by-step solution cannot be provided.

Why other options are wrong:\nSince the question is incomplete, it is not possible to determine why the other options are wrong.

Key Concepts:
- Thermodynamics
- Internal Energy

Memory Aids:
- N/A

Quick Tips:
- N/A

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Application +4 | -1

1 mole of gas occupying 3 L volume is expanded against a constant external pressure of 1 atm to a volume of 15 litre. The work done by the system is:

View answer and explanation

Correct option: A

1.215 \times {10^3}\mathrm{J}

Option A (

1.215 \times {10^3}\mathrm{J}

) is correct because it is calculated using the formula

W = -P_{ext} \Delta V

, converting liters to cubic meters.

Solution approach

Correct Answer + Core Reason: $1.215 \times {10^3}\mathrm{J}$ . The work done is calculated using the formula $W = -P_{ext} \Delta V$ .

Step-by-Step Solution:

Step 1: Identify the given values:
$P_{ext} = 1 \text{ atm} = 101325 \text{ Pa}$
$V_1 = 3 \text{ L} = 3 \times 10^{-3} \text{ m}^3$
$V_2 = 15 \text{ L} = 15 \times 10^{-3} \text{ m}^3$

Step 2: Calculate the change in volume:
$\Delta V = V_2 - V_1 = (15 - 3) \times 10^{-3} \text{ m}^3 = 12 \times 10^{-3} \text{ m}^3$

Step 3: Calculate the work done:\nW = -P_{ext} Δ V = -(101325 Pa) × (12 × 10^-3 m³) = -1215.9 J $\nSince the question asks for the work done *by* the system, we take the absolute value:$ 1215.9 J ≈ 1.215 × 10³ J $\n\n**Why other options are wrong:**\n- **Option B:** This value is incorrect; it likely involves an error in the unit conversion or calculation.\n- **Option C:** This value is incorrect; it likely involves an error in the unit conversion or calculation.\n- **Option D:** This value is incorrect; one of the other options is correct.\n\n**Key Concepts:**\n- Work done in expansion against constant external pressure\n- Unit conversion (L to m³)\n\n**Memory Aids:**\n-$ W = -P_{ext} Δ V $\n\n**Quick Tips:**\n- Ensure all units are in SI units (Pascals for pressure, m³ for volume, Joules for work). \n- Pay attention to the sign of the work done. Work done *by* the system is negative.$

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Understanding +4 | -1

If can ideal gas is subjected to isothermal free expansion then which of the following are incorrect?

View answer and explanation

Correct option: D

\mathrm{q}=0

Option D (q=0) is correct because in an isothermal free expansion, no work is done by the gas against external pressure, and since it's isothermal, there's no change in internal energy, hence no heat transfer.

Solution approach

Correct Answer + Core Reason: q=0. In isothermal free expansion, work done and heat exchanged are zero.

Step-by-Step Solution:

Step 1: Define isothermal free expansion. Isothermal means constant temperature ( $\Delta T = 0$ ). Free expansion means expansion against zero external pressure ( $P_{ext} = 0$ ).

Step 2: Recall the first law of thermodynamics: $\Delta U = q + w$ , where $\Delta U$ is the change in internal energy, $q$ is heat, and $w$ is work.

Step 3: For an ideal gas undergoing isothermal process, $\Delta U = 0$ because internal energy depends only on temperature. Therefore, $0 = q + w$ .

Step 4: Work done during expansion against external pressure is $w = -P_{ext} \Delta V$ . Since $P_{ext} = 0$ , $w = 0$ .

Step 5: Substituting $w = 0$ into the first law equation, $0 = q + 0$ , so $q = 0$ .

Step 6: For an ideal gas, enthalpy $H = U + PV$ . Since the process is isothermal and $\Delta U = 0$ , and $PV = nRT$ , then $\Delta(PV) = nR\Delta T = 0$ . Therefore, $\Delta H = \Delta U + \Delta(PV) = 0 + 0 = 0$ .

Step 7: Entropy change for an isothermal process is given by $\Delta S = nR \ln{\frac{V_2}{V_1}}$ . Since $V_2 > V_1$ in expansion, $\Delta S > 0$ .

Why other options are wrong:
- Option A: $\Delta S = 0$ is incorrect because entropy increases during free expansion.
- Option B: $\Delta U = 0$ is correct because internal energy depends only on temperature for an ideal gas in an isothermal process.
- Option C: $\Delta H = 0$ is correct because enthalpy is a function of temperature only for an ideal gas.

Key Concepts:
- First law of thermodynamics
- Isothermal process
- Free expansion
- Internal energy of an ideal gas
- Enthalpy of an ideal gas
- Entropy change

Memory Aids:
- Isothermal: ISO-Thermal = Same Temperature
- Free Expansion: No external pressure, no work done.

Quick Tips:
- In free expansion, $P_{ext} = 0$ , so $w = 0$ .
- For an ideal gas, $\Delta U = 0$ in an isothermal process.

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Remembering +4 | -1

The ratio of molar heats of vaporization and boiling point of a liquid is constant. This is known as

View answer and explanation

Correct option: D

Trouton’s rule

Option D (Trouton’s rule) is correct because it states that the ratio of molar heat of vaporization to the boiling point is approximately constant for many liquids.

Solution approach

Correct Answer + Core Reason: Trouton’s rule. Trouton's rule states that the molar heat of vaporization divided by the boiling point is approximately constant.

Step-by-Step Solution:

Step 1: Recall Trouton's rule.

Step 2: Trouton's rule states: $\frac{{\Delta H_{\text{{vap}}}}}}{{T_b}} \approx 85 \text{{ J mol}}^{-1} \text{{K}}^{-1}$ , where $\Delta H_{\text{{vap}}}$ is the molar heat of vaporization and $T_b$ is the boiling point in Kelvin.

Step 3: This implies that the ratio of molar heat of vaporization and boiling point is constant.

Why other options are wrong:
- Option A: Ostwald's rule relates to the ionization constant of weak electrolytes.
- Option B: Phase rule relates the number of degrees of freedom to the number of phases and components in a system.
- Option C: Van't Hoff rule describes the temperature dependence of equilibrium constants.

Key Concepts:
- Trouton's rule
- Molar heat of vaporization
- Boiling point

Memory Aids:
- Remember Trouton's rule relates vaporization enthalpy to boiling point.

Quick Tips:
- Recognize Trouton's rule when dealing with vaporization enthalpy and boiling point.

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Analyzing +4 | -1

He

,

{N}_2

and

{O}_3

are expanded adiabatically and their expansion curves between

\log P

and

\log V

are plotted under similar conditions. About the ratio of the slopes, which one is not correct?

View answer and explanation

Correct option: D

The slope of

{He}

is least steeper and for

{O}_3

is most steeper

Option D is correct because the slope depends on the adiabatic index (gamma), and He has the highest gamma, leading to the steepest slope, while O3 has the lowest.

Solution approach

Correct Answer + Core Reason: The slope of ${He}$ is least steeper and for ${O}_3$ is most steeper. The slope of the adiabatic curve in a $\log P$ vs $\log V$ plot is given by $-\gamma$ , where $\gamma$ is the adiabatic index. Since He is monoatomic, N2 is diatomic, and O3 is polyatomic, their $\gamma$ values are in the order He > N2 > O3, making the statement incorrect.

Step-by-Step Solution:

Step 1: Recall the relationship between pressure and volume in an adiabatic process: $PV^{\gamma} = constant$ , where $\gamma = \frac{C_P}{C_V}$ is the adiabatic index.

Step 2: Take the logarithm of both sides: $\log P + \gamma \log V = \log(constant)$ .

Step 3: Rearrange the equation to get the form of a linear equation (y = mx + c): $\log P = -\gamma \log V + \log(constant)$ .

Step 4: Identify the slope of the $\log P$ vs $\log V$ curve: The slope is $-\gamma$ .

Step 5: Determine the values of $\gamma$ for He, N2, and O3:
- He (monoatomic): $\gamma = \frac{5}{3} \approx 1.67$
- N2 (diatomic): $\gamma = \frac{7}{5} = 1.40$
- O3 (polyatomic): $\gamma$ is less than 1.40 but greater than 1. Since O3 is a triatomic non-linear molecule, it has more vibrational degrees of freedom, and its γ is closer to 1. We can approximate it as 1.3.

Step 6: Calculate the ratios of the slopes:
- He/O3: 1.67/1.3 = 1.28
- He/N2: 1.67/1.4 = 1.19
- N2/O3: 1.4/1.3 = 1.08

Step 7: Compare the calculated ratios with the given options:
- Option A: The ratio of slopes of $\log P$ vs $\log V$ curves for ${He}$ and ${O}_3$ is 1.25. This is approximately correct.
- Option B: The ratio of slopes of $\log P$ vs $\log V$ curves for ${He}$ and ${N}_2$ is 1.20. This is approximately correct.
- Option C: The ratio of slopes of $\log P$ vs $\log V$ curves for ${N}_2$ and ${O}_3$ is 1.05. This is approximately correct.
- Option D: The slope of ${He}$ is least steeper and for ${O}_3$ is most steeper. This is incorrect. Since the slope is $-\gamma$ , and He has the largest $\gamma$ , its slope is the *most* steep (most negative), and O3 has the smallest $\gamma$ , so its slope is the least steep (least negative).

Why other options are wrong:
- Option A: The ratio is approximately correct.
- Option B: The ratio is approximately correct.
- Option C: The ratio is approximately correct.

Key Concepts:
- Adiabatic process
- Adiabatic index ( $\gamma$ )
- Slope of a curve

Memory Aids:
- $PV^{\gamma} = constant$ (Adiabatic process)
- $\gamma = \frac{C_P}{C_V}$ (Adiabatic index)

Quick Tips:
- Remember that the adiabatic index ( $\gamma$ ) depends on the degrees of freedom of the gas molecules.

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Understanding +4 | -1

Enthalpy of neutralisation of acetic acid with

KOH

will be numerically

View answer and explanation

Correct option: C

< 57.2 kJ

Option C (< 57.2 kJ) is correct because acetic acid is a weak acid, and some energy is used for its ionization, resulting in a lower heat of neutralization than that of a strong acid.

Solution approach

Correct Answer + Core Reason: < 57.2 kJ. Acetic acid is a weak acid, so some energy is absorbed during its ionization.

Step-by-Step Solution:

Step 1: Recall the heat of neutralization for strong acid-strong base reactions: It is approximately -57.32 kJ/mol.

Step 2: Recognize that acetic acid (CH3COOH) is a weak acid.

Step 3: Understand that the neutralization of a weak acid with a strong base involves two steps:
- Ionization of the weak acid (endothermic)
- Neutralization of the resulting ions (exothermic)

Step 4: Conclusion: The overall heat of neutralization will be less negative (smaller magnitude) than that of a strong acid due to the energy absorbed during ionization.

Why other options are wrong:
- Option A: = 57.2 kJ is the approximate heat of neutralization for a strong acid-strong base reaction, not a weak acid-strong base reaction.
- Option B: > 57.2 kJ is incorrect because the heat of neutralization will be less due to the energy required for ionization of the weak acid.
- Option D: Unpredictable is incorrect because we can predict that the heat of neutralization will be less than 57.2 kJ.

Key Concepts:
- Heat of neutralization
- Weak acids and bases
- Ionization energy

Memory Aids:
- Weak acid/base \u2192 Less heat evolved

Quick Tips:
- Consider the strength of the acid and base when determining the heat of neutralization.

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Application +4 | -1

Two moles of helium gas expanded isothermally and irreversible at 27℃ from volume 1

d{m^3}

to 1

{m^3}

at constant pressure of 100 k Pa. Calculate the work done.

View answer and explanation

Correct option: B

99900 J

Option B (99900 J) is correct because it is calculated using the formula for work done in an irreversible isothermal expansion against constant pressure.

Solution approach

Correct Answer + Core Reason: 99900 J. Work done in an irreversible isothermal expansion against constant pressure is calculated using $W = -P_{ext}(V_2 - V_1)$ .

Step-by-Step Solution:

Step 1: Identify the given values:
- $n = 2 \text{{ moles}}$
- $T = 27^\circ C = 300 \text{{ K}}$
- $V_1 = 1 \text{{ dm}}^3 = 0.001 \text{{ m}}^3$
- $V_2 = 1 \text{{ m}}^3$
- $P_{ext} = 100 \text{{ kPa}} = 100 \times 10^3 \text{{ Pa}}$

Step 2: Use the formula for work done in an irreversible isothermal expansion against constant pressure:
- $W = -P_{ext}(V_2 - V_1)$

Step 3: Substitute the values:
- $W = -(100 \times 10^3 \text{{ Pa}})(1 \text{{ m}}^3 - 0.001 \text{{ m}}^3)$
- $W = -(100 \times 10^3 \text{{ Pa}})(0.999 \text{{ m}}^3)$

Step 4: Calculate the work done:
- $W = -99900 \text{{ J}}$

Step 5: Since the question asks for the magnitude of work done, we take the absolute value:
- $|W| = 99900 \text{{ J}}$

Why other options are wrong:
- Option A: This option is incorrect because it has units of kJ instead of J.
- Option C: This option is incorrect, likely due to using the reversible isothermal work formula instead of the irreversible one, or a calculation error.
- Option D: This option is incorrect because it has units of kJ instead of J, and the numerical value is also wrong.

Key Concepts:
- Irreversible isothermal expansion
- Work done against constant pressure: $W = -P_{ext}(V_2 - V_1)$
- Unit conversions (kPa to Pa, $dm^3$ to $m^3$ )

Memory Aids:
- Remember the formula for work done in an irreversible process.

Quick Tips:
- Always pay attention to the units and convert them if necessary.
- Distinguish between reversible and irreversible processes and use the appropriate formula.

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Analyzing +4 | -1

The bond dissociation energies of

N \equiv N, N=N, O=O \,\&\, N=O

are

946,418,498

and

607 \mathrm{~KJ/ mole}

respectively. If heat of formation of

{N}_2 {O}

is

82 \mathrm{~KJ/ mole}

then the magnitude of resonance energy of

{N}_2 {O}

in

\mathrm{KJ} / \mathrm{mole}

is ___

View answer and explanation

Correct option: C

88 \mathrm{~KJ} / \mathrm{mole}

Option C (88 kJ/mole) is correct because the resonance energy is calculated by subtracting the experimental heat of formation from the calculated heat of formation based on bond energies.

Solution approach

Correct Answer + Core Reason: 88 kJ/mole. Resonance energy is calculated by comparing the experimental heat of formation with the heat of formation calculated from bond energies.

Step-by-Step Solution:

Step 1: Draw the Lewis structure of N2O: N=N=O (major resonance structure).

Step 2: Calculate the heat of formation of N2O using bond energies:
$\Delta H_f = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products}$

Step 3: The reactants are N2 and O2. The products are N2O. The bonds broken are 1 N≡N and 1 O=O. The bonds formed are 1 N=N and 1 N=O.

Step 4: $\Delta H_f = [E(N \equiv N) + E(O=O)] - [E(N=N) + E(N=O)]$
$\Delta H_f = [946 + 498] - [418 + 607]$

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