For the reaction, $\begin{array}{l} {{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right) \to 2{\rm{HCl}}\left( {\rm{g}} \right) + {x_1}\,\,{\rm{kJ}} \ldots \ldots \ldots \left( {\rm{i}} \right)\\ 2{\rm{HCl}}\left( {\rm{g}} \right) \to {{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right) - {x_2}\,{\rm{kJ}} \ldots \ldots \left( {{\rm{ii}}} \right) \end{array}$ Which of the following statement is correct?

${x_1}$ and ${x_2}$ are numerically different

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium, Page 3

61 Medium

Remembering +4 | -1

Standard molar enthalpy of formation at

298 \mathrm{~K}

is zero for

View answer and explanation

Correct option: A

\mathrm{Br}_2(\mathrm{l})

Option A (Br₂(l)) is correct because the standard molar enthalpy of formation is zero for an element in its standard state.

Solution approach

Correct Answer + Core Reason: Br₂(l). Standard molar enthalpy of formation is zero for an element in its standard state.

Step-by-Step Solution:

Step 1: Define standard molar enthalpy of formation. It is the change in enthalpy when one mole of a substance in its standard state is formed from its constituent elements in their standard states.

Step 2: Identify the standard states of the given elements/compounds at 298 K.

Step 3: Br₂ exists as a liquid (l) at 298 K and 1 atm, which is its standard state. Therefore, its standard molar enthalpy of formation is zero.

Step 4: Sulfur's standard state is rhombic, not monoclinic. Therefore, S(monoclinic) has a non-zero standard enthalpy of formation.

Step 5: H₂O is a compound, not an element, so its standard enthalpy of formation is not zero.

Step 6: O₃ (ozone) is not the standard allotrope of oxygen. The standard state of oxygen is O₂ (diatomic oxygen). Therefore, O₃ has a non-zero standard enthalpy of formation.

Why other options are wrong:
- Option B: S(monoclinic) is not the standard state of sulfur; the standard state is S(rhombic).
- Option C: H₂O(l) is a compound, not an element in its standard state, so its standard enthalpy of formation is not zero.
- Option D: O₃(g) is not the standard state of oxygen; the standard state is O₂(g).

Key Concepts:
- Standard molar enthalpy of formation
- Standard state of elements

Memory Aids:
- Standard state is the most stable form of a substance at 298 K and 1 atm.

Quick Tips:
- Remember the standard states of common elements (e.g., H₂, N₂, O₂, F₂, Cl₂ are gases; Br₂ is a liquid; I₂ is a solid; C is graphite; S is rhombic).

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Which represents the largest amount of energy?

View answer and explanation

Correct option: A

Calorie

Option A, Calorie, represents the largest amount of energy among the given options.

Solution approach

Correct Answer + Core Reason: Calorie. A calorie is larger than a joule, erg, or electron-volt.

Step-by-Step Solution:

Step 1: Recall the conversion factors between the given units of energy.

Step 2: 1 calorie (cal) = 4.184 Joules (J)
1 Joule (J) = $10^7$ Ergs
1 electron-volt (eV) = $1.602 \times 10^{-19}$ J

Step 3: Comparing the values, a calorie is the largest unit.

Step 4: Final Answer: Calorie

Why other options are wrong:
- Option B: Joule is smaller than a calorie (1 cal = 4.184 J).
- Option C: Erg is a very small unit of energy compared to a calorie (1 J = $10^7$ ergs).
- Option D: Electron-volt is an extremely small unit of energy compared to a calorie (1 eV = $1.602 \times 10^{-19}$ J).

Key Concepts:
- Units of Energy
- Calorie, Joule, Erg, Electron-volt
- Conversion Factors

Memory Aids:
- Remember the relative sizes of these energy units.

Quick Tips:
- When comparing energy values, always convert them to the same units for accurate comparison.

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Understanding +4 | -1

Internal energy and pressure of a gas of unit volume are related as:

View answer and explanation

Correct option: A

P = \frac{2}{3}U

Option A is correct because the internal energy (U) of a gas is related to its pressure (P) and volume as

P = \frac{2}{3}U

for a monoatomic ideal gas.

Solution approach

Correct Answer + Core Reason: $P = \frac{2}{3}U$ . This is the correct relationship between pressure and internal energy for a gas of unit volume.

Step-by-Step Solution:

Step 1: Recall the kinetic theory of gases. The pressure of a gas is related to the average kinetic energy of its molecules.

Step 2: Relate kinetic energy to internal energy. For a monoatomic ideal gas, the internal energy (U) is equal to the total kinetic energy of the molecules.

Step 3: Use the formula for pressure in terms of kinetic energy. The pressure (P) of a gas is given by $P = \frac{2}{3} \frac{U}{V}$ , where U is the internal energy and V is the volume.

Step 4: Apply the condition of unit volume. Since the volume is unit volume (V = 1), the equation becomes $P = \frac{2}{3}U$ .

Why other options are wrong:
- Option B: This is not the correct relationship between pressure and internal energy.
- Option C: This is not the correct relationship between pressure and internal energy.
- Option D: This is not the correct relationship between pressure and internal energy.

Key Concepts:
- Kinetic Theory of Gases
- Internal Energy
- Pressure

Memory Aids:
- $P = \frac{2}{3} \frac{U}{V}$ for a monoatomic ideal gas

Quick Tips:
- Remember the relationship between pressure, volume, and internal energy from the kinetic theory of gases.

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The equilibrium constant

(K)

of a reaction may be written as,

View answer and explanation

Correct option: D

K=e^{-\Delta G^{\circ} / R T}

Option D (

K=e^{-\Delta G^{\circ} / R T}

) is correct because it accurately represents the relationship between the equilibrium constant (K) and the standard Gibbs free energy change (

\Delta G^{\circ}

).

Solution approach

Correct Answer + Core Reason: $K=e^{-\Delta G^{\circ} / R T}$ . This equation directly links the equilibrium constant to the standard Gibbs free energy change.

Step-by-Step Solution:

Step 1: Recall the relationship between the equilibrium constant (K) and the standard Gibbs free energy change ( $\Delta G^{\circ}$ ), the gas constant (R), and temperature (T).

Step 2: The equation is: $\Delta G^{\circ} = -RT \ln K$

Step 3: Rearranging the equation to solve for K, we get: $K = e^{-\Delta G^{\circ} / RT}$

Why other options are wrong:
- Option A: $K=e^{-\Delta H^{\circ} / R T}$ is incorrect because it relates K to the standard enthalpy change, not the Gibbs free energy change.
- Option B: $K=e^{-\Delta G / R T}$ is incorrect because it uses the non-standard Gibbs free energy change, not the standard one.
- Option C: $K=e^{-\Delta H / R T}$ is incorrect because it relates K to the enthalpy change, not the Gibbs free energy change.

Key Concepts:
- Equilibrium Constant
- Gibbs Free Energy

Memory Aids:
- Remember that a negative $\Delta G^{\circ}$ leads to a large K, indicating a spontaneous reaction.

Quick Tips:
- Always use the standard Gibbs free energy change ( $\Delta G^{\circ}$ ) when calculating the equilibrium constant.

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Analyzing +4 | -1

Consider the following reactions :

\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})+x \mathrm{~kJ}

\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+y \mathrm{~kJ}

The heat of formation of

\mathrm{CO}(g)

is :

View answer and explanation

Correct option: C

(y-x) \mathrm{kJ} / \mathrm{mol}

Option C is correct. The heat of formation of CO(g) can be calculated by subtracting the second reaction from the first, resulting in (y-x) kJ/mol.

Solution approach

Correct Answer + Core Reason: $(y-x) \mathrm{kJ} / \mathrm{mol}$ . The heat of formation is calculated by manipulating the given equations.

Step-by-Step Solution:

Step 1: Write the target equation for the heat of formation of CO(g): $\mathrm{C}(s) + \frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g) + z \mathrm{kJ}$ , where z is the heat of formation.

Step 2: Write the given equations:
Equation 1: $\mathrm{C}(s)+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+x \mathrm{kJ}$
Equation 2: $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)+y \mathrm{kJ}$

Step 3: Manipulate the equations to obtain the target equation. Subtract Equation 2 from Equation 1:
[ $\mathrm{C}(s)+\mathrm{O}_2(g)$ ] - [ $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g)$ ] $\longrightarrow$ [ $\mathrm{CO}_2(g)+x \mathrm{kJ}$ ] - [ $\mathrm{CO}_2(g)+y \mathrm{kJ}$ ]

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Understanding +4 | -1

When hydrogen and oxygen burn to form water in an oxyhydrogen torch, the entropy change is:

View answer and explanation

Correct option: A

Negative

Option A (Negative) is correct because the formation of water from hydrogen and oxygen involves a decrease in the number of gas molecules, leading to a decrease in entropy.

Solution approach

Correct Answer + Core Reason: Negative. The reaction reduces the number of gaseous molecules, decreasing entropy.

Step-by-Step Solution:

Step 1: Identify the reaction: $2H_2(g) + O_2(g) \longrightarrow 2H_2O(l)$

Step 2: Analyze the change in the number of moles of gas: 3 moles of gas are converted into 0 moles of gas (liquid water has negligible entropy compared to gases).

Step 3: Since the number of gaseous molecules decreases, the disorder decreases, and hence the entropy change is negative.

Step 4: Conclusion: Entropy decreases.

Why other options are wrong:
- Option B: Positive - This would be true if the reaction increased the number of gaseous molecules, which is not the case here.
- Option C: Zero - This would be true if there was no change in the number of gaseous molecules, which is not the case here.
- Option D: May be positive or negative - While entropy changes can be positive or negative depending on the reaction, in this specific reaction, it is definitively negative.

Key Concepts:
- Entropy and its relation to disorder
- Entropy change in chemical reactions
- Phase changes and entropy

Memory Aids:
- Gases have higher entropy than liquids or solids.
- Reactions that decrease the number of gas molecules usually have a negative entropy change.

Quick Tips:
- Always consider the change in the number of moles of gas when determining the sign of entropy change in a reaction.

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Which of the following relation is correct
regarding reversible adiabatic process?

View answer and explanation

Correct option: C

\mathrm{T}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}^{\gamma-1}=\mathrm{T}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}^{\gamma-1}

Option C (

\mathrm{T}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}^{\gamma-1}=\mathrm{T}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}^{\gamma-1}

) is correct because it represents the correct relationship between temperature and volume in a reversible adiabatic process.

Solution approach

Correct Answer + Core Reason: $\mathrm{T}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}^{\gamma-1}=\mathrm{T}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}^{\gamma-1}$ . This equation describes the relationship between temperature and volume in a reversible adiabatic process.

Step-by-Step Solution:

Step 1: Recall the equation for a reversible adiabatic process relating temperature and volume.

Step 2: The correct equation is $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$ .

Step 3: This equation is derived from the adiabatic condition $PV^{\gamma} = constant$ and the ideal gas law $PV = nRT$ .

Step 4: Combining these equations, we can eliminate pressure and obtain the relationship between temperature and volume.

Step 5: The final equation is $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$ .

Why other options are wrong:
- Option A: $\mathrm{T}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}^\gamma=\mathrm{T}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}^\gamma$ is incorrect because the exponent should be $\gamma - 1$ .
- Option B: $P_i V_i=P_f V_f$ is incorrect because it represents Boyle's law, which applies to isothermal processes, not adiabatic processes.
- Option D: $P_i^\gamma V_i=P_f^\gamma V_f$ is incorrect. The correct relation involving P and V is $P_i V_i^\gamma = P_f V_f^\gamma$ .

Key Concepts:
- Reversible adiabatic process
- Adiabatic index ( $\gamma$ )
- Ideal gas law

Memory Aids:
- Remember that in adiabatic processes, no heat is exchanged with the surroundings.
- The exponent of V in the T-V relation is always ( $\gamma - 1$ ).

Quick Tips:
- Always remember the correct exponents in the adiabatic process equations.
- Distinguish between isothermal and adiabatic processes.

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Remembering +4 | -1

Which of the following represents the first law
of thermodynamics

View answer and explanation

Correct option: A

\Delta U=q+W

The correct answer is A (

\Delta U=q+W

) because it accurately represents the first law of thermodynamics, stating that the change in internal energy equals heat added plus work done on the system.

Solution approach

Correct Answer + Core Reason: $\Delta U=q+W$ . This is the standard mathematical expression of the first law of thermodynamics.

Step-by-Step Solution:

Step 1: State the first law of thermodynamics: The change in internal energy of a system is equal to the heat added to the system plus the work done on the system.

Step 2: Express mathematically: $\Delta U = q + W$ , where $\Delta U$ is the change in internal energy, $q$ is the heat added to the system, and $W$ is the work done *on* the system. Note that some conventions define W as work done *by* the system, in which case the equation becomes $\Delta U = q - W$ .

Why other options are wrong:
- Option B: $\Delta U = p \Delta V$ : This represents work done at constant pressure, not the first law itself. It's missing the heat term.
- Option C: $\Delta U = \Delta H + p \Delta V$ : This is incorrect. The correct relationship is $\Delta H = \Delta U + p \Delta V$ , where $\Delta H$ is the change in enthalpy.
- Option D: $\Delta H = q + W$ : This is incorrect. Enthalpy change is defined as $\Delta H = \Delta U + p \Delta V$ , and at constant pressure, $\Delta H = q_p$ .

Key Concepts:
- First law of thermodynamics
- Internal energy (U)
- Heat (q)
- Work (W)
- Enthalpy (H)

Memory Aids:
- Energy cannot be created or destroyed, only transferred or converted.

Quick Tips:
- Pay attention to the sign conventions for work (done on vs. done by the system).

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Application +4 | -1

For a reaction,

A\left( l \right) \to 4B\left( g \right)

if

\Delta U = 4.5\,kJ

and

\Delta S = 60\,J{K^{ - 1}}

at

{27^0}C

then

\Delta G

will be

View answer and explanation

Correct option: C

–3.5 kJ

Option C (–3.5 kJ) is correct because it is calculated using the formula

\Delta G = \Delta U + \Delta(PV) - T\Delta S = \Delta U + \Delta nRT - T\Delta S

, where

\Delta n

is the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: –3.5 kJ. $\Delta G$ is calculated using $\Delta G = \Delta H - T\Delta S$ and $\Delta H = \Delta U + \Delta n_g RT$ .

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta U = 4.5 \text{ kJ}$ , $\Delta S = 60 \text{ J/K} = 0.06 \text{ kJ/K}$ , and $T = 27^{\circ}C = 300 \text{ K}$ . The reaction is $A(l) \to 4B(g)$ .

Step 2: Calculate the change in the number of moles of gas: $\Delta n_g = 4 - 0 = 4$ .

Step 3: Calculate $\Delta H$ : $\Delta H = \Delta U + \Delta n_g RT = 4.5 \text{ kJ} + 4 \times (8.314 \times 10^{-3} \text{ kJ/mol K}) \times 300 \text{ K} = 4.5 \text{ kJ} + 9.9768 \text{ kJ} \approx 4.5 + 10 = 14.5 \text{ kJ}$ . This is incorrect. We should use $\Delta H = \Delta U + P\Delta V = \Delta U + \Delta n RT = 4.5 + 4(8.314/1000)(300) = 4.5 + 9.9768 = 14.4768 kJ$

Step 4: Calculate $\Delta G$ : $\Delta G = \Delta H - T\Delta S = 14.4768 \text{ kJ} - 300 \text{ K} \times 0.06 \text{ kJ/K} = 14.4768 \text{ kJ} - 18 \text{ kJ} = -3.5232 \text{ kJ}$ .

Step 5: Round to the correct number of significant figures: $\Delta G \approx -3.5 \text{ kJ}$ .

Final Answer: –3.5 kJ

Why other options are wrong:
- Option A: – 2.5 kJ is incorrect. This value might be obtained by not including the $\Delta n RT$ term in the calculation of $\Delta H$ .
- Option B: –1.5 kJ is incorrect. This value might be obtained by using an incorrect value for $\Delta n$ or $\Delta S$ .
- Option D: –6.5 kJ is incorrect. This value might be obtained by incorrectly adding the $\Delta H$ and $T\Delta S$ terms instead of subtracting them.

Key Concepts:
- Gibbs free energy: $\Delta G = \Delta H - T\Delta S$
- Enthalpy change: $\Delta H = \Delta U + \Delta n_g RT$

Memory Aids:
- Remember the relationship between Gibbs free energy, enthalpy, entropy, and temperature: $G = H - TS$ .

Quick Tips:
- Always convert temperature to Kelvin when dealing with thermodynamic calculations.
- Pay attention to the units of $\Delta U$ , $\Delta S$ , and $R$ to ensure consistency.

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Understanding +4 | -1

Consider an endothermic reaction

X \to Y

with the activation energies, respectively in general

View answer and explanation

Correct option: A

{E_b} < {E_f}

Option A (

{E_b} < {E_f}

) is correct because in an endothermic reaction, the energy of the products is higher than the energy of the reactants, thus the forward activation energy is greater than the backward activation energy.

Solution approach

Correct Answer + Core Reason: ${E_b} < {E_f}$ . For an endothermic reaction, the activation energy for the forward reaction is greater than the activation energy for the reverse reaction.

Step-by-Step Solution:

Step 1: Define endothermic reaction: A reaction that absorbs heat, meaning the products have higher energy than the reactants.

Step 2: Visualize the energy diagram: In an endothermic reaction, the energy level of the products is higher than that of the reactants. The activation energy for the forward reaction ( $E_f$ ) is the energy difference between the reactants and the transition state, while the activation energy for the backward reaction ( $E_b$ ) is the energy difference between the products and the transition state.

Step 3: Compare $E_f$ and $E_b$ : Since the products have higher energy than the reactants, the energy difference between the reactants and the transition state ( $E_f$ ) must be greater than the energy difference between the products and the transition state ( $E_b$ ).

Step 4: Conclusion: Therefore, ${E_b} < {E_f}$ .

Why other options are wrong:
- Option B: ${E_b} > {E_f}$ : This is incorrect because it describes an exothermic reaction.
- Option C: ${E_b} = {E_f}$ : This is incorrect because it would mean there is no net energy change in the reaction.
- Option D: None: This is incorrect because option A is correct.

Key Concepts:
- Endothermic Reactions
- Activation Energy
- Energy Diagrams

Memory Aids:
- Endothermic: Energy In (products higher than reactants)
- Exothermic: Energy Out (reactants higher than products)

Quick Tips:
- Draw an energy diagram to visualize the activation energies for forward and reverse reactions.

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Analyzing +4 | -1

The heats of neutralization of

C{H_3}COOH,\,\,\,HCOOH,\,\,HCN

and

{H_2}S

are

- 13.2,\, - \,13.4,\, - 2.9

and - 3.8 kcal per equivalent respectively. Arrange the increasing order of strength.

View answer and explanation

Correct option: A

HCCOH > C{H_3}COOH > {H_2}S > HCN

Option A (

HCCOH > C{H_3}COOH > {H_2}S > HCN

) is correct because the strength of an acid is inversely proportional to its heat of neutralization; lower the heat of neutralization, stronger the acid.

Solution approach

Correct Answer + Core Reason: $HCCOH > C{H_3}COOH > {H_2}S > HCN$ . Acid strength is inversely proportional to the heat of neutralization.

Step-by-Step Solution:

Step 1: The heats of neutralization are given as:
$CH_3COOH$ : -13.2 kcal/equivalent\nHCOOH: -13.4 kcal/equivalent\nH2S: -3.8 kcal/equivalent\nHCN: -2.9 kcal/equivalent

Step 2: The strength of an acid is inversely proportional to its heat of neutralization. A stronger acid will have a higher heat of neutralization (more negative value) because more heat is released when it reacts with a base.

Step 3: Comparing the values, we can arrange the acids in increasing order of strength:\nHCN < H2S < CH3COOH < HCOOH

Step 4: Therefore, the correct order is: HCCOH > C{H_3}COOH > {H_2}S > HCN

Why other options are wrong:
- **Option B ( $C{H_3}COOH > HCOOH > {H_2}S > HCN$ ): This is incorrect because it swaps the positions of HCOOH and CH3COOH.
- Option C ( ${H_2}S > HCOOH > C{H_3}COOH > HCN$ ): This is incorrect because it incorrectly orders the acids based on their heats of neutralization.
- Option D ( $HCOOH > {H_2}S > C{H_3}COOH > HCN$ ): This is incorrect because it incorrectly orders the acids based on their heats of neutralization.

Key Concepts:
- Heat of neutralization
- Acid strength

Memory Aids:
- Remember that stronger acids have higher (more negative) heats of neutralization.

Quick Tips:**
- The heat of neutralization is a measure of the heat released when an acid and a base react to form salt and water.
- Stronger acids release more heat upon neutralization.

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Analyzing +4 | -1

At

25^{\circ} \mathrm{C}

, the enthalpy of the following processes are given

\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =78 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =-242 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

\mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=436 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=249 \,\mathrm{kJ} \,\mathrm{mol}^{-1}

What would be the value of

X

for the following reaction? _____ (Nearest integer)

\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=\mathrm{X} \,\mathrm{kJ} \,\mathrm{mol}^{-1}

View answer and explanation

Correct option: A

499

The correct answer is 499. By manipulating the given equations and applying Hess's Law, we can calculate the enthalpy change for the desired reaction.

Solution approach

Correct Answer + Core Reason: 499. Hess's Law is applied to calculate the enthalpy change for the reaction by manipulating the given reactions.

Step-by-Step Solution:
Step 1: Identify the target reaction: $\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=\mathrm{X} \mathrm{~kJ} \mathrm{~mol}^{-1}$
Step 2: Write down the given reactions and their enthalpy changes:
(1) $\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =78 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(2) $\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ} =-242 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(3) $\mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=436 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(4) $\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{O}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=249 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Step 3: Manipulate the equations to obtain the target reaction:\nReverse equation (2): $\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = 242 \mathrm{~kJ} \mathrm{~mol}^{-1}$ \nDivide equation (1) by 2: $\frac{1}{2}\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = 39 \mathrm{~kJ} \mathrm{~mol}^{-1}$ \nDivide equation (3) by 2: $\frac{1}{2}\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = 218 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Step 4: Add the manipulated equations:
$\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = 242 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\frac{1}{2}\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = 39 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\frac{1}{2}\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = 218 \mathrm{~kJ} \mathrm{~mol}^{-1}$ \nAdding these gives: $\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g}) + \mathrm{OH}(\mathrm{g}) \Delta \mathrm{H}^{\circ} = (242 + 39 + 218) \mathrm{~kJ} \mathrm{~mol}^{-1} = 499 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Why other options are wrong:
- Option B: Incorrect calculation. The enthalpy change is not zero.
- Option C: Incorrect calculation. The enthalpy change is not zero.
- Option D: Incorrect calculation. The enthalpy change is not zero.

Key Concepts:
- Hess's Law
- Enthalpy change of reaction
- Manipulating thermochemical equations

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

Quick Tips:
- Carefully manipulate the given equations to match the target equation.
- Remember to change the sign of $\Delta H$ when reversing a reaction.

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Understanding +4 | -1

A closed flask contains water in all its three states, solids, liquid and vapour at 0℃. In this situation the average KE of the water molecule will be:

View answer and explanation

Correct option: D

Same in all the three states

Option D (Same in all the three states) is correct because the average KE of water molecules depends only on temperature, which is constant at 0°C.

Solution approach

Correct Answer + Core Reason: Same in all the three states. The average kinetic energy of molecules depends only on temperature.

Step-by-Step Solution:

Step 1: Identify the key concept: Average kinetic energy is directly proportional to temperature.

Step 2: Recall the formula: $KE_{avg} = \frac{3}{2}kT$ , where k is the Boltzmann constant and T is the temperature.

Step 3: Since the temperature is constant at 0°C for all three states (solid, liquid, and vapor), the average kinetic energy will be the same.

Step 4: Conclusion: The average KE is the same in all three states.

Why other options are wrong:
- Option A: Maximum in vapour state: This is incorrect because the temperature is the same for all states.
- Option B: Maximum in solid state: This is incorrect because the temperature is the same for all states.
- Option C: Greater in the liquid than in vapour state: This is incorrect because the temperature is the same for all states.

Key Concepts:
- Kinetic Molecular Theory
- Average Kinetic Energy
- Temperature

Memory Aids:
- Kinetic energy is directly proportional to temperature.

Quick Tips:
- Remember that phase changes occur at constant temperature.

74 Medium

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Understanding +4 | -1

The internal energy change when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversible path and returns to state A by an irreversible path, what would be the net change in internal energy?

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because internal energy is a state function, meaning the change in internal energy depends only on the initial and final states, not the path taken. Since the system returns to its initial state, the net change in internal energy is zero.

Solution approach

Correct Answer + Core Reason: Zero. Internal energy is a state function.

Step-by-Step Solution:

Step 1: Define a state function as a property that depends only on the initial and final states, not the path taken.

Step 2: Recognize that internal energy (E) is a state function.

Step 3: Since the system returns to its initial state, the change in internal energy is zero, regardless of the path taken.

Why other options are wrong:
- Option A: 40 kJ is incorrect because it only considers the change in internal energy from A to B, not the entire cycle.
- Option B: >40 kJ is incorrect because the net change must be zero for a state function.
- Option C: <40 kJ is incorrect because the net change must be zero for a state function.

Key Concepts:
- State function
- Internal energy (E)

Memory Aids:
- State functions depend only on initial and final states.

Quick Tips:
- For any cyclic process, the change in a state function is zero.

75 Medium

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Remembering +4 | -1

For the reversible process, the value of

\Delta S

is given by the expression

View answer and explanation

Correct option: B

q_{(\text {rev })} \div \mathrm{T}

Option B (

q_{(\text {rev })} \div \mathrm{T}

) is correct because the change in entropy for a reversible process is defined as the reversible heat transfer divided by the absolute temperature.

Solution approach

Correct Answer + Core Reason: $q_{(\text {rev })} \div \mathrm{T}$ . This is the fundamental definition of entropy change for a reversible process.

Step-by-Step Solution:

Step 1: Recall the definition of entropy change for a reversible process.

Step 2: The entropy change ( $\Delta S$ ) for a reversible process is given by the equation: $\Delta S = \frac{{q_{\text{(rev)}}}}{T}$ , where $q_{\text{(rev)}}$ is the heat transferred reversibly and T is the absolute temperature.

Why other options are wrong:
- Option A: This is incorrect because it multiplies the heat by the temperature instead of dividing.
- Option C: This is incorrect because it relates to enthalpy change and temperature change, which is not the direct definition of entropy change for a reversible process.
- Option D: This is incorrect because it divides the temperature by the heat, which is the inverse of the correct relationship.

Key Concepts:
- Entropy change
- Reversible process

Memory Aids:
- $\Delta S = \frac{{q_{\text{rev}}}}{T}$

Quick Tips:
- Remember that this formula applies only to reversible processes.

76 Medium

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Understanding +4 | -1

Work done by the system on surroundings is:

View answer and explanation

Correct option: B

Negative

Option B (Negative) is correct because, by convention, work done *by* the system on the surroundings is considered negative, indicating energy leaving the system.

Solution approach

Correct Answer + Core Reason: Negative. Work done by the system is considered negative.

Step-by-Step Solution:

Step 1: By convention, work done *by* the system on the surroundings is considered negative.

Step 2: This is because the system is losing energy to the surroundings.

Step 3: Mathematically, $w = -P\Delta V$ when the system expands.

Why other options are wrong:
- Option A: Positive: This is incorrect because work done *on* the system is positive, not by the system.
- Option C: Zero: This is incorrect because work is done if there is a change in volume against an external pressure.
- Option D: None of these: This is incorrect because work done by the system is negative.

Key Concepts:
- Work done by the system
- Sign convention for work

Memory Aids:
- Work *by* the system is negative.
- Work *on* the system is positive.

Quick Tips:
- Remember the sign convention for work.

77 Medium

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Understanding +4 | -1

For the reaction,

\begin{array}{l} {\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {x_1}\,\,\mathrm{kJ} \ldots \ldots \ldots \left( \mathrm{i} \right)\\ 2\mathrm{HCl}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) - {x_2}\,\mathrm{kJ} \ldots \ldots \left( {\mathrm{ii}} \right) \end{array}

Which of the following statement is correct?

View answer and explanation

Correct option: A

{x_1}

  and 

{x_2}

  are numerically equal

Option A (

{x_1}

  and 

{x_2}

  are numerically equal) is correct because the forward and reverse reactions have the same magnitude of enthalpy change, but opposite signs.

Solution approach

Correct Answer + Core Reason: ${x_1}$   and  ${x_2}$   are numerically equal. The forward and reverse reactions have the same magnitude of enthalpy change, but opposite signs.

Step-by-Step Solution:

Step 1: Analyze the given reactions:
${\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{HCl}\left( \mathrm{g} \right) + {x_1}\,\mathrm{kJ}$ (i)
$2\mathrm{HCl}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) - {x_2}\,\mathrm{kJ}$ (ii)

Step 2: Reaction (ii) is the reverse of reaction (i).

Step 3: The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to the enthalpy change for the forward reaction. Therefore, $x_1 = x_2$ .

Why other options are wrong:
- Option B: ${x_1}$ and ${x_2}$ are numerically equal, not different.
- Option C: ${x_1} - {x_2} = 0$ , so ${x_1} - {x_2} > 0$ is incorrect.
- Option D: ${x_1} - {x_2} = 0$ , so ${x_1} - {x_2} < 0$ is incorrect.

Key Concepts:
- Enthalpy change
- Forward and reverse reactions

Memory Aids:
- Reverse reactions have the opposite sign of $\Delta H$ .

Quick Tips:
- Remember that the magnitude of $\Delta H$ is the same for forward and reverse reactions.

78 Medium

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Understanding +4 | -1

Net work done by the system in a cyclic process is equal to:

View answer and explanation

Correct option: D

q

Option D (q) is correct because in a cyclic process, the change in internal energy is zero, so the net work done is equal to the heat absorbed by the system.

Solution approach

Correct Answer + Core Reason: q. In a cyclic process, the net work done is equal to the heat exchanged with the surroundings.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta U = q - w$ , where $\Delta U$ is the change in internal energy, q is the heat added to the system, and w is the work done by the system.

Step 2: In a cyclic process, the system returns to its initial state, so the change in internal energy is zero: $\Delta U = 0$ .

Step 3: Therefore, $0 = q - w$ , which means $w = q$ .

Step 4: The net work done by the system is equal to the heat absorbed by the system.

Why other options are wrong:
- Option A: This is incorrect because while the change in internal energy is zero, the net work done is not necessarily zero.
- Option B: This is incorrect because the change in internal energy is zero in a cyclic process.
- Option C: This is incorrect because enthalpy is not directly related to the net work done in a cyclic process.

Key Concepts:
- First law of thermodynamics
- Cyclic process
- Internal energy

Memory Aids:
- Cyclic process: $\Delta U = 0$ , so w = q

Quick Tips:
- Remember that in a cyclic process, the system returns to its initial state, so all state functions have a change of zero.

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Remembering +4 | -1

The Kirchhoff’s equation gives the effect of …..on heat of reaction.

View answer and explanation

Correct option: B

Temperature

Option B (Temperature) is correct because Kirchhoff's equation describes how the heat of reaction changes with temperature.

Solution approach

Correct Answer + Core Reason: Temperature. Kirchhoff's equation relates the change in enthalpy of a reaction to the change in temperature.

Step-by-Step Solution:

Step 1: Recall Kirchhoff's equation.

Step 2: Kirchhoff's equation is given by: $\frac{{d(\Delta H)}}{{dT}} = \Delta C_p$ , where $\Delta H$ is the enthalpy change, $T$ is the temperature, and $\Delta C_p$ is the change in heat capacity at constant pressure.

Step 3: This equation shows the effect of temperature on the heat of reaction.

Why other options are wrong:
- Option A: Pressure does affect the equilibrium constant, but not directly through Kirchhoff's equation.
- Option C: Volume is not directly related to the heat of reaction through Kirchhoff's equation.
- Option D: Molecularity is related to the reaction mechanism, not the heat of reaction's temperature dependence.

Key Concepts:
- Kirchhoff's equation
- Heat of reaction
- Temperature dependence of enthalpy

Memory Aids:
- Kirchhoff's equation relates enthalpy change to temperature change.

Quick Tips:
- Remember that Kirchhoff's equation deals with the temperature dependence of the heat of reaction.

80 Medium

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Understanding +4 | -1

The bond energy of

{H_2}

is 104.3 kcal

mo{l^{ - 1}}.

. It means : (N=Av.no.)

View answer and explanation

Correct option: C

104.3 kcal heat is evolved during combination of 2N atoms of H to from N molecules of

{H_2}

Heat of formation of H atom

Option C (104.3 kcal heat is evolved during combination of 2N atoms of H to from N molecules of

{H_2}

) is correct because bond energy is the energy released when a bond is formed.

Solution approach

Correct Answer + Core Reason: 104.3 kcal heat is evolved during combination of 2N atoms of H to from N molecules of ${H_2}$ Heat of formation of H atom. Bond energy is the energy released when a bond is formed.

Step-by-Step Solution:

Step 1: Define bond energy.\nBond energy is the amount of energy required to break one mole of bonds of a particular type in gaseous molecules. It is also the amount of energy released when one mole of bonds of a particular type is formed from gaseous atoms.

Step 2: Analyze the given information.\nThe bond energy of ${H_2}$ is 104.3 kcal $mo{l^{ - 1}}$ .

Step 3: Relate bond energy to bond formation and breaking.\nBreaking a bond requires energy input (endothermic), while forming a bond releases energy (exothermic).

Step 4: Evaluate the statements.\nStatement A: 104.3 kcal heat is required to break up N bonds in N molecules of ${H_2}$ . This statement is incorrect because it should refer to N molecules, not N bonds.\nStatement B: 104.3 kcal heat is required to break up N molecules to 2N atoms of H. This statement is correct.\nStatement C: 104.3 kcal heat is evolved during combination of 2N atoms of H to from N molecules of ${H_2}$ . This statement is correct because bond formation is an exothermic process.\nStatement D: $= \frac{1}{2} \times \mathrm{bond\,\, energy\,\, of\,\, H}-\mathrm{H}$ . This statement is incorrect. The heat of formation of H atom is half of the bond dissociation energy of H2.

Step 5: Considering the question asks for what it *means*, not what is required, C is the best answer.

Why other options are wrong:
- Option A: 104.3 kcal heat is required to break up N bonds in N molecules of ${H_2}$ . This is incorrect because the statement is about N molecules, not N bonds.
- Option B: 104.3 kcal heat is required to break up N molecules to 2N atoms of H. While technically correct regarding breaking the bonds, the question asks what the bond energy *means*, not what it requires to break bonds.
- Option D: $= \frac{1}{2} \times \mathrm{bond\,\, energy\,\, of\,\, H}-\mathrm{H}$ . This is incorrect because it is not a complete statement and doesn't fully explain the meaning of bond energy.

Key Concepts:
- Bond energy
- Bond formation and breaking
- Endothermic and exothermic processes
- Avogadro's number

Memory Aids:
- Breaking bonds requires energy (endothermic).
- Forming bonds releases energy (exothermic).

Quick Tips:
- Remember that bond energy is always a positive value, representing the energy required to break a bond or the energy released when a bond is formed.

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Understanding +4 | -1

One mole of a gas absorbs 200 J of heat at constant volume. Its temperature rises from 298K To 308K. The change in internal energy is:

View answer and explanation

Correct option: A

200\mathrm{J}

The correct answer is A (200J) because at constant volume, all the heat absorbed goes into increasing the internal energy of the gas (q = ∆U).

Solution approach

Correct Answer + Core Reason: $200\mathrm{J}$ . At constant volume, the heat absorbed equals the change in internal energy.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta U = q + W$

Step 2: Apply the condition of constant volume: At constant volume, no work is done, so $W = 0$ .

Step 3: Simplify the equation: $\Delta U = q + 0 = q$

Step 4: Substitute the given value: $q = 200\mathrm{J}$ , so $\Delta U = 200\mathrm{J}$

Why other options are wrong:
- Option B: $- 200\mathrm{J}$ : This is incorrect because the heat absorbed is positive, so the change in internal energy is also positive.
- Option C: $200 \times \frac{{308}}{{298}}\mathrm{J}$ : This is incorrect because the temperature change is not needed to calculate the change in internal energy at constant volume.
- Option D: $200 \times \frac{{298}}{{308}}\mathrm{J}$ : This is incorrect because the temperature change is not needed to calculate the change in internal energy at constant volume.

Key Concepts:
- First law of thermodynamics
- Internal energy (U)
- Heat (q)
- Work (W)
- Constant volume process

Memory Aids:
- Constant volume = No work
- $\Delta U = q$ (at constant volume)

Quick Tips:
- Remember that at constant volume, all heat absorbed goes into increasing internal energy.

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Application +4 | -1

A system is provided 50 J of heat and work done on the system is 10 J. what is the change in internal energy

View answer and explanation

Correct option: B

60 J

Option B (60 J) is correct because the change in internal energy is the sum of heat added to the system and work done on the system, according to the first law of thermodynamics.

Solution approach

Correct Answer + Core Reason: 60 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 values: Heat added to the system, $q = 50 \text{ J}$ , and work done on the system, $w = 10 \text{ J}$ .

Step 2: Apply the first law of thermodynamics: $\Delta U = q + w$ .

Step 3: Substitute the values into the equation: $\Delta U = 50 \text{ J} + 10 \text{ J}$ .

Step 4: Calculate the change in internal energy: $\Delta U = 60 \text{ J}$ .

Step 5: Final Answer: 60 J.

Why other options are wrong:
- Option A (40 J): This is incorrect as it likely involves subtracting the work done from the heat added, instead of adding them.
- Option C (30 J): This is incorrect and doesn't follow from a correct application of the first law of thermodynamics.
- Option D (50 J): This is incorrect as it only considers the heat added and neglects the work done on the system.

Key Concepts:
- First Law of Thermodynamics: $\Delta U = q + w$
- Internal Energy, Heat, Work

Memory Aids:
- $\Delta U = q + w$ (Internal energy change equals heat plus work)

Quick Tips:
- Remember the sign conventions: Heat added to the system is positive, heat released by the system is negative. Work done on the system is positive, work done by the system is negative.

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Application +4 | -1

The values of ∆H and ∆S for the reaction.

{C_{\left( {\mathrm{graphite}} \right)}} + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{CO}\left( \mathrm{g} \right)

Are 170 kJ and 170

\mathrm{J}{\mathrm{K}^{ - 1}}

respectively. this reaction will be spontaneous at:

View answer and explanation

Correct option: D

1110 K

Option D (1110 K) is correct because the reaction becomes spontaneous when T > ∆H/∆S, which calculates to be 1000 K.

Solution approach

Correct Answer + Core Reason: 1110 K. The reaction is spontaneous when $\Delta G < 0$ , which occurs when $T > \frac{\Delta H}{\Delta S}$ .

Step-by-Step Solution:

Step 1: Identify the given values: $\Delta H = 170 \text{{ kJ}} = 170000 \text{{ J}}$ , $\Delta S = 170 \text{{ J/K}}$

Step 2: Determine the condition for spontaneity: $\Delta G = \Delta H - T\Delta S < 0$

Step 3: Rearrange the inequality: $T > \frac{\Delta H}{\Delta S}$

Step 4: Substitute the values: $T > \frac{170000 \text{{ J}}}{170 \text{{ J/K}}}$

Step 5: Calculate the temperature: $T > 1000 \text{{ K}}$

Step 6: From the given options, the temperature at which the reaction is spontaneous is 1110 K, as it is the only option greater than 1000 K.

Step 7: Final Answer: 1110 K

Why other options are wrong:
- Option A (510 K): This is incorrect because 510 K is less than 1000 K, meaning $\Delta G$ would be positive and the reaction would not be spontaneous.
- Option B (710 K): This is incorrect because 710 K is less than 1000 K, meaning $\Delta G$ would be positive and the reaction would not be spontaneous.
- Option C (910 K): This is incorrect because 910 K is less than 1000 K, meaning $\Delta G$ would be positive and the reaction would not be spontaneous.

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

Memory Aids:
- For a reaction to be spontaneous, the temperature must be high enough to overcome the enthalpy change, considering the entropy change.

Quick Tips:
- Always convert $\Delta H$ to Joules if $\Delta S$ is in J/K.
- Remember that a reaction is spontaneous when $\Delta G < 0$ .

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Remembering +4 | -1

Internal energy of an ideal gas depends on:

View answer and explanation

Correct option: B

Temperature

Option B (Temperature) is correct because the internal energy of an ideal gas depends only on its temperature, as there are no intermolecular forces.

Solution approach

Correct Answer + Core Reason: Temperature. The internal energy of an ideal gas is solely dependent on its temperature.

Step-by-Step Solution:

Step 1: Ideal Gas Definition: An ideal gas is a theoretical gas with no intermolecular forces and negligible volume of gas particles.

Step 2: Internal Energy of Ideal Gas: The internal energy (U) of an ideal gas is a function of temperature (T) only. It is independent of pressure and volume.

Step 3: Mathematical Representation: $U = f(T)$

Step 4: Conclusion: Therefore, the internal energy of an ideal gas depends only on temperature.

Why other options are wrong:
- Option A: Pressure: The internal energy of an ideal gas does not depend on pressure.
- Option C: Volume: The internal energy of an ideal gas does not depend on volume.
- Option D: None of these: This is incorrect because the internal energy does depend on temperature.

Key Concepts:
- Internal Energy (U)
- Ideal Gas
- Kinetic Molecular Theory

Memory Aids:
- Ideal Gas = Temperature Only

Quick Tips:
- Remember that for an ideal gas, internal energy is only a function of temperature.

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Application +4 | -1

A carnot engine operates between temperature T and 400 K (T>400 K). If efficiency of engine is 25%, the temperature T is:

View answer and explanation

Correct option: C

533.3 K

Option C (533.3 K) is correct because the efficiency of a Carnot engine is given by

\eta = 1 - \frac{T_c}{T_h}

, where

T_c

is the cold reservoir temperature and

T_h

is the hot reservoir temperature. Solving for

T_h

with the given efficiency yields approximately 533.3 K.

Solution approach

Correct Answer + Core Reason: 533.3 K. The temperature T is calculated using the Carnot engine efficiency formula.

Step-by-Step Solution:

Step 1: Carnot Engine Efficiency Formula: The efficiency ( $\eta$ ) of a Carnot engine is given by $\eta = 1 - \frac{T_c}{T_h}$ , where $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir (both in Kelvin).

Step 2: Given values: $\eta = 25\% = 0.25$ and $T_c = 400 \text{{ K}}$ . We need to find $T_h = T$ .

Step 3: Substitute the values into the equation: $0.25 = 1 - \frac{400}{T}$

Step 4: Rearrange the equation to solve for $T$ : $\frac{400}{T} = 1 - 0.25 = 0.75$

Step 5: Solve for $T$ : $T = \frac{400}{0.75} = 533.33 \text{{ K}}$

Why other options are wrong:
- Option A: 400 K. This is the temperature of the cold reservoir, not the hot reservoir.
- Option B: 500 K. This is incorrect; the correct calculation gives 533.3 K.
- Option D: 600 K. This is incorrect; the correct calculation gives 533.3 K.

Key Concepts:
- Carnot Engine
- Efficiency
- Thermodynamics

Memory Aids:
- Remember the Carnot efficiency formula: $\eta = 1 - \frac{T_c}{T_h}$

Quick Tips:
- Always use Kelvin for temperatures in thermodynamic calculations.

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Remembering +4 | -1

Which one of the following is used to remove most of plutonium from spent nuclear fuel?

View answer and explanation

Correct option: B

{O_2}{F_2}

Option B (

{O_2}{F_2}

) is correct because oxygen difluoride is a strong oxidizing agent used to remove plutonium as

PuF_6

.

Solution approach

Correct Answer + Core Reason: ${O_2}{F_2}$ . Oxygen difluoride is a strong fluorinating agent that can convert plutonium to plutonium hexafluoride, which is volatile and can be separated.

Step-by-Step Solution:

Step 1: Understand the problem: The problem asks for a reagent used to remove plutonium from spent nuclear fuel.

Step 2: Recall the properties of plutonium: Plutonium is a radioactive element that needs to be separated from other components of spent nuclear fuel for reprocessing or disposal.

Step 3: Consider the properties of the given reagents:
- $Cl{F_3}$ : Chlorine trifluoride is a strong fluorinating agent, but not as effective as $O_2F_2$ for plutonium removal.
- ${O_2}{F_2}$ : Oxygen difluoride is a very strong oxidizing and fluorinating agent. It can react with plutonium to form plutonium hexafluoride ( $PuF_6$ ), which is a volatile compound that can be easily separated.
- ${I_2}{O_5}$ : Iodine pentoxide is an oxidizing agent, but not strong enough to fluorinate plutonium.
- $Br{O_3}$ : Bromate ion is an oxidizing agent, but not suitable for plutonium removal.

Step 4: Choose the best reagent: Oxygen difluoride ( $O_2F_2$ ) is the most effective reagent for removing plutonium from spent nuclear fuel because it can convert plutonium to a volatile hexafluoride.

Why other options are wrong:
- Option A: $ClF_3$ is a fluorinating agent but less effective than $O_2F_2$ for plutonium.
- Option C: $I_2O_5$ is an oxidizing agent, but not strong enough to fluorinate plutonium.
- Option D: $BrO_3$ is an oxidizing agent, but not suitable for plutonium removal.

Key Concepts:
- Nuclear Fuel Reprocessing
- Fluorination
- Oxidizing Agents
- Volatile Compounds

Memory Aids:
- Remember that $O_2F_2$ is a very strong oxidizing agent.

Quick Tips:
- When dealing with nuclear fuel reprocessing, consider the volatility of the products formed.

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Understanding +4 | -1

For which process energy will be absorbed?

View answer and explanation

Correct option: D

Separation of an electron from a neutral atom

Option D (Separation of an electron from a neutral atom) is correct because energy is always required to remove an electron from a neutral atom due to the attractive force between the nucleus and the electron.

Solution approach

Correct Answer + Core Reason: Separation of an electron from a neutral atom. Removing an electron from a neutral atom requires overcoming the electrostatic attraction between the electron and the positively charged nucleus, thus absorbing energy.

Step-by-Step Solution:

Step 1: Consider the process of ionization, which involves removing an electron from an atom.

Step 2: Ionization requires energy to overcome the electrostatic attraction between the electron and the nucleus.

Step 3: Therefore, energy will be absorbed in the separation of an electron from a neutral atom.

Why other options are wrong:
- Option A: Separation of an electron from an electron is incorrect because electrons repel each other, so energy would be released, not absorbed.
- Option B: Separation of proton from a neutron is incorrect because this involves nuclear forces, and while energy is involved, it's not the primary process described in the question.
- Option C: Separation of a neutron from neutron is incorrect because this involves nuclear forces, and while energy is involved, it's not the primary process described in the question.

Key Concepts:
- Ionization Energy
- Electrostatic Attraction

Memory Aids:
- Removing an electron requires energy.

Quick Tips:
- Remember that ionization energy is always positive.

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Understanding +4 | -1

Which is correct about the heat of combustion?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because heat of combustion is always exothermic, and its value changes with temperature.

Solution approach

Correct Answer + Core Reason: All of the above. Heat of combustion is always exothermic and its value changes with temperature.

Step-by-Step Solution:

Step 1: Define heat of combustion. Heat of combustion is the heat evolved when 1 mole of a substance is completely burnt in oxygen.

Step 2: Consider the nature of combustion. Combustion is always an exothermic process, releasing heat.

Step 3: Consider the temperature dependence. The heat of reaction, including heat of combustion, varies with temperature according to Kirchhoff's equation.

Step 4: Conclusion: Since combustion is always exothermic and its value changes with temperature, "All of the above" is the correct answer.

Why other options are wrong:
- Option A: The combustion be exothermic in some cases and endothermic in other cases: Combustion is always exothermic.
- Option B: Heat of combustion is always exothermic: While true, it's not the complete answer.
- Option C: Its value change with temperature: While true, it's not the complete answer.

Key Concepts:
- Heat of Combustion
- Exothermic Reactions
- Temperature Dependence of Reaction Enthalpy

Memory Aids:
- Combustion = Burning = Heat Released (Exothermic)

Quick Tips:
- Remember that combustion reactions are always exothermic.

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Analyzing +4 | -1

Assertion: When ideal gas expand from

P_1, V_1, T_1

to

P_2, V_2, T_2

in multiple steps, work done is high where number of steps are high.
Reason: Work is path function.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.

Option B is correct because both the assertion and reason are true, but the reason doesn't fully explain why more steps lead to more work; it's about approaching reversibility.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. The work done in multiple steps is higher because it approaches a reversible process, and work is a path function.

Step-by-Step Solution:

Step 1: Understand the Assertion: When an ideal gas expands, the work done depends on the process. More steps usually mean the process is closer to reversible.

Step 2: Understand the Reason: Work is a path function, meaning its value depends on the path taken during the process.

Step 3: Relate Assertion and Reason: While the reason is true, it doesn't fully explain the assertion. The assertion is true because a multi-step process approximates a reversible process, which yields maximum work. The path dependency of work is relevant but not the primary explanation.

Step 4: Conclusion: Both statements are correct, but the reason doesn't directly explain why more steps lead to more work; it's about approaching reversibility.

Why other options are wrong:
- Option A: The reason is not the correct explanation.
- Option C: The assertion is correct.
- Option D: Both statements are correct.

Key Concepts:
- Work done in thermodynamic processes
- Reversible vs. irreversible processes
- Path function

Memory Aids:
- Reversible processes do maximum work.
- Work depends on the path taken.

Quick Tips:
- Remember that reversible processes are ideal and often involve many small steps.

90 Medium

90 of 244Question permalink

Application +4 | -1

Minimum work is obtained when 1 kg of … gas expanded under 500 kPa to 200 kPa pressure at 0℃.

View answer and explanation

Correct option: A

Chlorine

Option A (Chlorine) is correct because minimum work is obtained when the gas with the highest molar mass expands, as it requires more energy to expand due to stronger intermolecular forces. Chlorine has the highest molar mass among the given options.

Solution approach

Correct Answer + Core Reason: Chlorine. Minimum work is obtained when the gas with the highest molar mass expands.

Step-by-Step Solution:

Step 1: Work done in expansion: The work done during the expansion of a gas depends on several factors, including the initial and final pressures, the temperature, and the nature of the gas.

Step 2: Isothermal Reversible Expansion: For an isothermal reversible expansion, the work done is given by $W = -nRT \ln{\frac{V_2}{V_1}} = -nRT \ln{\frac{P_1}{P_2}}$ . Since the pressure change and temperature are the same for all gases, the work done is proportional to the number of moles (n).

Step 3: Number of moles and molar mass: For a given mass (1 kg), the number of moles is inversely proportional to the molar mass ( $n = \frac{m}{M}$ ). Therefore, the gas with the highest molar mass will have the smallest number of moles.

Step 4: Minimum Work: Since the work done is proportional to the number of moles, the gas with the smallest number of moles (highest molar mass) will require the minimum work for expansion.

Step 5: Molar masses of the gases:
- Chlorine (Cl₂): 70.9 g/mol
- Oxygen (O₂): 32 g/mol
- Nitrogen (N₂): 28 g/mol
- Methane (CH₄): 16 g/mol

Step 6: Conclusion: Chlorine has the highest molar mass, so it will have the smallest number of moles for 1 kg of gas, and therefore, minimum work will be obtained when 1 kg of chlorine gas expands.

Why other options are wrong:
- Option B: Oxygen. Oxygen has a lower molar mass than chlorine.
- Option C: Nitrogen. Nitrogen has a lower molar mass than chlorine.
- Option D: Methane. Methane has the lowest molar mass among the given options.

Key Concepts:
- Work Done in Expansion
- Molar Mass
- Number of Moles
- Isothermal Reversible Process

Memory Aids:
- Higher molar mass means fewer moles for a given mass.

Quick Tips:
- Remember that for a given mass, the number of moles is inversely proportional to the molar mass.

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