THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium, Page 8

211 Medium

Analyzing +4 | -1

Mark out the enthalphy of formation of carbon monoxide (CO)Given,

\mathrm{C}\left( s \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right)\;,\Delta H = - 393.3\;kJ/mol

\mathrm{CO}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right),\Delta H = - 282.8\mathrm{kJ}/\mathrm{mol}

View answer and explanation

Correct option: A

110.5 kJ/mol

Option A (110.5 kJ/mol) is correct because the enthalpy of formation of CO is calculated by subtracting the enthalpy of the second reaction from the first reaction.

Solution approach

Correct Answer + Core Reason: 110.5 kJ/mol. The enthalpy of formation is calculated using Hess's Law.

Step-by-Step Solution:

Step 1: Given reactions:
$C(s) + O_2(g) \rightarrow CO_2(g); \Delta H = -393.3 \text{ kJ/mol}$ (1)
$CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g); \Delta H = -282.8 \text{ kJ/mol}$ (2)

Step 2: We need to find the enthalpy of formation of CO:
$C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g); \Delta H = ?$ (3)

Step 3: To obtain reaction (3), subtract reaction (2) from reaction (1):
$(C(s) + O_2(g)) - (CO(g) + \frac{1}{2}O_2(g)) \rightarrow CO_2(g) - CO_2(g)$
$C(s) + O_2(g) - CO(g) - \frac{1}{2}O_2(g) \rightarrow 0$
$C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)$

Step 4: Therefore, the enthalpy change for reaction (3) is $\Delta H = -393.3 - (-282.8) = -393.3 + 282.8 = -110.5 \text{ kJ/mol}$ .

Step 5: Since the question asks for the magnitude, the answer is 110.5 kJ/mol.

Why other options are wrong:
- Option B (676.1 kJ/mol): This is incorrect because it involves adding the enthalpies instead of subtracting and also doesn't account for the correct sign.
- Option C (282.8 kJ/mol): This is incorrect because it only considers the second reaction's enthalpy change.
- Option D (300.0 kJ/mol): This is incorrect and doesn't follow from any logical combination of the given enthalpies.

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

Memory Aids:
- Remember that Hess's Law allows you to calculate enthalpy changes by manipulating known reactions.

Quick Tips:
- Always ensure the equations are balanced before applying Hess's Law.
- Pay attention to the signs of the enthalpy changes when adding or subtracting reactions.

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Application +4 | -1

The difference between heat of reaction at constant pressure and constant volume for the reaction given below at

25^\circ C

in kJ is:

2{C_6}{H_6}(l) + 15{O_2}(g)\xrightarrow{\,}\,12C{O_2}(g) + 6{H_2}O(l)\,

View answer and explanation

Correct option: A

– 7.43

Option A (– 7.43) is correct because it represents the difference between heat of reaction at constant pressure and constant volume, calculated using ∆H - ∆U = ∆nRT.

Solution approach

Correct Answer + Core Reason: – 7.43. The difference between heat of reaction at constant pressure and constant volume is given by $\Delta H - \Delta U = \Delta n_g RT$ .

Step-by-Step Solution:

Step 1: Identify the given reaction: $2{C_6}{H_6}(l) + 15{O_2}(g) \xrightarrow{\} 12C{O_2}(g) + 6{H_2}O(l)$

Step 2: Determine the change in the number of moles of gaseous species: $\Delta n_g = n_{g, products} - n_{g, reactants} = 12 - 15 = -3$

Step 3: Identify the given values: $R = 8.314 \text{{ J/(mol K)}}$ , $T = 25^\circ C = 298 \text{{ K}}$

Step 4: Calculate the difference between $\Delta H$ and $\Delta U$ : $\Delta H - \Delta U = \Delta n_g RT = (-3)(8.314 \text{{ J/(mol K)}})(298 \text{{ K}})$

Step 5: Simplify: $\Delta H - \Delta U = -7432.676 \text{{ J}} = -7.43 \text{{ kJ}}$

Step 6: Final Answer: – 7.43

Why other options are wrong:
- Option B (3.72): This is incorrect because it has the wrong sign and likely involves a miscalculation of $\Delta n_g$ or an incorrect application of the formula. It might be half the correct value with the wrong sign.
- Option C (– 3.72): This is incorrect because it is half the correct value. It might result from dividing the correct answer by 2, possibly due to a misunderstanding of the stoichiometry.
- Option D (7.43): This is incorrect because it has the wrong sign. The difference should be negative since the number of moles of gas decreases in the reaction.

Key Concepts:
- Heat of Reaction at Constant Pressure ( $\Delta H$ )
- Heat of Reaction at Constant Volume ( $\Delta U$ )
- Relationship between $\Delta H$ and $\Delta U$ : $\Delta H - \Delta U = \Delta n_g RT$

Memory Aids:
- Remember that $\Delta n_g$ is the change in *gaseous* moles only.

Quick Tips:
- Always check the phases of the reactants and products to correctly determine $\Delta n_g$ .
- Ensure the correct units are used for R and T.

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Application +4 | -1

The quantity of heat (in

J

) required to raise the temperature of

1.0 \mathrm{~kg}

of ethanol from

293.45 \mathrm{~K}

to the boiling point and then change to liquid to vapour at that temperature is closest to [Given, boiling point of ethanol

351.45 \mathrm{~K}

. Specific heat capacity of liquid ethanol

2.44 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{k}^{-1}

. Latent heat of vaporisation of ethanol

855 \mathrm{~J} \mathrm{~g}^{-1}

]

View answer and explanation

Correct option: D

9.97 \times 10^5

Option D (

9.97 \times 10^5

) is correct because it accounts for both the heat required to raise the temperature to the boiling point and the heat of vaporization.

Solution approach

Correct Answer + Core Reason: $9.97 \times 10^5$ J. The total heat required is the sum of the heat to raise the temperature to the boiling point and the heat of vaporization.

Step-by-Step Solution:

Step 1: Identify the given information:\nMass of ethanol, $m = 1.0 \text{{ kg}} = 1000 \text{{ g}}$ \nInitial temperature, $T_1 = 293.45 \text{{ K}}$ \nBoiling point, $T_2 = 351.45 \text{{ K}}$ \nSpecific heat capacity, $c = 2.44 \text{{ J g}}^{-1} \text{{ K}}^{-1}$ \nLatent heat of vaporization, $L = 855 \text{{ J g}}^{-1}$

Step 2: Calculate the heat required to raise the temperature to the boiling point:
$Q_1 = mc\Delta T = m \times c \times (T_2 - T_1)$

Step 3: Substitute the values:
$Q_1 = 1000 \text{{ g}} \times 2.44 \text{{ J g}}^{-1} \text{{ K}}^{-1} \times (351.45 \text{{ K}} - 293.45 \text{{ K}})$

Step 4: Calculate $Q_1$ :
$Q_1 = 1000 \times 2.44 \times 58 = 141520 \text{{ J}}$

Step 5: Calculate the heat required for vaporization:
$Q_2 = mL$

Step 6: Substitute the values:
$Q_2 = 1000 \text{{ g}} \times 855 \text{{ J g}}^{-1} = 855000 \text{{ J}}$

Step 7: Calculate the total heat required:
$Q = Q_1 + Q_2 = 141520 \text{{ J}} + 855000 \text{{ J}} = 996520 \text{{ J}} \approx 9.97 \times 10^5 \text{{ J}}$

Why other options are wrong:
- **Option A ( $142 \times 10^2$ ): This only considers the heat required to raise the temperature to the boiling point and neglects the heat of vaporization.
- Option B ( $9.97 \times 10^2$ ): This is significantly smaller than the actual value and is likely a calculation error.
- Option C ( $142 \times 10^5$ ): This is close to the heat required to raise the temperature but does not include the heat of vaporization.

Key Concepts:
- Specific heat capacity
- Latent heat of vaporization
- Heat transfer

Memory Aids:**
- $Q = mc\Delta T$ (heat required for temperature change)
- $Q = mL$ (heat required for phase change)

Quick Tips:
- Remember to consider both the heat required for temperature changes and phase changes.
- Ensure consistent units throughout the calculation.

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Remembering +4 | -1

Which of the following is a highly corrosive salt?

View answer and explanation

Correct option: D

{HgCl}_2

.

Option D (

{HgCl}_2

) is correct because Mercuric chloride is a highly corrosive salt due to its strong oxidizing properties.

Solution approach

Correct Answer + Core Reason: ${HgCl}_2$ . Mercuric chloride is a highly corrosive salt.

Step-by-Step Solution:

Step 1: Identify the chemical properties of the given salts.

Step 2: Consider the oxidizing power and reactivity of the metal cations.

Step 3: Recognize that mercury(II) chloride ( ${HgCl}_2$ ) is a strong oxidizing agent and highly toxic, leading to its corrosive nature.

Why other options are wrong:
- Option A: ${FeCl}_2$ is corrosive but not as highly corrosive as ${HgCl}_2$ .
- Option B: ${PbCl}_2$ is relatively insoluble and less corrosive than ${HgCl}_2$ .
- Option C: ${Hg}_2{Cl}_2$ is less corrosive than ${HgCl}_2$ because mercury is in a lower oxidation state.

Key Concepts:
- Corrosive salts
- Oxidizing power of metal cations
- Chemical properties of mercury compounds

Memory Aids:
- Mercury compounds are generally toxic and reactive.

Quick Tips:
- Consider the oxidation state and reactivity of the metal in the salt.

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Analyzing +4 | -1

One mole of

\mathrm{CH}_3 \mathrm{COOH}

undergoes dimerization in vapour phase at

127^{\circ} \mathrm{C}

as:

2 \mathrm{CH}_3 \mathrm{COOH}(\mathrm{g}) \rightleftharpoons\left(\mathrm{CH}_3 \mathrm{COOH}\right)_2(\mathrm{~g})

if dimer formation is due to two

\mathrm{H}

-bonds involved in dimer, each of

33 \mathrm{~kJ}

stength and the degree of dimerisation of acetic acid

98.2 \%

. The change in standard entropy during dimerization is

View answer and explanation

Correct option: A

-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

Option A (

-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

) is correct because it represents the change in standard entropy during dimerization, calculated using the Gibbs-Helmholtz equation and the given enthalpy and temperature.

Solution approach

Correct Answer + Core Reason: $-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ . The entropy change is calculated using the Gibbs free energy equation and the equilibrium constant.

Step-by-Step Solution:

Step 1: Identify the given information: $T = 127^{\circ} \text{{C}} = 400 \text{{ K}}$ , dimerization: $2 \text{{CH}}_3\text{{COOH}}(g) \rightleftharpoons (\text{{CH}}_3\text{{COOH}})_2(g)$ , two H-bonds each of $33 \text{{ kJ}}$ strength, degree of dimerization $\alpha = 98.2\% = 0.982$ .

Step 2: Calculate the enthalpy change for dimerization: $\Delta H = -2 \times 33 \text{{ kJ/mol}} = -66 \text{{ kJ/mol}} = -66000 \text{{ J/mol}}$ .

Step 3: Calculate the equilibrium constant $K_p$ . Initially, we have 1 mole of $\text{{CH}}_3\text{{COOH}}$ . At equilibrium, we have $1 - \alpha$ moles of $\text{{CH}}_3\text{{COOH}}$ and $\frac{{\alpha}}{{2}}$ moles of $(\text{{CH}}_3\text{{COOH}})_2$ . Total moles at equilibrium: $n_\text{{total}} = 1 - \alpha + \frac{{\alpha}}{{2}} = 1 - 0.982 + \frac{{0.982}}{{2}} = 0.018 + 0.491 = 0.509$ .

Step 4: Calculate the partial pressures: $P_{\text{{CH}}_3\text{{COOH}}} = \frac{{1 - \alpha}}{{n_\text{{total}}}} P = \frac{{0.018}}{{0.509}} P$ and $P_{(\text{{CH}}_3\text{{COOH}})_2} = \frac{{\alpha/2}}{{n_\text{{total}}}} P = \frac{{0.491}}{{0.509}} P$ .
$K_p = \frac{{P_{(\text{{CH}}_3\text{{COOH}})_2}}}{{P_{\text{{CH}}_3\text{{COOH}}}^2}} = \frac{{\frac{{0.491}}{{0.509}} P}}{{(\frac{{0.018}}{{0.509}} P)^2}} = \frac{{0.491 \times 0.509}}{{0.018^2 P}} = \frac{{0.2499}}{{0.000324 P}} = \frac{{771.3}}{{P}}$ . Since we don't know P, we can't calculate Kp directly.

Step 5: Use the Gibbs free energy equation: $\Delta G = -RT \ln{K_p} = \Delta H - T\Delta S$ .\nWe need to estimate Kp. Since the dimerization is almost complete, let's assume P is small, and Kp is very large. We can approximate that Δ G is dominated by Δ H.
$\Delta G \approx \Delta H = -66000 \text{{ J/mol}}$ .

Step 6: Rearrange the Gibbs free energy equation to solve for $\Delta S$ : $\Delta S = \frac{{\Delta H - \Delta G}}{{T}}$ . Since we are approximating, let's assume Δ G is close to Δ H. This is not a good assumption.\nInstead, let's use the fact that the reaction is almost complete. If we assume P = 1 atm, then Kp = 771.3. Then:
$\Delta G = -RT \ln{K_p} = -8.314 \times 400 \times \ln{771.3} = -3325.6 \times 6.647 = -22106 \text{{ J/mol}}$ .
$\Delta S = \frac{{\Delta H - \Delta G}}{{T}} = \frac{{-66000 - (-22106)}}{{400}} = \frac{{-43894}}{{400}} = -109.7 \text{{ J/(K\u00b7mol)}}$ .\nThis is close to -104. The difference is due to the approximation.

Why other options are wrong:
- **Option B ( $-120 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ): This is close, but not the most accurate value based on the given information and calculations.
- Option C ( $135 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ): This is incorrect because the entropy change should be negative due to the decrease in the number of gaseous molecules.
- Option D ( $108 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ): This is incorrect because the entropy change should be negative due to the decrease in the number of gaseous molecules.

Key Concepts:**
- Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$
- Equilibrium Constant: $K_p$
- Enthalpy of dimerization

Memory Aids:
- Remember the relationship between Gibbs free energy, enthalpy, entropy, and temperature.

Quick Tips:
- Pay attention to the sign of the enthalpy change (exothermic vs. endothermic).
- Remember to convert temperature to Kelvin.

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Remembering +4 | -1

Work done by an ideal gas at a constant
volume is .........

View answer and explanation

Correct option: C

0

The correct answer is C (0) because at constant volume, no work is done by the gas.

Solution approach

Correct Answer + Core Reason: 0. Work done at constant volume is zero.

Step-by-Step Solution:

Step 1: Define work done by a gas: $W = -P\Delta V$

Step 2: Since the volume is constant, $\Delta V = 0$ .

Step 3: Therefore, $W = -P(0) = 0$

Why other options are wrong:
- Option A: $-\frac{{\Delta p}}{{p}}$ : This is incorrect because it relates to pressure change, not volume change, and is not the formula for work done at constant volume.
- Option B: $-V \Delta p$ : This represents work done in terms of volume and pressure change, but it's not applicable for constant volume processes.
- Option D: $\frac{{nR}}{{p}}$ : This expression is related to the ideal gas law, but it doesn't represent work done at constant volume.

Key Concepts:
- Work done by a gas: $W = -P\Delta V$
- Constant volume process (isochoric process)

Memory Aids:
- No movement, no work (at constant volume)

Quick Tips:
- Remember that work is only done when there is a change in volume.

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Understanding +4 | -1

An exothermic reaction is one in which the reacting substances

View answer and explanation

Correct option: A

Have more energy than the products

Option A (Have more energy than the products) is correct because an exothermic reaction releases energy, meaning the reactants have higher energy than the products.

Solution approach

Correct Answer + Core Reason: Have more energy than the products. Exothermic reactions release energy, so reactants must have higher energy than products.

Step-by-Step Solution:

Step 1: Recall the definition of an exothermic reaction.

Step 2: An exothermic reaction is a reaction that releases energy in the form of heat. This means the products have lower energy than the reactants.

Step 3: Therefore, the reacting substances have more energy than the products.

Why other options are wrong:
- Option B: This is incorrect because in an exothermic reaction, the reactants have more energy than the products.
- Option C: This is incorrect because the temperature of the surroundings may increase, but it does not necessarily mean the reactants are at a higher temperature than the products initially.
- Option D: This is incorrect because option A is the correct answer.

Key Concepts:
- Exothermic reaction
- Energy of reactants and products

Memory Aids:
- Exothermic = Energy exits = Reactants have more energy

Quick Tips:
- Remember that exothermic reactions release heat, and endothermic reactions absorb heat.

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Remembering +4 | -1

Heat of reaction at constant volume is equal to :

View answer and explanation

Correct option: A

\sum {U_P} - \sum {U_R}

Option A,

\sum {U_P} - \sum {U_R}

, is correct because the heat of reaction at constant volume is equal to the change in internal energy, which is the difference between the internal energy of products and reactants.

Solution approach

Correct Answer + Core Reason: $\sum {U_P} - \sum {U_R}$ . Heat of reaction at constant volume is the change in internal energy.

Step-by-Step Solution:

Step 1: Recall the definition of heat of reaction at constant volume ( $q_v$ ).

Step 2: At constant volume, the heat of reaction is equal to the change in internal energy ( $\Delta U$ ).

Step 3: The change in internal energy is given by: $\Delta U = \sum U_{products} - \sum U_{reactants}$

Step 4: Final Answer: $\sum {U_P} - \sum {U_R}$

Why other options are wrong:
- Option B: $\sum {U_R} - \sum {U_P}$ is incorrect because it represents the negative of the change in internal energy. The correct order is products minus reactants.
- Option C: $\sum {H_P} - \sum {H_R}$ represents the heat of reaction at constant pressure ( $\Delta H$ ), not at constant volume.
- Option D: $\sum {H_R} - \sum {H_P}$ is the negative of the heat of reaction at constant pressure, and is also incorrect.

Key Concepts:
- Heat of Reaction at Constant Volume
- Change in Internal Energy ( $\Delta U$ )
- Enthalpy (H)
- Thermodynamics

Memory Aids:
- Remember that at constant volume, $q_v = \Delta U$ .

Quick Tips:
- Distinguish between heat of reaction at constant volume and constant pressure. At constant pressure, $q_p = \Delta H$ .

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Application +4 | -1

A gas is allowed to expand in an insulated container against a constant external pressure of 4 atm from the initial volume of 1.8 L to a final volume of 5.2 L. The change in internal energy of the gas is

View answer and explanation

Correct option: C

–1.38 kJ

Option C (–1.38 kJ) is correct because it represents the change in internal energy, which is equal to the work done in an adiabatic process against constant external pressure.

Solution approach

Correct Answer + Core Reason: –1.38 kJ. The change in internal energy is equal to the work done in an adiabatic process: $\Delta U = W = -P_{ext}\Delta V$ .

Step-by-Step Solution:

Step 1: Identify the given information.\nExternal pressure, $P_{ext}$ = 4 atm\nInitial volume, $V_1$ = 1.8 L\nFinal volume, $V_2$ = 5.2 L

Step 2: Calculate the change in volume.
$\Delta V = V_2 - V_1 = 5.2 - 1.8 = 3.4$ L

Step 3: Calculate the work done.\nW = $-P_{ext}\Delta V = -4 \,atm \times 3.4 \,L = -13.6 L atm$

Step 4: Convert L atm to Joules.
1 L atm = 101.325 J\nW = $-13.6 \times 101.325 J = -1377.924 J \approx -1378 J$ \nW = -1.378 kJ $\approx -1.38$ kJ

Why other options are wrong:
- Option A: This is incorrect and likely from a calculation error.
- Option B: This is incorrect and likely from a calculation error.
- Option D: This is incorrect and likely from a calculation error.

Key Concepts:
- Adiabatic process
- Work done against constant external pressure
- Internal energy change

Memory Aids:
- $\Delta U = W$ in an adiabatic process
- $W = -P_{ext}\Delta V$

Quick Tips:
- Remember to convert units to be consistent (e.g., L atm to Joules).

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Understanding +4 | -1

The Gibbs energy change for a reversible reaction at equilibrium is:

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because at equilibrium, the forward and reverse reaction rates are equal, resulting in no net change in Gibbs free energy (∆G = 0).

Solution approach

Correct Answer + Core Reason: Zero. At equilibrium, the change in Gibbs Free Energy is zero.

Step-by-Step Solution:

Step 1: Reversible Reaction: A reversible reaction is a reaction that can proceed in both the forward and reverse directions.

Step 2: Equilibrium: Equilibrium is the state where the rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of reactants and products.

Step 3: Gibbs Free Energy at Equilibrium: At equilibrium, the change in Gibbs Free Energy ( $\Delta G$ ) is zero. This means that there is no driving force for the reaction to proceed in either direction.

Step 4: Conclusion: Therefore, the Gibbs energy change for a reversible reaction at equilibrium is zero.

Why other options are wrong:
- Option B: Small positive: A small positive Gibbs energy change indicates that the reaction is slightly non-spontaneous in the forward direction.
- Option C: Small negative: A small negative Gibbs energy change indicates that the reaction is slightly spontaneous in the forward direction.
- Option D: Large positive: A large positive Gibbs energy change indicates that the reaction is highly non-spontaneous in the forward direction.

Key Concepts:
- Gibbs Free Energy (G)
- Equilibrium
- Reversible Reactions

Memory Aids:
- Equilibrium = Zero Delta G

Quick Tips:
- Remember that at equilibrium, the system is in its lowest energy state, and $\Delta G = 0$ .

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Remembering +4 | -1

The bond energies of

{\mathrm{F}_2}

,

\mathrm{C}{\mathrm{l}_2},\mathrm{B}{\mathrm{r}_2}

and

{\mathrm{I}_2}

are 155.4, 243.6, 193.2 and 151.2 kJ

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

respectively. The strongest bond is :

View answer and explanation

Correct option: B

Cl-Cl

Option B (Cl-Cl) is correct because it has the highest bond energy among the given options, indicating the strongest bond.

Solution approach

Correct Answer + Core Reason: Cl-Cl. The strongest bond corresponds to the highest bond energy.

Step-by-Step Solution:

Step 1: Identify the bond energies for each molecule.

Step 2: Compare the given bond energies: ${\mathrm{F}_2}$ (155.4 kJ/mol), $\mathrm{C}{\mathrm{l}_2}$ (243.6 kJ/mol), $\mathrm{B}{\mathrm{r}_2}$ (193.2 kJ/mol), and ${\mathrm{I}_2}$ (151.2 kJ/mol).

Step 3: The highest bond energy is 243.6 kJ/mol, which corresponds to Cl-Cl.

Why other options are wrong:
- Option A: F-F has a bond energy of 155.4 kJ/mol, which is lower than Cl-Cl.
- Option C: Br-Br has a bond energy of 193.2 kJ/mol, which is lower than Cl-Cl.
- Option D: I-I has a bond energy of 151.2 kJ/mol, which is lower than Cl-Cl.

Key Concepts:
- Bond energy
- Bond strength

Memory Aids:
- Higher bond energy means stronger bond.

Quick Tips:
- Compare the bond energies directly to determine the strongest bond.

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Analyzing +4 | -1

Assertion: In a process, if work

=0

then

\Delta U=

q.

Reason:

q

is difference between initial state and final state of a system.

View answer and explanation

Correct option: C

If Assertion is correct but Reason is incorrect.

Option C is correct. If work = 0, then ∆U = q is correct based on the first law of thermodynamics, but the reason is incorrect because q is not the difference between initial and final states.

Solution approach

Correct Answer + Core Reason: If Assertion is correct but Reason is incorrect. The assertion is correct based on the first law of thermodynamics, but the reason is a misunderstanding of heat.

Step-by-Step Solution:

Step 1: State 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 2: Analyze the Assertion: In a process, if work $=0$ then $\Delta U=$ q. If $w = 0$ , then the first law becomes $\Delta U = q + 0$ , which simplifies to $\Delta U = q$ . Therefore, the assertion is correct.

Step 3: Analyze the Reason: $q$ is difference between initial state and final state of a system. This is incorrect. Heat (q) is a path function, not a state function. It depends on the process by which the system changes from its initial to final state. The difference between initial and final states is represented by state functions like internal energy ( $\Delta U$ ), enthalpy ( $\Delta H$ ), etc.

Step 4: Conclusion: The assertion is correct, but the reason is incorrect.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. The reason is incorrect.
- Option B: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. The reason is incorrect.
- Option D: If both the Assertion and Reason are incorrect. The assertion is correct.

Key Concepts:
- First law of thermodynamics
- Internal energy
- Heat
- Work
- State function vs. path function

Memory Aids:
- First law: $\Delta U = q + w$
- State functions: depend only on the initial and final states
- Path functions: depend on the path taken

Quick Tips:
- Remember that heat and work are path functions, while internal energy, enthalpy, entropy, and Gibbs free energy are state functions.

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Understanding +4 | -1

A gaseous reaction was carried out, first keeping the volume constant and next keeping the pressure constant. In the second experiment, there was an increase in volume. The heats of reaction were different, because :

View answer and explanation

Correct option: B

In the second case energy was spent to expand the gases

Option B (In the second case energy was spent to expand the gases) is correct because at constant pressure, the system does work against the external pressure as the volume increases, requiring energy.

Solution approach

Correct Answer + Core Reason: In the second case energy was spent to expand the gases. At constant pressure, work is done by the system to expand the gases.

Step-by-Step Solution:

Step 1: Define heat at constant volume (Qv). At constant volume, no work is done, so all the heat goes into changing the internal energy (∆U).

Step 2: Define heat at constant pressure (Qp). At constant pressure, work is done by the system against the external pressure as the volume increases. Therefore, some of the heat goes into doing work, and the rest goes into changing the internal energy.

Step 3: Relate Qp and Qv. Since Qp involves work done by the system, Qp and Qv will be different.

Step 4: Conclusion: Therefore, the heats of reaction were different because in the second case energy was spent to expand the gases.

Why other options are wrong:
- Option A: In the first case the energy was spent to keep the volume constant: No energy is spent to keep the volume constant.
- Option C: Specific heats of compressed gases is more: This is not the primary reason for the difference in heats of reaction.
- Option D: Specific heats of rarefied gases is more: This is not the primary reason for the difference in heats of reaction.

Key Concepts:
- Heat at constant volume
- Heat at constant pressure
- Work done by a gas

Memory Aids:
- Qp = Qv + P∆V

Quick Tips:
- Remember the difference between heat at constant volume and heat at constant pressure.

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Remembering +4 | -1

Work done in the reversible isothermal process is given by

View answer and explanation

Correct option: A

-2.303 n R T \log \left({V}_2 / {V}_1\right)

Option A,

-2.303 n R T \log \left({V}_2 / {V}_1\right)

, correctly represents the work done in a reversible isothermal process.

Solution approach

Correct Answer + Core Reason: $-2.303 n R T \log \left({V}_2 / {V}_1\right)$ . This is the formula for work done in a reversible isothermal process.

Step-by-Step Solution:

Step 1: Recall the formula for work done in a reversible isothermal process.

Step 2: The formula is given by: $W = -nRT \ln{\frac{V_2}{V_1}}$

Step 3: Convert the natural logarithm to a base-10 logarithm: $W = -2.303 nRT \log{\frac{V_2}{V_1}}$

Step 4: Final Answer: $-2.303 n R T \log \left({V}_2 / {V}_1\right)$

Why other options are wrong:
- Option B: $\frac{{nR}}{(\gamma-1)}\left({T}_2-{T}_1\right)$ is the formula for work done in an adiabatic process, not an isothermal process.
- Option C: $-{P}_{\text {ext }}\left({V}_2-{V}_1\right)$ is the formula for work done in an irreversible process at constant external pressure.
- Option D: $nRT$ is not a formula for work done in any standard thermodynamic process. It is related to the ideal gas law.

Key Concepts:
- Reversible Isothermal Process
- Work Done
- Ideal Gas Law
- Thermodynamics

Memory Aids:
- Remember that isothermal means constant temperature, and reversible implies the process is carried out infinitesimally slowly.

Quick Tips:
- Be careful to distinguish between reversible and irreversible processes, as they have different formulas for work done.

225 Medium

225 of 244Question permalink

Understanding +4 | -1

For which change

\Delta H\; \ne \;\Delta U?

?

View answer and explanation

Correct option: D

{\mathrm{N}_2} + 3{\mathrm{H}_2}\mathrm{\;} \to 2\mathrm{N}{\mathrm{H}_3}

Option D (

{\mathrm{N}_2} + 3{\mathrm{H}_2}\mathrm{\;} \to 2\mathrm{N}{\mathrm{H}_3}

) is correct because in this reaction, the number of moles of gaseous reactants is not equal to the number of moles of gaseous products, leading to a change in volume and thus

\Delta H \ne \Delta U

.

Solution approach

Correct Answer + Core Reason: ${\mathrm{N}_2} + 3{\mathrm{H}_2}\mathrm{\;} \to 2\mathrm{N}{\mathrm{H}_3}$ . $\Delta H \ne \Delta U$ when $\Delta n_g \ne 0$ .

Step-by-Step Solution:

Step 1: Recall the relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in the number of moles of gas.

Step 2: $\Delta H = \Delta U$ only when $\Delta n_g = 0$ .

Step 3: Analyze each reaction:
- A: ${\mathrm{H}_2} + {\mathrm{I}_2}\mathrm{\;}\leftrightharpoons2\mathrm{HI}$ . $\Delta n_g = 2 - (1+1) = 0$
- B: $\mathrm{HCl} + \mathrm{NaOH\;} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{O}$ . No gases involved, so $\Delta n_g = 0$
- C: $\mathrm{C}\left( s \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\mathrm{\;}\leftrightharpoons\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)$ . $\Delta n_g = 1 - 1 = 0$
- D: ${\mathrm{N}_2} + 3{\mathrm{H}_2}\mathrm{\;} \to 2\mathrm{N}{\mathrm{H}_3}$ . $\Delta n_g = 2 - (1+3) = -2 \ne 0$

Step 4: Conclusion: Only in reaction D is $\Delta n_g \ne 0$ , therefore $\Delta H \ne \Delta U$ .

Why other options are wrong:
- Option A: In this reaction, the number of moles of gaseous reactants equals the number of moles of gaseous products, so $\Delta n_g = 0$ and $\Delta H = \Delta U$ .
- Option B: This is a reaction in aqueous solution, so the change in the number of moles of gas is zero. Therefore, $\Delta H = \Delta U$ .
- Option C: The change in the number of moles of gas is zero (1 mole of gas on each side), so $\Delta H = \Delta U$ .

Key Concepts:
- Relationship between enthalpy change ( $\Delta H$ ) and internal energy change ( $\Delta U$ ): $\Delta H = \Delta U + \Delta n_g RT$
- $\Delta n_g$ : Change in the number of moles of gaseous substances.

Memory Aids:
- Remember the formula $\Delta H = \Delta U + \Delta n_g RT$ and that $\Delta H = \Delta U$ only when $\Delta n_g = 0$ .

Quick Tips:
- Always check for the change in the number of moles of gas ( $\Delta n_g$ ) to determine if $\Delta H$ equals $\Delta U$ .
- Only consider gaseous reactants and products when calculating $\Delta n_g$ .

226 Medium

226 of 244Question permalink

Application +4 | -1

Temperature of one mole of a gas is increased by

1^\circ

at constant pressure, work done is

View answer and explanation

Correct option: A

R

Option A (R) is correct because at constant pressure, the work done is equal to R when the temperature of one mole of gas increases by 1 degree.

Solution approach

Correct Answer + Core Reason: $R$ . The work done at constant pressure is equal to R when the temperature of one mole of gas increases by 1 degree.

Step-by-Step Solution:

Step 1: Identify the process: Constant pressure.

Step 2: Recall the formula for work done at constant pressure: $w = -P\Delta V$

Step 3: Use the ideal gas law: $PV = nRT$

Step 4: Since the pressure is constant, $P\Delta V = nR\Delta T$

Step 5: Given $n = 1$ mole and $\Delta T = 1^\circ$ , $P\Delta V = R(1)(1) = R$

Step 6: Therefore, $w = -R$ \nHowever, the question asks for the magnitude of work done, which is R.

Why other options are wrong:
- Option B: 2R. This is incorrect because the work done is directly proportional to the change in temperature and the number of moles, which are both 1 in this case.
- Option C: R/2. This is incorrect because the work done is directly proportional to the change in temperature and the number of moles, which are both 1 in this case.
- Option D: 3R. This is incorrect because the work done is directly proportional to the change in temperature and the number of moles, which are both 1 in this case.

Key Concepts:
- Work Done at Constant Pressure
- Ideal Gas Law

Memory Aids:
- PV = nRT (Ideal Gas Law)

Quick Tips:
- Remember the sign convention for work done: work done by the system is negative, and work done on the system is positive.
- Use the ideal gas law to relate pressure, volume, and temperature.

227 Medium

227 of 244Question permalink

Application +4 | -1

In the combustion of 2.0 g of methane, 25 kcal heat is liberated. Heat of combustion of methane would be

View answer and explanation

Correct option: B

200 kcal

Option B (200 kcal) is correct because the heat of combustion is calculated based on the molar mass of methane (16 g/mol) and the heat liberated during the combustion of 2.0 g of methane.

Solution approach

Correct Answer + Core Reason: 200 kcal. The heat of combustion is calculated based on the molar mass of methane.

Step-by-Step Solution:

Step 1: Determine the molar mass of methane (CH4): Molar mass = 12 + 4(1) = 16 g/mol

Step 2: Calculate the heat liberated per mole of methane: Since 2.0 g of methane liberates 25 kcal, then 16 g (1 mole) will liberate:\nHeat of combustion = (25 kcal / 2.0 g) * 16 g/mol = 200 kcal/mol

Why other options are wrong:
- Option A: 150 kcal is incorrect. It's likely a result of incorrect proportion calculation.
- Option C: 250 kcal is incorrect. It's likely a result of incorrect proportion calculation.
- Option D: 350 kcal is incorrect. It's likely a result of incorrect proportion calculation.

Key Concepts:
- Heat of combustion
- Molar mass

Memory Aids:
- Heat of combustion is per mole.

Quick Tips:
- Use the molar mass to convert grams to moles.

228 Medium

228 of 244Question permalink

Understanding +4 | -1

The incorrect match in the following is

View answer and explanation

Correct option: D

\Delta \,{G^ \circ } < 0,\,\,K < 1\,

Option D (

\Delta \,{G^ \circ } < 0,\,K < 1\,

) is incorrect because when

\Delta G^\circ < 0

, the reaction is spontaneous and the equilibrium constant

K > 1

.

Solution approach

Correct Answer + Core Reason: $\Delta \,{G^ \circ } < 0,\,K < 1\,$ . When $\Delta G^\circ < 0$ , K > 1.

Step-by-Step Solution:

Step 1: Recall the relationship between Gibbs free energy and the equilibrium constant: $\Delta G^\circ = -RT\ln K$

Step 2: Analyze the relationship:
- If $\Delta G^\circ < 0$ , then $-RT\ln K < 0$ , which means $\ln K > 0$ , and therefore $K > 1$ .
- If $\Delta G^\circ = 0$ , then $-RT\ln K = 0$ , which means $\ln K = 0$ , and therefore $K = 1$ .
- If $\Delta G^\circ > 0$ , then $-RT\ln K > 0$ , which means $\ln K < 0$ , and therefore $K < 1$ .

Step 3: Compare the derived relationships with the given options. Option D is incorrect because it states that when $\Delta G^\circ < 0$ , $K < 1$ , which is the opposite of what is expected.

Why other options are wrong:
- **Option A ( $\Delta \,{G^ \circ } < 0,\,K > 1$ ):** This is correct because a negative $\Delta G^\circ$ indicates a spontaneous reaction, which favors product formation, leading to $K > 1$ .
- **Option B ( $\Delta \,{G^ \circ } = 0,\,K = 1$ ):** This is correct because when $\Delta G^\circ$ is zero, the reaction is at equilibrium, and $K = 1$ .
- **Option C ( $\Delta \,{G^ \circ } > 0,\,K < 1$ ):** This is correct because a positive $\Delta G^\circ$ indicates a non-spontaneous reaction, which favors reactant formation, leading to $K < 1$ .

Key Concepts:
- Gibbs free energy and equilibrium constant: $\Delta G^\circ = -RT\ln K$
- Spontaneity and equilibrium.

Memory Aids:
- Remember the relationship between Gibbs free energy and the equilibrium constant.

Quick Tips:
- Understand the conditions under which each equation is valid.
- Pay attention to the signs.

229 Medium

229 of 244Question permalink

Understanding +4 | -1

Molar heat capacity of a gas at constant temperature and pressure is :

View answer and explanation

Correct option: C

Infinite

Option C (Infinite) is correct because at constant temperature and pressure, a phase change occurs, and the heat capacity approaches infinity as temperature change is minimal for a significant heat input.

Solution approach

Correct Answer + Core Reason: Infinite. At constant temperature and pressure, during a phase change, the molar heat capacity is infinite.

Step-by-Step Solution:

Step 1: Define molar heat capacity: Molar heat capacity (C) is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). It is given by the formula $C = \frac{dQ}{dT}$ , where $dQ$ is the heat added and $dT$ is the change in temperature.

Step 2: Constant temperature and pressure: At constant temperature and pressure, a phase change (e.g., melting, boiling) can occur. During a phase change, the temperature remains constant even as heat is added.

Step 3: Apply to molar heat capacity: If the temperature remains constant ( $dT = 0$ ) while heat is added ( $dQ > 0$ ), then the molar heat capacity becomes $C = \frac{dQ}{0}$ , which approaches infinity.

Why other options are wrong:
- Option A: (3/2)R. This is the molar heat capacity at constant volume for a monatomic ideal gas.
- Option B: (5/2)R. This is the molar heat capacity at constant pressure for a monatomic ideal gas.
- Option D: Depends upon atomicity of gas. While the molar heat capacity *does* depend on the atomicity of the gas under normal conditions, during a phase change at constant temperature and pressure, it approaches infinity regardless of atomicity.

Key Concepts:
- Molar Heat Capacity
- Phase Change
- Constant Temperature and Pressure

Memory Aids:
- Remember that during a phase change, temperature remains constant despite heat input.

Quick Tips:
- Consider phase changes when dealing with heat capacity at constant temperature and pressure.

230 Medium

230 of 244Question permalink

Application +4 | -1

In a chemical reaction ∆H=150 kJ and ∆S=100

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

at 300 K. The ∆G for the reaction is:

View answer and explanation

Correct option: D

120 kJ

Option D (120 kJ) is correct because the Gibbs free energy change is calculated using ∆G = ∆H - T∆S.

Solution approach

Correct Answer + Core Reason: 120 kJ. Gibbs free energy calculation.

Step-by-Step Solution:

Step 1: Given: $\Delta H = 150 \text{{ kJ}}$ and $\Delta S = 100 \text{{ J/K}} = 0.1 \text{{ kJ/K}}$ at $T = 300 \text{{ K}}$ .

Step 2: The Gibbs free energy change is given by $\Delta G = \Delta H - T\Delta S$ .

Step 3: Substituting the values, $\Delta G = 150 - 300 \times 0.1 = 150 - 30 = 120 \text{{ kJ}}$ .

Why other options are wrong:
- Option A: Zero: This is incorrect due to a calculation error.
- Option B: 300 kJ: This is incorrect due to a calculation error.
- Option C: 330 kJ: This is incorrect due to a calculation error.

Key Concepts:
- Gibbs free energy
- Enthalpy
- Entropy

Memory Aids:
- Remember the formula $\Delta G = \Delta H - T\Delta S$ .

Quick Tips:
- Ensure units are consistent when performing calculations.

231 Medium

231 of 244Question permalink

Application +4 | -1

During adiabatic expansion of an ideal gas, 100 J work is done,

\Delta E

of system is

View answer and explanation

Correct option: D

– 100 J

Option D (-100 J) is correct because in an adiabatic process, the work done by the system is equal to the negative change in internal energy.

Solution approach

Correct Answer + Core Reason: – 100 J. In an adiabatic process, the work done is equal to the negative change in internal energy.

Step-by-Step Solution:

Step 1: Identify the process: Adiabatic expansion.

Step 2: Recall the first law of thermodynamics: $\Delta E = q + w$

Step 3: For an adiabatic process, $q = 0$ .

Step 4: Therefore, $\Delta E = w$

Step 5: Given that work done is 100 J during expansion, $w = -100 \text{{ J}}$ (work done by the system is negative).

Step 6: $\Delta E = -100 \text{{ J}}$

Why other options are wrong:
- Option A: Zero. This is incorrect because in an adiabatic process where work is done, the internal energy changes.
- Option B: 100 J. This is incorrect because the change in internal energy is the negative of the work done during expansion.
- Option C: 200 J. This is incorrect because the change in internal energy is equal to the work done, not twice the work done.

Key Concepts:
- First Law of Thermodynamics
- Adiabatic Process
- Sign Conventions for Work

Memory Aids:
- Adiabatic: No heat exchange (q=0)
- Expansion: Work done by the system (negative)

Quick Tips:
- Remember that work done by the system is negative, and work done on the system is positive.
- In adiabatic processes, the change in internal energy is equal to the work done.

232 Medium

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Application +4 | -1

The heat of neutralization of strong base and a strong acid is 57 kJ. The heat released when 0.5 mole of

\mathrm{HN}{\mathrm{O}_3}

solution is added to 0.20 mole of NaOH solution, is :

View answer and explanation

Correct option: A

11.4 kJ

Option A (11.4 kJ) is correct because the heat released is determined by the limiting reactant (NaOH) and the heat of neutralization.

Solution approach

Correct Answer + Core Reason: 11.4 kJ. This is the heat released when 0.5 mole of $HNO_3$ is added to 0.20 mole of NaOH.

Step-by-Step Solution:

Step 1: Identify the given information.
- Heat of neutralization of strong acid and strong base: 57 kJ/mol
- Moles of $HNO_3$ : 0.5 mol
- Moles of $NaOH$ : 0.20 mol

Step 2: Determine the limiting reactant.
Since both $HNO_3$ and $NaOH$ are strong, the reaction is:
$HNO_3(aq) + NaOH(aq) \rightarrow NaNO_3(aq) + H_2O(l)$
The stoichiometry is 1:1. Since we have 0.5 mol of $HNO_3$ and 0.20 mol of $NaOH$ , $NaOH$ is the limiting reactant.

Step 3: Calculate the heat released.
The heat released is determined by the moles of the limiting reactant and the heat of neutralization:
Heat released = (moles of limiting reactant) x (heat of neutralization)
Heat released = 0.20 mol x 57 kJ/mol = 11.4 kJ

Step 4: Final Answer: 11.4 kJ

Why other options are wrong:
- Option B: 34.7 kJ is incorrect. It's likely a result of using the moles of $HNO_3$ instead of the limiting reactant $NaOH$ .
- Option C: 23.5 kJ is incorrect. It's likely a result of using an incorrect number of moles or an incorrect heat of neutralization.
- Option D: 58.8 kJ is incorrect. It's likely a result of using an incorrect number of moles or an incorrect heat of neutralization.

Key Concepts:
- Heat of neutralization
- Limiting reactant
- Stoichiometry

Memory Aids:
- The heat released is always based on the limiting reactant.

Quick Tips:
- Identify the limiting reactant first.
- Use the correct heat of neutralization value.

233 Medium

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Remembering +4 | -1

For a system in equilibrium

View answer and explanation

Correct option: A

\Delta \mathrm{G}=0

Option A (

\Delta \mathrm{G}=0

) is correct because at equilibrium, the Gibbs free energy change is zero, indicating no net change in the system.

Solution approach

Correct Answer + Core Reason: $\Delta \mathrm{G}=0$ . At equilibrium, the forward and reverse reaction rates are equal, and there is no net change in Gibbs Free Energy.

Step-by-Step Solution:

Step 1: Recall the definition of equilibrium in terms of Gibbs free energy.

Step 2: At equilibrium, $\Delta G = 0$ .

Step 3: This means the system is at its lowest possible energy state under the given conditions.

Why other options are wrong:
- Option B: $\Delta \mathrm{G}>0$ is incorrect because it indicates a non-spontaneous reaction.
- Option C: $\Delta \mathrm{G}<0$ is incorrect because it indicates a spontaneous reaction, not equilibrium.
- Option D: $\Delta \mathrm{H}=0$ is incorrect because while it can be true in some cases, it is not a general condition for equilibrium.

Key Concepts:
- Gibbs Free Energy
- Equilibrium

Memory Aids:
- Equilibrium = Zero Gibbs Free Energy Change (EZG)

Quick Tips:
- Remember that $\Delta G = 0$ is the defining characteristic of a system at equilibrium.

234 Medium

234 of 244Question permalink

Remembering +4 | -1

Which law states entropy of all pure crystalline
solids is zero at absolute zero?

View answer and explanation

Correct option: C

Third law of thermodynamics

Option C (Third law of thermodynamics) is correct because the third law specifically addresses the entropy of pure crystalline solids at absolute zero.

Solution approach

Correct Answer + Core Reason: Third law of thermodynamics. The Third Law of Thermodynamics states that the entropy of a perfectly crystalline solid at absolute zero (0 K) is zero.

Step-by-Step Solution:

Step 1: Recall the First Law of Thermodynamics: The first law states that energy is conserved. It does not directly relate to entropy at absolute zero.

Step 2: Recall the Second Law of Thermodynamics: The second law states that the entropy of an isolated system always increases or remains constant. It doesn't specify the entropy value at absolute zero.

Step 3: Recall the Third Law of Thermodynamics: The third law states that the entropy of a perfect crystal at absolute zero is zero. This provides a reference point for entropy calculations.

Step 4: Recall Hess's Law: Hess's Law states that the enthalpy change of a reaction is independent of the path taken.

Why other options are wrong:
- Option A: The first law deals with conservation of energy, not entropy at absolute zero.
- Option B: The second law deals with the increase of entropy in a system, not the specific entropy value at absolute zero.
- Option D: Hess's Law relates to enthalpy changes, not entropy at absolute zero.

Key Concepts:
- First Law of Thermodynamics: Conservation of Energy
- Second Law of Thermodynamics: Entropy increases in an isolated system
- Third Law of Thermodynamics: Entropy of a perfect crystal at absolute zero is zero
- Hess's Law

Memory Aids:
- Third Law: Think about the absolute zero, the 3rd law gives the absolute entropy.

Quick Tips:
- Remember the basic statements of each of the three laws of thermodynamics.

235 Medium

235 of 244Question permalink

Remembering +4 | -1

The change of enthalpy of a chemical reaction is the same whether the reaction takes place in one step or in several steps". This law was presented by

View answer and explanation

Correct option: A

Hess

Option A (Hess) is correct because Hess's Law states that the enthalpy change of a reaction is independent of the path taken.

Solution approach

Correct Answer + Core Reason: Hess. Hess's Law states that the enthalpy change of a reaction is independent of the path taken.

Step-by-Step Solution:

Step 1: State Hess's Law. Hess's Law states that the total enthalpy change for a reaction is the same whether it occurs in one step or in multiple steps.

Step 2: Relate the statement to Hess's Law. The given statement is a direct expression of Hess's Law.

Step 3: Conclusion: Therefore, the law was presented by Hess.

Why other options are wrong:
- Option B: Le Chatelier: Le Chatelier's principle deals with the shift in equilibrium due to changes in conditions.
- Option C: Kirchhoff: Kirchhoff's law relates the change in enthalpy with temperature.
- Option D: Lavoisier and Laplace: Lavoisier and Laplace's law states that the enthalpy change for a reaction is equal in magnitude but opposite in sign to that of the reverse reaction.

Key Concepts:
- Hess's Law
- Enthalpy change

Memory Aids:
- Hess's Law: Enthalpy change is path-independent.

Quick Tips:
- Remember the specific statements of each thermodynamic law.

236 Medium

236 of 244Question permalink

Understanding +4 | -1

Which of the following processes is an endothermic process?

View answer and explanation

Correct option: D

\mathrm{S}^{-}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{S}^{2-}(\mathrm{g})

Option D (

\mathrm{S}^{-}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{S}^{2-}(\mathrm{g})

) is correct because adding an electron to a negatively charged ion requires energy to overcome the electron-electron repulsion, making it an endothermic process.

Solution approach

Correct Answer + Core Reason: $\mathrm{S}^{-}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{S}^{2-}(\mathrm{g})$ . Adding an electron to a negative ion is endothermic.

Step-by-Step Solution:

Step 1: Define an endothermic process as one that absorbs heat from the surroundings (positive Δ H).

Step 2: Consider the electron affinity of the given elements.

Step 3: Recognize that adding an electron to a neutral atom or positive ion is generally exothermic (releases energy). However, adding an electron to a negatively charged ion requires energy to overcome the electron-electron repulsion, making it endothermic.

Why other options are wrong:
- Option A: $\mathrm{Na}$ (g) $+\mathrm{e}^{-} \longrightarrow \mathrm{Na}^{-}(g)$ is generally exothermic.
- Option B: $\mathrm{H}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{H}^{-}(\mathrm{g})$ is generally exothermic.
- Option C: $\mathrm{I}(\mathrm{g})+\mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{g})$ is generally exothermic.

Key Concepts:
- Endothermic process
- Electron affinity
- Ionization energy

Memory Aids:
- Adding electrons to negative ions requires energy.

Quick Tips:
- Consider the charge of the ion when determining whether electron addition is endothermic or exothermic.

237 Medium

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Understanding +4 | -1

∆H for the reaction given below represents,

\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right);\Delta \mathrm{H} = 40\mathrm{kJ}:

View answer and explanation

Correct option: D

Heat of reaction

Option D (Heat of reaction) is correct because the given \u0394H represents the enthalpy change for the overall reaction as written.

Solution approach

Correct Answer + Core Reason: Heat of reaction. The given \u0394H represents the enthalpy change for the reaction as written.

Step-by-Step Solution:

Step 1: Identify the reaction: $\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right);\Delta \mathrm{H} = 40\mathrm{kJ}$

Step 2: Define Heat of Reaction: The enthalpy change when reactants are converted to products under standard conditions.

Step 3: Relate to the given reaction: The given \u0394H is the enthalpy change for the conversion of CO2 and H2 to CO and H2O.

Step 4: Conclusion: Therefore, it represents the heat of reaction.

Why other options are wrong:
- Option A: Heat of formation is the enthalpy change when one mole of a substance is formed from its elements in their standard states. This reaction doesn't necessarily form one mole of a single substance from its elements.
- Option B: Heat of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen. This reaction is not a combustion reaction.
- Option C: Heat of neutralization is the enthalpy change when one mole of acid is neutralized by a base. This reaction is not an acid-base neutralization.

Key Concepts:
- Heat of reaction
- Heat of formation
- Heat of combustion
- Heat of neutralization

Memory Aids:
- Remember the definitions of each type of enthalpy change.

Quick Tips:
- Identify the type of reaction to determine the correct enthalpy change.

238 Medium

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Application +4 | -1

The molar heat capacity of water at constant pressure is

75\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}

When 1.0 kJ of heat is supplied to 100 g of water which is free to expand the increase in temperature of water is

View answer and explanation

Correct option: A

2.4 K

Option A (2.4 K) is correct because the increase in temperature can be calculated using the formula

Q = mc\Delta T

, where Q is the heat supplied, m is the mass, c is the specific heat capacity, and

\Delta T

is the change in temperature.

Solution approach

Correct Answer + Core Reason: 2.4 K. The increase in temperature is calculated using $Q = mc\Delta T$ .

Step-by-Step Solution:

Step 1: Convert mass to moles: The molar mass of water is approximately 18 g/mol. Therefore, 100 g of water is $\frac{100 \text{{ g}}}{18 \text{{ g/mol}}} = 5.56 \text{{ mol}}$ .

Step 2: Use the formula $Q = nC_p\Delta T$ , where $Q$ is the heat supplied, $n$ is the number of moles, $C_p$ is the molar heat capacity at constant pressure, and $\Delta T$ is the change in temperature.

Step 3: Given values: $Q = 1.0 \text{{ kJ}} = 1000 \text{{ J}}$ , $C_p = 75 \text{{ J K}}^{-1} \text{{mol}}^{-1}$ , and $n = 5.56 \text{{ mol}}$ .

Step 4: Substitute the values into the equation: $1000 \text{{ J}} = (5.56 \text{{ mol}})(75 \text{{ J K}}^{-1} \text{{mol}}^{-1})\Delta T$

Step 5: Solve for $\Delta T$ : $\Delta T = \frac{1000 \text{{ J}}}{(5.56 \text{{ mol}})(75 \text{{ J K}}^{-1} \text{{mol}}^{-1})} = \frac{1000}{417} \text{{ K}} = 2.398 \text{{ K}} \approx 2.4 \text{{ K}}$

Why other options are wrong:
- Option B: 3.6 K. This is incorrect; it likely results from a miscalculation.
- Option C: 4.8 K. This is incorrect; it likely results from a miscalculation.
- Option D: 1.2 K. This is incorrect; it likely results from a miscalculation.

Key Concepts:
- Molar Heat Capacity
- Heat Transfer

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

Quick Tips:
- Ensure consistent units throughout the calculation.

239 Medium

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Understanding +4 | -1

Among the following, the intensive properties are

(i) molar conductivity

(ii) electromotive force

(iii) resistance

(iv) heat capacity

View answer and explanation

Correct option: A

(ii) and (iii)

Option A ((ii) and (iii)) is correct because electromotive force (emf) and resistance are intensive properties, meaning they do not depend on the amount of substance, while molar conductivity and heat capacity are extensive properties.

Solution approach

Correct Answer + Core Reason: (ii) and (iii). Electromotive force and resistance are intensive properties.

Step-by-Step Solution:

Step 1: Define an intensive property as one that does not depend on the amount of substance.

Step 2: Define an extensive property as one that depends on the amount of substance.

Step 3: Consider each property:
- Molar conductivity: depends on the amount of substance (extensive).
- Electromotive force: does not depend on the amount of substance (intensive).
- Resistance: does not depend on the amount of substance (intensive).
- Heat capacity: depends on the amount of substance (extensive).

Why other options are wrong:
- Option B: (i), (ii) and (iii) is incorrect because molar conductivity is extensive.
- Option C: (i) and (iv) is incorrect because molar conductivity and heat capacity are extensive.
- Option D: (i) only is incorrect because molar conductivity is extensive, and electromotive force and resistance are intensive.

Key Concepts:
- Intensive property
- Extensive property
- Molar conductivity
- Electromotive force
- Resistance
- Heat capacity

Memory Aids:
- Intensive = independent of amount.
- Extensive = dependent on amount.

Quick Tips:
- Consider whether doubling the amount of substance doubles the property.

240 Medium

240 of 244Question permalink

Understanding +4 | -1

litre-atmosphere is equal to:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because 1 litre-atmosphere is equivalent to 101.3 J, 24.206 cal, and

101.3 \times {10^7}\mathrm{erg}

.

Solution approach

Correct Answer + Core Reason: All of these. 1 litre-atmosphere is equal to 101.3 J, 24.206 cal, and $101.3 \times {10^7}\mathrm{erg}$ .

Step-by-Step Solution:

Step 1: Define litre-atmosphere: 1 litre-atmosphere (L atm) is a unit of energy or work.

Step 2: Conversion to Joules: 1 L atm = 101.325 J (approximately 101.3 J).

Step 3: Conversion to Calories: 1 L atm = 24.217 cal (approximately 24.206 cal).

Step 4: Conversion to Ergs: 1 L atm = $101.325 \times 10^7$ ergs (approximately $101.3 \times 10^7$ ergs).

Step 5: Conclusion: All the given options are correct conversions of 1 litre-atmosphere.

Why other options are wrong:
- Option A: 101.3 J is a correct value for 1 L atm, but it's not the only correct option.
- Option B: 24.206 cal is a correct value for 1 L atm, but it's not the only correct option.
- Option C: $101.3 \times {10^7}\mathrm{erg}$ is a correct value for 1 L atm, but it's not the only correct option.

Key Concepts:
- Units of Energy
- Conversion Factors
- Litre-atmosphere

Memory Aids:
- 1 L atm ≈ 101.3 J ≈ 24.2 cal ≈ $10^9$ ergs (approximate values)

Quick Tips:
- Remember the conversion factors between different energy units.

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