The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of $5\,\,{\rm{d}}{{\rm{m}}^3}$ isothermally until the volume become $25\,\,{\rm{d}}{{\rm{m}}^3}$ is:

$ + 2.81 \times {10^3}{\rm{J}}$

$2.01 \times {10^{ - 3}}{\rm{J}}$

$ + 2.01 \times {10^{ - 6}}{\rm{J}}$

Given that standard heat enthalpy of ${\rm{C}}{{\rm{H}}_4},{{\rm{C}}_2}{{\rm{H}}_4}$and ${{\rm{C}}_3}{{\rm{H}}_8}$are −17.9,12.5,−24.8 kcal/mol. The ∆? for ${\rm{C}}{{\rm{H}}_4} + {{\rm{C}}_2}{{\rm{H}}_4} \to {{\rm{C}}_3}{{\rm{H}}_8}$ is :

$ - 19.4{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

$ - 55.2{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

$ - 30.2{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

$55.2{\rm{kcal mo}}{{\rm{l}}^{ - 1}}$

${C_{{\rm{graphite}}}} + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);$ ∆H= -94.05 kcal ${\rm{mo}}{{\rm{l}}^{ - 1}}$ ${C_{{\rm{diamond}}}} + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);$ ∆H= -94.05 kcal ${\rm{mo}}{{\rm{l}}^{ - 1}}$ therfore :

${C_{{\rm{diamond}}}} \to {C_{{\rm{graphite}}}};\Delta H_{298{\rm{K}}}^0 = + 450{\rm{cal mo}}{{\rm{l}}^{ - 1}}$

${C_{{\rm{graphite}}}} \to {C_{{\rm{diamond}}}};\Delta H_{298{\rm{K}}}^0 = - 450{\rm{cal mo}}{{\rm{l}}^{ - 1}}$

Diamond is harder than graphite

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium, Page 2

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Understanding +4 | -1

The volume of gas is reduced to half from its original volume. The specific heat will be

View answer and explanation

Correct option: C

Remain constant

Option C (Remain constant) is correct because specific heat is an intensive property and does not depend on the amount of substance.

Solution approach

Correct Answer + Core Reason: Remain constant. Specific heat is an intensive property and is independent of the amount of substance.

Step-by-Step Solution:

Step 1: Define specific heat: Specific heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.

Step 2: Specific heat is an intensive property, meaning it does not depend on the amount of substance.

Step 3: Therefore, changing the volume of a gas does not change its specific heat.

Why other options are wrong:
- Option A: Reduced to half. This is incorrect because specific heat is an intensive property.
- Option B: Be doubled. This is incorrect because specific heat is an intensive property.
- Option D: Increase four times. This is incorrect because specific heat is an intensive property.

Key Concepts:
- Specific heat
- Intensive properties
- Extensive properties

Memory Aids:
- Intensive properties are independent of the amount of substance.

Quick Tips:
- Remember the difference between intensive and extensive properties.

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Application +4 | -1

Assuming that water vapours are ideal gas. The change in internal energy (∆U) when 1 mol of water is vaporized at 1 bar pressure and 100C. (Given molar enthalpy of vaporization at 1 bar and 373 K is 41 kJ

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

and R=8.3 J

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

{K^{ - 1}}

) will be:

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it is calculated by subtracting the work done during vaporization from the molar enthalpy of vaporization.

Solution approach

Correct Answer + Core Reason: $37.904\mathrm{kJ mo}{\mathrm{l}^{ - 1}}$ . The change in internal energy is calculated by subtracting the work done during vaporization from the enthalpy of vaporization.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H = 41 \text{{ kJ/mol}}$ , $P = 1 \text{{ bar}}$ , $T = 100^\circ \text{{C}} = 373 \text{{ K}}$ , $R = 8.3 \text{{ J/(mol K)}}$

Step 2: Use the relationship between enthalpy change ( $\Delta H$ ) and internal energy change ( $\Delta U$ ): $\Delta H = \Delta U + P\Delta V$

Step 3: Since water vapor is assumed to be an ideal gas, $P\Delta V = \Delta n_{gas}RT$ . Here, $\Delta n_{gas} = 1$ (1 mole of gas is formed from liquid).

Step 4: Substitute the values: $P\Delta V = (1) \times (8.3 \text{{ J/(mol K)}}) \times (373 \text{{ K}}) = 3100 \text{{ J/mol}} = 3.1 \text{{ kJ/mol}}$

Step 5: Calculate $\Delta U$ : $\Delta U = \Delta H - P\Delta V = 41 \text{{ kJ/mol}} - 3.1 \text{{ kJ/mol}} = 37.9 \text{{ kJ/mol}}$

Step 6: Final Answer: $37.904\mathrm{kJ mo}{\mathrm{l}^{ - 1}}$

Why other options are wrong:
- Option A: This is the value of $\Delta H$ , not $\Delta U$ . It neglects the work done during vaporization.
- Option B: This is incorrect and doesn't relate to any logical step in the calculation.
- Option C: This is incorrect and doesn't relate to any logical step in the calculation.

Key Concepts:
- Enthalpy change, internal energy change, ideal gas law
- Relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + P\Delta V$

Memory Aids:
- Remember that $\Delta H$ includes the energy required to change the volume of the system against a constant pressure.

Quick Tips:
- Always convert all units to a consistent set (e.g., J to kJ) before performing calculations.
- Remember to consider the change in the number of moles of gas when calculating $P\Delta V$ .

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Application +4 | -1

∆H for

\mathrm{CaC}{\mathrm{O}_3}\left( s \right)

\to \mathrm{CaO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)

is 176 kJ

mo{l^{ - 1}}

at 1240 K. The ∆U for the change is equal to:

View answer and explanation

Correct option: B

165.6 kJ

The

\Delta U

for the change is 165.6 kJ, calculated using the formula

\Delta H = \Delta U + \Delta n_g RT

.

Solution approach

Correct Answer + Core Reason: 165.6 kJ. Using the relationship $\Delta H = \Delta U + \Delta n_g RT$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $\mathrm{CaC}{\mathrm{O}_3}\left( s \right) \to \mathrm{CaO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)$ .

Step 2: Identify the given information: $\Delta H = 176 \mathrm{~kJ/mol}$ and $T = 1240 \mathrm{~K}$ .

Step 3: Calculate the change in the number of moles of gas: $\Delta n_g = n_{products}(g) - n_{reactants}(g) = 1 - 0 = 1$ .

Step 4: Use the relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + \Delta n_g RT$ .

Step 5: Rearrange the equation to solve for $\Delta U$ : $\Delta U = \Delta H - \Delta n_g RT$ .

Step 6: Substitute the values: $\Delta U = 176000 \mathrm{~J/mol} - (1) (8.314 \mathrm{~J/mol \cdot K}) (1240 \mathrm{~K})$ .

Step 7: Calculate $\Delta U$ : $\Delta U = 176000 - 10309.36 = 165690.64 \mathrm{~J/mol} \approx 165.7 \mathrm{~kJ/mol}$ .

Why other options are wrong:
- Option A: 160 kJ is incorrect due to calculation errors or neglecting the $\Delta n_g RT$ term.
- Option C: 186.3 kJ is incorrect, likely from adding the $\Delta n_g RT$ term instead of subtracting.
- Option D: 180.0 kJ is incorrect due to calculation errors.

Key Concepts:
- Enthalpy Change ( $\Delta H$ )
- Internal Energy Change ( $\Delta U$ )
- Change in Number of Moles of Gas ( $\Delta n_g$ )

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

Quick Tips:
- Ensure consistent units (J vs. kJ) throughout the calculation.
- Pay attention to the sign of $\Delta n_g$ .

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Understanding +4 | -1

Choose intensive properties in the following:
(i) density

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

(ii) specific volume

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

(iii) Heat capacity
(iv) mole

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

(v) concentration

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

(vi) volume

View answer and explanation

Correct option: C

i, ii, v

Option C (i, ii, v) is correct because density, specific volume, and concentration are intensive properties, independent of the amount of substance.

Solution approach

Correct Answer + Core Reason: i, ii, v. Intensive properties do not depend on the amount of substance.

Step-by-Step Solution:

Step 1: Define intensive and extensive properties.\nIntensive properties: Properties that do not depend on the amount of substance (e.g., temperature, pressure, density).\nExtensive properties: Properties that depend on the amount of substance (e.g., mass, volume, heat capacity).

Step 2: Analyze each property.
(i) Density: Mass per unit volume. It's an intensive property.
(ii) Specific volume: Volume per unit mass. It's an intensive property.
(iii) Heat capacity: The amount of heat required to raise the temperature of a substance by 1 degree Celsius. It's an extensive property.
(iv) Mole: Amount of substance. It's an extensive property.
(v) Concentration: Amount of solute per unit volume of solution. It's an intensive property.
(vi) Volume: The amount of space occupied by a substance. It's an extensive property.

Step 3: Identify the intensive properties.\nBased on the analysis, density, specific volume, and concentration are intensive properties.

Step 4: Select the correct option.\nThe correct option is C, which includes i, ii, and v.

Why other options are wrong:
- Option A: Includes 'iv' (mole), which is an extensive property.
- Option B: Includes 'vi' (volume) and 'iii' (heat capacity), which are extensive properties.
- Option D: Includes 'iii' (heat capacity) and 'vi' (volume), which are extensive properties.

Key Concepts:
- Intensive properties
- Extensive properties

Memory Aids:
- Intensive properties are Independent of amount.
- Extensive properties are dependent on amount.

Quick Tips:
- Remember the definitions of intensive and extensive properties. Think about whether changing the amount of substance changes the property.

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Analyzing +4 | -1

Assertion : Many endothermic reactions that are not spontaneous at room temperature become spontaneous at high temperature.
Reason : Entropy of the system increases with increase in temperature.

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 correct, but the reason doesn't fully explain the assertion. The spontaneity is determined by

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

, and at high temperatures, the

T\Delta S

term becomes dominant.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. While both statements are true, the primary reason for spontaneity at high temperatures is the dominance of the $T\Delta S$ term in the Gibbs free energy equation.

Step-by-Step Solution:

Step 1: Recall the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$ .

Step 2: Understand the conditions for spontaneity: $\Delta G < 0$ for a spontaneous process.

Step 3: Analyze endothermic reactions: $\Delta H > 0$ .

Step 4: Consider the effect of temperature on the $T\Delta S$ term: As temperature increases, the magnitude of the $T\Delta S$ term increases.

Step 5: For many endothermic reactions, $\Delta S > 0$ (entropy increases). At low temperatures, $\Delta H$ may be larger than $T\Delta S$ , resulting in $\Delta G > 0$ (non-spontaneous).

Step 6: As temperature increases, $T\Delta S$ becomes larger. Eventually, $T\Delta S$ can become greater than $\Delta H$ , resulting in $\Delta G < 0$ (spontaneous).

Step 7: The assertion is correct: Many endothermic reactions become spontaneous at high temperatures.

Step 8: The reason is correct: Entropy of the system increases with an increase in temperature. However, this is not the *complete* explanation. The key is the *dominance* of the $T\Delta S$ term at high temperatures.

Why other options are wrong:
- Option A: The reason is not the *complete* explanation, so this option is wrong.
- Option C: The reason is correct, so this option is wrong.
- Option D: Both the assertion and reason are correct, so this option is wrong.

Key Concepts:
- Gibbs free energy equation
- Spontaneity
- Endothermic reactions
- Entropy and temperature relationship

Memory Aids:
- $\Delta G = \Delta H - T\Delta S$
- Spontaneous: $\Delta G < 0$

Quick Tips:
- Remember that the spontaneity of a reaction depends on both enthalpy and entropy changes.
- At high temperatures, the entropy term becomes more important.

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Understanding +4 | -1

A thermodynamic quantity is that:

View answer and explanation

Correct option: C

Quantity which depends only on the state of the system

Option C (Quantity which depends only on the state of the system) is correct because a thermodynamic quantity is a state function, meaning its value depends only on the initial and final states of the system, not the path taken.

Solution approach

Correct Answer + Core Reason: Quantity which depends only on the state of the system. Thermodynamic quantities are state functions.

Step-by-Step Solution:

Step 1: Define thermodynamic quantity.\nA thermodynamic quantity is a property of a system that can be measured or calculated and is used to describe the state of the system.

Step 2: Explain state functions.\nA state function is a property whose value does not depend on the path taken to reach a specific state, but only on the initial and final states. Examples include internal energy, enthalpy, entropy, and Gibbs free energy.

Step 3: Conclusion: A thermodynamic quantity depends only on the state of the system.

Why other options are wrong:
- Option A: Which is used in thermochemistry is incorrect because while thermodynamic quantities are used in thermochemistry, this is not the defining characteristic.
- Option B: Which obeys all the laws of thermodynamics is incorrect because all physical quantities obey the laws of thermodynamics, not just thermodynamic quantities.
- Option D: Quantity which is used in measuring thermal change is incorrect because while thermodynamic quantities are used to measure thermal changes, this is not the defining characteristic.

Key Concepts:
- Thermodynamic quantity
- State function

Memory Aids:
- State functions depend only on the state of the system.

Quick Tips:
- Remember the definition of a state function and examples like internal energy, enthalpy, entropy, and Gibbs free energy.

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Analyzing +4 | -1

\begin{aligned} & \mathrm{HA}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-}+\mathrm{q}_1 \mathrm{~kJ} \\ & \mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{q}_2 \mathrm{~kJ} \end{aligned}

The enthalpy of ionisation of

\mathrm{HA}

is

View answer and explanation

Correct option: C

\left(\mathrm{q_2-q}_1\right)

Option C (

(q_2-q_1)

) is correct because the enthalpy of ionization is the difference between the heat released during the neutralization of a strong acid and the heat released during the neutralization of the weak acid.

Solution approach

Correct Answer + Core Reason: $\left(\mathrm{q_2-q_1}\right)$ . The enthalpy of ionization is the difference between the enthalpy of neutralization of a strong acid and a weak acid.

Step-by-Step Solution:

Step 1: Write the given equations:
Equation 1: $\mathrm{HA}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-}+\mathrm{q}_1 \mathrm{~kJ}$
Equation 2: $\mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{q}_2 \mathrm{~kJ}$

Step 2: The enthalpy of ionization of HA is the energy required to ionize HA into $\mathrm{H}^+$ and $\mathrm{A}^-$ :
$\mathrm{HA} \rightarrow \mathrm{H}^{+} + \mathrm{A}^{-} \qquad \Delta H_{ionization} = ?$

Step 3: We can obtain this equation by subtracting Equation 1 from Equation 2:
$\mathrm{H}^{+}+\mathrm{OH}^{-} - (\mathrm{HA}+\mathrm{OH}^{-}) \rightarrow \mathrm{H}_2 \mathrm{O} - (\mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-}) \qquad \Delta H = q_2 - q_1$

Step 4: Simplify the equation:
$\mathrm{H}^{+} - \mathrm{HA} \rightarrow - \mathrm{A}^{-} \qquad \Delta H = q_2 - q_1$
$\mathrm{HA} \rightarrow \mathrm{H}^{+} + \mathrm{A}^{-} \qquad \Delta H_{ionization} = q_2 - q_1$

Step 5: Therefore, the enthalpy of ionization of HA is $(q_2 - q_1)$ kJ.

Why other options are wrong:
- Option A: $\left(\mathrm{q_1+q}_2\right)$ is incorrect because it involves adding the heats of reaction instead of subtracting.
- Option B: $\left(\mathrm{q_1-q}_2\right)$ is incorrect because it subtracts in the wrong order.
- Option D: $-\left(\mathrm{q_1+q}_2\right)$ is incorrect because it involves adding the heats of reaction and changing the sign.

Key Concepts:
- Enthalpy of ionization
- Hess's Law
- Neutralization reactions

Memory Aids:
- Enthalpy of ionization is the energy required to ionize a substance.

Quick Tips:
- Manipulate the given equations to obtain the desired equation.
- Remember to change the sign of $\Delta H$ when reversing an equation.
- Subtract the heat of neutralization of weak acid from strong acid to get enthalpy of ionization.

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Understanding +4 | -1

The volume of gas is reduced to half form its original volume. The specific heat will be

View answer and explanation

Correct option: C

Remain constant

Option C (Remain constant) is correct because specific heat is an intensive property and does not depend on the amount of substance or volume.

Solution approach

Correct Answer + Core Reason: Remain constant. Specific heat is an intensive property and does not depend on the volume of the gas.

Step-by-Step Solution:

Step 1: Define specific heat: Specific heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin).

Step 2: Intensive vs. Extensive Properties: Specific heat is an intensive property, meaning it does not depend on the amount of substance. Volume, mass, and energy are extensive properties, which do depend on the amount of substance.

Step 3: Analyze the situation: Reducing the volume of the gas does not change the amount of heat required to raise the temperature of 1 gram of the gas by 1 degree Celsius.

Why other options are wrong:
- Option A: Recuce to half. Specific heat is an intensive property; it does not depend on the volume.
- Option B: Be doubled. Specific heat is an intensive property; it does not depend on the volume.
- Option D: Increase four times. Specific heat is an intensive property; it does not depend on the volume.

Key Concepts:
- Specific Heat
- Intensive Properties
- Extensive Properties

Memory Aids:
- Intensive properties: Independent of amount (e.g., temperature, pressure, density, specific heat).
- Extensive properties: Dependent on amount (e.g., mass, volume, energy).

Quick Tips:
- Remember the difference between intensive and extensive properties.

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Remembering +4 | -1

If water is formed from

{H^ + }

ions and

O{H^ - },

the heat of formation of water is :

View answer and explanation

Correct option: A

- 13.7\mathrm{kcal}

Option A (-13.7 kcal) is correct because the heat of formation of water from its ions is negative, indicating an exothermic process.

Solution approach

Correct Answer + Core Reason: -13.7 kcal. The formation of water from H+ and OH- ions is an exothermic process.

Step-by-Step Solution:

Step 1: The reaction is ${H^ + } + O{H^ - } \to {H_2}O$ .

Step 2: This is a neutralization reaction, and the heat of neutralization of a strong acid and a strong base is approximately -13.7 kcal/mol.

Step 3: Therefore, the heat of formation of water from its ions is -13.7 kcal.

Why other options are wrong:
- Option B: +13.7 kcal is incorrect because the reaction is exothermic, so the heat of formation is negative.
- Option C: -63.4 kcal is incorrect because this value is too large for the heat of formation of water from its ions.
- Option D: +63.4 kcal is incorrect because the reaction is exothermic, so the heat of formation is negative and the value is too large.

Key Concepts:
- Heat of formation
- Neutralization reaction
- Exothermic reaction

Memory Aids:
- Strong acid + Strong base neutralization ≈ -13.7 kcal/mol

Quick Tips:
- Remember the approximate heat of neutralization for strong acids and bases.

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Application +4 | -1

Theenthalpy change for the transition of liquid water to steam is

\Delta {H_v}ap = 37.3kJmo{l^{( - 1)}}

at 373 K. The entropy change for the process is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the entropy change is calculated using ∆S = ∆H/T for a phase transition at constant temperature.

Solution approach

Correct Answer + Core Reason: $100\mathrm{J mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}$ . Entropy change for phase transition.

Step-by-Step Solution:

Step 1: Given: $\Delta H_{vap} = 37.3 \text{{ kJ/mol}} = 37300 \text{{ J/mol}}$ and $T = 373 \text{{ K}}$ .

Step 2: The entropy change for the process is given by $\Delta S = \frac{{\Delta H}}{T}$ .

Step 3: Substituting the values, $\Delta S = \frac{{37300}}{{373}} = 100 \text{{ J/mol/K}}$ .

Why other options are wrong:
- Option A: $132.5\mathrm{J mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}$ : This is incorrect due to a calculation error.
- Option C: $135.3\mathrm{J mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}$ : This is incorrect due to a calculation error.
- Option D: $75.5\mathrm{J mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}$ : This is incorrect due to a calculation error.

Key Concepts:
- Entropy change
- Enthalpy change
- Phase transition

Memory Aids:
- Remember the formula $\Delta S = \frac{{\Delta H}}{T}$ .

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

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Application +4 | -1

For vaporization of water at 1 atmospheric pressure, the values of ∆H and ∆S are 40.63 kJ

mo{l^{ - 1}}

and 108.8

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

respectively. The temperature when Gibbs energy change (∆G) for this transformation will be zero, is:

View answer and explanation

Correct option: C

373.4 K

Option C (373.4 K) is correct because it is the temperature at which the Gibbs free energy change (∆G) is zero, calculated using the formula ∆G = ∆H - T∆S.

Solution approach

Correct Answer + Core Reason: 373.4 K. The temperature at which Gibbs energy change is zero is calculated using the formula $T = \frac{{\Delta H}}{{\Delta S}}$ .

Step-by-Step Solution:

Step 1: Identify the given values: $\Delta H = 40.63 \text{{ kJ/mol}} = 40630 \text{{ J/mol}}$ and $\Delta S = 108.8 \text{{ J/(K\cdot mol)}}$ .

Step 2: Use the formula $\Delta G = \Delta H - T\Delta S$ . We want to find the temperature $T$ when $\Delta G = 0$ .

Step 3: Set $\Delta G = 0$ and solve for $T$ : $0 = \Delta H - T\Delta S \Rightarrow T = \frac{{\Delta H}}{{\Delta S}}$ .

Step 4: Substitute the given values: $T = \frac{{40630 \text{{ J/mol}}}}{{108.8 \text{{ J/(K\cdot mol)}}}} = 373.4 \text{{ K}}$ .

Step 5: Final Answer: $T = 373.4 \text{{ K}}$ .

Why other options are wrong:
- Option A (273.4 K): This is incorrect as it doesn't use the correct formula for calculating the temperature at which Gibbs free energy change is zero.
- Option B (393.4 K): This is incorrect as it doesn't use the correct formula for calculating the temperature at which Gibbs free energy change is zero.
- Option D (293.4 K): This is incorrect as it doesn't use the correct formula for calculating the temperature at which Gibbs free energy change is zero.

Key Concepts:
- Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$
- Phase transition temperature calculation.

Memory Aids:
- Remember the Gibbs Free Energy equation: \

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Remembering +4 | -1

Heat of reaction at constant volume is equal to

View answer and explanation

Correct option: A

\Delta U

Option A (

\Delta U

) is correct because heat of reaction at constant volume is equal to the change in internal energy.

Solution approach

Correct Answer + Core Reason: $\Delta U$ . Heat of reaction at constant volume is equal to the change in internal energy.

Step-by-Step Solution:

Step 1: Define heat of reaction at constant volume: The heat of reaction at constant volume ( $q_v$ ) is the heat absorbed or released during a reaction when the volume is kept constant.

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

Step 3: Consider constant volume: At constant volume, no work is done ( $w = 0$ ). Therefore, $\Delta U = q_v$ .

Why other options are wrong:
- Option B: $\Delta G$ . $\Delta G$ is the change in Gibbs free energy, which is related to spontaneity at constant temperature and pressure.
- Option C: $\Delta \mathrm{{H}}$ . $\Delta H$ is the change in enthalpy, which is equal to the heat of reaction at constant pressure.
- Option D: $\Delta S$ . $\Delta S$ is the change in entropy, which is a measure of disorder.

Key Concepts:
- First law of thermodynamics
- Internal energy
- Heat of reaction
- Constant volume conditions
- Enthalpy
- Gibbs free energy
- Entropy

Memory Aids:
- $\Delta U = q_v$ (at constant volume)
- $\Delta H = q_p$ (at constant pressure)

Quick Tips:
- Remember that the heat of reaction at constant volume is equal to the change in internal energy, and the heat of reaction at constant pressure is equal to the change in enthalpy.

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Application +4 | -1

If 50 calorie are added to a system and system does work of 30 calorie on surroundings, the change in internal energy of system is:

View answer and explanation

Correct option: A

20 cal

Option A (20 cal) is correct because the change in internal energy is calculated by subtracting the work done by the system from the heat added to the system.

Solution approach

Correct Answer + Core Reason: 20 cal. The change in internal energy is calculated using the first law of thermodynamics.

Step-by-Step Solution:

Step 1: Identify the given information:
- Heat added to the system, $q = 50\, \mathrm{cal}$
- Work done by the system, $w = 30\, \mathrm{cal}$

Step 2: Apply the first law of thermodynamics: $\Delta U = q - w$ , where $\Delta U$ is the change in internal energy.

Step 3: Substitute the values: $\Delta U = 50\, \mathrm{cal} - 30\, \mathrm{cal} = 20\, \mathrm{cal}$

Why other options are wrong:
- Option B: 50 cal is the heat added, not the change in internal energy.
- Option C: 40 cal is incorrect. It likely involves an incorrect calculation.
- Option D: 30 cal is the work done, not the change in internal energy.

Key Concepts:
- First law of thermodynamics

Memory Aids:
- $\Delta U = q - w$ (Work done BY the system is positive).

Quick Tips:
- Pay attention to the sign conventions for heat and work.

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Understanding +4 | -1

Which compound will absorb the maximum amount of heat when dissolved in the same amount of water? (Integral heats of solution at

{25^0}c

in kcal/mol of each solute are given in brackets):

View answer and explanation

Correct option: C

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}\left( {\Delta H = + 16.08} \right)

Option C is correct. Ammonium nitrate absorbs the maximum amount of heat because it has the largest positive

\Delta H

value, indicating an endothermic process where heat is absorbed from the surroundings.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}\left( {\Delta H = + 16.08} \right)$ . The compound with the largest positive $\Delta H$ absorbs the most heat.

Step-by-Step Solution:
Step 1: Understand the concept of heat of solution: The heat of solution ( $\Delta H$ ) is the enthalpy change when one mole of a substance is dissolved in a solvent. A positive $\Delta H$ indicates an endothermic process (heat is absorbed), and a negative $\Delta H$ indicates an exothermic process (heat is released).
Step 2: Analyze the given $\Delta H$ values:
- HCl: $\Delta H = -17.74$ kcal/mol (exothermic)
- HNO3: $\Delta H = -7.85$ kcal/mol (exothermic)
- NH4NO3: $\Delta H = +16.08$ kcal/mol (endothermic)
- NaCl: $\Delta H = +1.02$ kcal/mol (endothermic)
Step 3: Determine the compound that absorbs the maximum amount of heat: The compound with the largest positive $\Delta H$ value will absorb the most heat. In this case, it is NH4NO3 with $\Delta H = +16.08$ kcal/mol.

Why other options are wrong:
- Option A: HCl releases heat (exothermic) when dissolved.
- Option B: HNO3 releases heat (exothermic) when dissolved.
- Option D: NaCl absorbs heat, but less than NH4NO3.

Key Concepts:
- Heat of solution ( $\Delta H$ )
- Endothermic and exothermic processes

Memory Aids:
- Positive $\Delta H$ = endothermic (heat absorbed)
- Negative $\Delta H$ = exothermic (heat released)

Quick Tips:
- When comparing heat absorption, look for the largest positive $\Delta H$ value.

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Analyzing +4 | -1

A: An ideal crystal has more entropy than a real crystal.
R: An ideal crystal has more disorder.

View answer and explanation

Correct option: D

If both A and R are incorrect

Option D is correct because an ideal crystal has less entropy and more order than a real crystal due to the absence of defects.

Solution approach

Correct Answer + Core Reason: If both A and R are incorrect. An ideal crystal has less entropy and more order.

Step-by-Step Solution:

Step 1: Define ideal crystal: An ideal crystal is a hypothetical crystal with perfect order and no defects.

Step 2: Define real crystal: A real crystal contains defects such as vacancies, impurities, and dislocations.

Step 3: Entropy and disorder: Entropy is a measure of disorder. Higher disorder means higher entropy.

Step 4: Compare entropy: An ideal crystal has perfect order, so it has lower entropy than a real crystal.

Step 5: Analyze Assertion: Assertion states that an ideal crystal has more entropy than a real crystal, which is incorrect.

Step 6: Analyze Reason: Reason states that an ideal crystal has more disorder, which is also incorrect.

Step 7: Conclusion: Both the assertion and the reason are incorrect.

Why other options are wrong:
- Option A: Incorrect because both A and R are incorrect.
- Option B: Incorrect because both A and R are incorrect.
- Option C: Incorrect because both A and R are incorrect.
- Option D: Correct because both A and R are incorrect.

Key Concepts:
- Entropy
- Ideal crystal
- Real crystal
- Disorder

Memory Aids:
- Ideal = Ordered = Low Entropy
- Real = Defects = High Entropy

Quick Tips:
- Remember that ideal systems are theoretical and often represent the lowest possible entropy state.

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Application +4 | -1

The maximum work done in expanding 16 g oxygen at 300 K and occupying a volume of

5\,\,\mathrm{d}{\mathrm{m}^3}

isothermally until the volume become

25\,\,\mathrm{d}{\mathrm{m}^3}

is:

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the maximum work done calculated using the isothermal reversible expansion formula.

Solution approach

Correct Answer + Core Reason: $2.01 \times {10^3}\mathrm{J}$ . The maximum work done during isothermal expansion is calculated using the formula $W = -nRT \ln{\frac{V_2}{V_1}}$ .

Step-by-Step Solution:

Step 1: Identify the given values: $m = 16 \text{{ g}}$ , $T = 300 \text{{ K}}$ , $V_1 = 5 \text{{ dm}}^3$ , $V_2 = 25 \text{{ dm}}^3$ .

Step 2: Calculate the number of moles of oxygen ( $O_2$ ): $n = \frac{m}{M} = \frac{16 \text{{ g}}}{32 \text{{ g/mol}}} = 0.5 \text{{ mol}}$

Step 3: Use the formula for maximum work done in an isothermal reversible expansion: $W = -nRT \ln{\frac{V_2}{V_1}}$

Step 4: Substitute the values: $W = -(0.5 \text{{ mol}})(8.314 \text{{ J/(mol K)}})(300 \text{{ K}}) \ln{\frac{25 \text{{ dm}}^3}{5 \text{{ dm}}^3}}$

Step 5: Simplify: $W = -(0.5)(8.314)(300) \ln{5} = -1247.1 \times 1.6094 \approx -2007.9 \text{{ J}}$

Step 6: Since we are looking for the magnitude of work done, we take the absolute value: $|W| = 2007.9 \text{{ J}} \approx 2.01 \times 10^3 \text{{ J}}$

Step 7: Final Answer: $2.01 \times {10^3}\mathrm{J}$

Why other options are wrong:
- **Option B ( $+ 2.81 \times {10^3}\mathrm{J}$ ): This is incorrect due to a miscalculation in the natural logarithm or incorrect application of the formula. It might arise from using a wrong value for R or T.
- Option C ( $2.01 \times {10^{ - 3}}\mathrm{J}$ ): This is incorrect because it is significantly smaller than the actual work done. This could be due to a mistake in the units or a miscalculation of the number of moles.
- Option D ( $+ 2.01 \times {10^{ - 6}}\mathrm{J}$ ): This is incorrect as it is an extremely small value, indicating a significant error in the calculation, possibly involving incorrect units or a misunderstanding of the formula.

Key Concepts:**
- Isothermal Reversible Expansion
- Work Done: $W = -nRT \ln{\frac{V_2}{V_1}}$
- Ideal Gas Constant (R = 8.314 J/(mol K))

Memory Aids:
- Isothermal: Constant Temperature
- Work done *by* the system is negative, work done *on* the system is positive.

Quick Tips:
- Ensure the correct units are used for all variables.
- Remember to use the natural logarithm ( $\ln$ ) in the formula.
- Pay attention to the sign of the work done; expansion implies work done *by* the system.

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Application +4 | -1

Given that standard heat enthalpy of

\mathrm{C}{\mathrm{H}_4},{\mathrm{C}_2}{\mathrm{H}_4}

and

{\mathrm{C}_3}{\mathrm{H}_8}

are −17.9,12.5,−24.8 kcal/mol. The ∆? for

\mathrm{C}{\mathrm{H}_4} + {\mathrm{C}_2}{\mathrm{H}_4} \to {\mathrm{C}_3}{\mathrm{H}_8}

is :

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it represents the enthalpy change calculated by subtracting the sum of the enthalpies of the reactants from the enthalpy of the product.

Solution approach

Correct Answer + Core Reason: $- 19.4\mathrm{kcal mo}{\mathrm{l}^{ - 1}}$ . The enthalpy change is calculated using the formula $\Delta H = \sum H_{products} - \sum H_{reactants}$ .

Step-by-Step Solution:

Step 1: Identify the given information.
$\Delta H_f(CH_4) = -17.9$ kcal/mol
$\Delta H_f(C_2H_4) = 12.5$ kcal/mol
$\Delta H_f(C_3H_8) = -24.8$ kcal/mol\nReaction: $\mathrm{C}{\mathrm{H}_4} + {\mathrm{C}_2}{\mathrm{H}_4} \to {\mathrm{C}_3}{\mathrm{H}_8}$

Step 2: Write the formula for enthalpy change.
$\Delta H = \sum H_{products} - \sum H_{reactants}$
$\Delta H = [H(C_3H_8)] - [H(CH_4) + H(C_2H_4)]$

Step 3: Substitute the given values.
$\Delta H = [-24.8] - [-17.9 + 12.5]$

Step 4: Calculate the enthalpy change.
$\Delta H = -24.8 - (-5.4)$
$\Delta H = -24.8 + 5.4$
$\Delta H = -19.4\mathrm{kcal mo}{\mathrm{l}^{ - 1}}$

Why other options are wrong:
- Option A: This is likely a calculation error. It's much larger in magnitude than the correct answer, suggesting a misunderstanding of how to apply the formula.
- Option B: This is likely a calculation error.
- Option C: This is the positive value of what you get if you subtract the product from the reactants. The sign is important in enthalpy calculations.

Key Concepts:
- Enthalpy change calculation
- Standard enthalpies of formation

Memory Aids:
- Remember the formula: Products minus Reactants

Quick Tips:
- Pay attention to the stoichiometric coefficients in the balanced equation (in this case, all are 1).
- Ensure correct units are used.

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Understanding +4 | -1

If latent heat of vaporization is

L

at boiling point

T\left( \mathrm{K} \right)

then entropy of vaporisation is

View answer and explanation

Correct option: B

L{T^{ - 1}}

Option B (

L{T^{ - 1}}

) is correct because entropy change is defined as heat absorbed divided by temperature.

Solution approach

Correct Answer + Core Reason: $L{T^{ - 1}}$ . Entropy change is heat absorbed divided by temperature.

Step-by-Step Solution:

Step 1: Define entropy of vaporization.

Step 2: Entropy of vaporization is the change in entropy when a liquid is converted to a gas at its boiling point.

Step 3: The formula for entropy change is $\Delta S = \frac{q}{T}$ , where q is the heat absorbed and T is the temperature.

Step 4: In this case, the heat absorbed is the latent heat of vaporization (L). Therefore, the entropy of vaporization is $\Delta S = \frac{L}{T} = L{T^{ - 1}}$ .

Why other options are wrong:
- Option A: $LT$ is incorrect because it multiplies latent heat and temperature, which is not the correct formula for entropy of vaporization.
- Option C: $T{L^{ - 1}}$ is incorrect because it divides temperature by latent heat, which is the inverse of the correct formula.
- Option D: None of these is incorrect because option B is the correct answer.

Key Concepts:
- Entropy of vaporization
- Latent heat
- Entropy change

Memory Aids:
- $\Delta S = \frac{q}{T}$ (Entropy change = Heat/Temperature)

Quick Tips:
- Remember the formula for entropy change.

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Understanding +4 | -1

The enthalpy of formation of HI is 30.4 kJ. Which statement is false according to this observation?

View answer and explanation

Correct option: C

HI is a stable compound

Option C (HI is a stable compound) is false because a positive enthalpy of formation indicates that HI is an endothermic and therefore unstable compound.

Solution approach

Correct Answer + Core Reason: HI is a stable compound. A positive enthalpy of formation indicates an unstable compound.

Step-by-Step Solution:

Step 1: Understand the meaning of enthalpy of formation.\nEnthalpy of formation ( $\Delta H_f$ ) is the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states.

Step 2: Analyze the given information.\nThe enthalpy of formation of HI is 30.4 kJ, which is a positive value.

Step 3: Relate enthalpy of formation to stability.\nA positive enthalpy of formation indicates that the formation of HI is an endothermic process, meaning it requires energy input. Endothermic compounds are generally less stable than their constituent elements.

Step 4: Evaluate the statements.\nStatement A: HI is an endothermic compound - This is true because its enthalpy of formation is positive.\nStatement B: For the reaction, ${\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right) \to 2\mathrm{HI}\left( \mathrm{g} \right);\Delta H = 60.8\mathrm{kJ}$ - This is true because the enthalpy change for the formation of 2 moles of HI is twice the enthalpy of formation of 1 mole of HI (2 * 30.4 kJ = 60.8 kJ).\nStatement D: HI is an unstable compound - This is true because its enthalpy of formation is positive.\nStatement C: HI is a stable compound - This is false because its enthalpy of formation is positive, indicating it's unstable.

Why other options are wrong:
- Option A: HI is an endothermic compound. This is a true statement as the enthalpy of formation is positive.
- Option B: For the reaction, ${\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right) \to 2\mathrm{HI}\left( \mathrm{g} \right);\Delta H = 60.8\mathrm{kJ}$ . This is a true statement as it represents the enthalpy change for the formation of 2 moles of HI.
- Option D: HI is an unstable compound. This is a true statement as the enthalpy of formation is positive.

Key Concepts:
- Enthalpy of formation
- Endothermic and exothermic reactions
- Stability of compounds

Memory Aids:
- Positive $\Delta H_f$ = Endothermic = Less Stable
- Negative $\Delta H_f$ = Exothermic = More Stable

Quick Tips:
- Remember that a positive enthalpy of formation indicates an endothermic and unstable compound, while a negative enthalpy of formation indicates an exothermic and stable compound.

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Analyzing +4 | -1

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

∆H= -94.05 kcal

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

{C_{\mathrm{diamond}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);

∆H= -94.05 kcal

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

therfore :

View answer and explanation

Correct option: D

Graphite is the stabler allotrope

The correct answer is Graphite is the stabler allotrope. Since the combustion of graphite releases slightly more heat than the combustion of diamond, graphite is at a lower energy state and therefore more stable.

Solution approach

Correct Answer + Core Reason: Graphite is the stabler allotrope. The heat released during the combustion of graphite is slightly more negative, indicating a lower energy state.

Step-by-Step Solution:
Step 1: Analyze the given information: Both reactions produce the same product (CO2) and have similar, but not identical, $\Delta H$ values.
Step 2: Calculate the enthalpy change for the conversion of diamond to graphite: Subtract the enthalpy of formation of CO2 from diamond from the enthalpy of formation of CO2 from graphite: $\Delta H_{graphite \to diamond} = \Delta H_{graphite} - \Delta H_{diamond} = -94.05 - (-94.50) = +0.45 kcal/mol$ . This means that the conversion of graphite to diamond is endothermic, and the reverse (diamond to graphite) is exothermic.
Step 3: Determine the stability: Since the combustion of graphite is slightly more exothermic, it implies that graphite has a lower energy state than diamond. Therefore, graphite is the more stable allotrope.

Why other options are wrong:
- Option A: The enthalpy change for diamond to graphite is negative, not positive. The correct value is -450 cal/mol or -0.45 kcal/mol.
- Option B: The enthalpy change for graphite to diamond is positive, not negative. The correct value is +450 cal/mol or +0.45 kcal/mol.
- Option C: While diamond is harder than graphite, this is not directly derived from the given thermochemical data. This is a separate physical property.

Key Concepts:
- Enthalpy of formation
- Hess's Law (implicitly used)
- Allotropes
- Stability and energy states

Memory Aids:
- Lower energy state = more stable.

Quick Tips:
- When comparing stability based on heats of formation, the allotrope with the more negative (or less positive) heat of formation is more stable.

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Understanding +4 | -1

A heat engine absorbs heat

{Q_1}

at temperature

{T_1}

and heat

{Q_2}

at temperature

\;{T_2}

, work done by the engine is (

{Q_1}

+

{Q_2}

) this data

View answer and explanation

Correct option: D

Does not violates 1st law of thermodynamics

Option D (Does not violates 1st law of thermodynamics) is correct because the first law of thermodynamics states that energy is conserved, and the work done by the engine can be equal to the sum of the heat absorbed.

Solution approach

Correct Answer + Core Reason: Does not violates 1st law of thermodynamics. The first law of thermodynamics states that energy is conserved.

Step-by-Step Solution:

Step 1: State the First Law of Thermodynamics: The first law of thermodynamics states that energy is conserved, i.e., $\Delta U = Q - W$ , where $\Delta U$ is the change in internal energy, Q is the heat absorbed by the system, and W is the work done by the system.

Step 2: Apply to the Heat Engine: In a heat engine, the work done is equal to the net heat absorbed: $W = Q_1 + Q_2$ .

Step 3: Analyze the given data: The engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_2$ , and the work done by the engine is $W = Q_1 + Q_2$ .

Step 4: Conclusion: Since the work done is equal to the sum of the heat absorbed, the first law of thermodynamics is not violated.

Why other options are wrong:
- Option A: Violates 1st law of thermodynamics is incorrect because the given data satisfies the first law.
- Option B: Violates 1st law of thermodynamics if $Q_1$ is –ve is incorrect because even if $Q_1$ is negative (heat released), the first law can still be satisfied if the net heat absorbed equals the work done.
- Option C: Violates 1st law of thermodynamics if $Q_2$ is -ve is incorrect for the same reason as option B.

Key Concepts:
- First Law of Thermodynamics
- Heat Engine
- Conservation of Energy

Memory Aids:
- First Law: Energy is conserved

Quick Tips:
- Remember that the first law of thermodynamics is a statement of energy conservation.

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Understanding +4 | -1

Which of the following have same units?(i) work (ii) Heat
(iii) Energy (iv) Entropy

View answer and explanation

Correct option: A

(i),(ii) and (iii)

Option A ((i),(ii) and (iii)) is correct because work, heat, and energy all have the same units of energy, typically Joules (J).

Solution approach

Correct Answer + Core Reason: (i),(ii) and (iii). Work, heat, and energy all have the same units.

Step-by-Step Solution:

Step 1: Define the terms and their units.\nWork: The energy transferred when a force causes displacement. Units: Joules (J).\nHeat: The energy transferred due to a temperature difference. Units: Joules (J).\nEnergy: The capacity to do work. Units: Joules (J).\nEntropy: A measure of disorder or randomness. Units: Joules per Kelvin (J/K).

Step 2: Compare the units.\nWork, heat, and energy all have the same units (Joules).\nEntropy has different units (Joules per Kelvin).

Step 3: Conclusion: Work, heat, and energy have the same units.

Why other options are wrong:
- Option B: (i),(ii) and (iv) is incorrect because entropy has different units than work and heat.
- Option C: (ii),(iii) and (iv) is incorrect because entropy has different units than heat and energy.
- Option D: (iii) and (iv) is incorrect because energy and entropy have different units, and it excludes work and heat.

Key Concepts:
- Work
- Heat
- Energy
- Entropy

Memory Aids:
- Work, heat, and energy are all forms of energy.

Quick Tips:
- Remember the units of common physical quantities.

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Application +4 | -1

The enthalpy changes for two reactions are given by the equations.

2C{r_{\left( g \right)}} + \frac{3}{2}{O_{2\left( g \right)}} \to C{r_2}{O_{3\left( g \right)}},\,\,\Delta H = - 1130kJ\,\,\,\,\,\,{C_{\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \to C{O_{\left( g \right)}},\,\,\Delta H = - 110kJ\,\,\,\,

What is the enthalpy change (in kJ) for the reaction?

3C\left( s \right) + C{r_2}{O_3}\left( s \right) \to 2Cr\left( s \right) + 3CO\left( g \right)

View answer and explanation

Correct option: B

+ 800kJ

Option B (+ 800kJ) is correct because the enthalpy change for the reaction is calculated by manipulating the given equations using Hess's Law.

Solution approach

Correct Answer + Core Reason: + 800kJ. The enthalpy change is calculated by manipulating the given equations using Hess's Law.

Step-by-Step Solution:

Step 1: Write the target reaction: $3C(s) + Cr_2O_3(s) \to 2Cr(s) + 3CO(g)$

Step 2: Write the given reactions:
- (1) $2Cr(s) + \frac{3}{2}O_2(g) \to Cr_2O_3(s), \Delta H = -1130kJ$
- (2) $C(s) + \frac{1}{2}O_2(g) \to CO(g), \Delta H = -110kJ$

Step 3: Manipulate the equations to obtain the target reaction: Reverse equation (1) and multiply equation (2) by 3:
- $Cr_2O_3(s) \to 2Cr(s) + \frac{3}{2}O_2(g), \Delta H = +1130kJ$
- $3C(s) + \frac{3}{2}O_2(g) \to 3CO(g), \Delta H = -330kJ$

Step 4: Add the manipulated equations:
$3C(s) + Cr_2O_3(s) \to 2Cr(s) + 3CO(g), \Delta H = 1130 - 330 = 800kJ$

Why other options are wrong:
- Option A: -1460kJ is incorrect. It's likely a result of incorrect manipulation of the given equations.
- Option C: -800kJ is incorrect. It's likely a result of incorrect manipulation of the given equations.
- Option D: -450kJ is incorrect. It's likely a result of incorrect manipulation of the given equations.

Key Concepts:
- Hess's Law
- Enthalpy change

Memory Aids:
- Manipulate equations to match the target reaction.

Quick Tips:
- Reverse the sign of \u0394H when reversing a reaction.
- Multiply \u0394H by the same factor when multiplying a reaction.

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Application +4 | -1

2.1 g of Fe combines with S evolving 3.77 kJ. The heat of formation of FeS in kJ/mole is

View answer and explanation

Correct option: B

- 100.5

Option B (

- 100.5

) is correct because it represents the heat released per mole of FeS formed, calculated from the given data.

Solution approach

Correct Answer + Core Reason: $- 100.5$ . The heat of formation is calculated by finding the heat evolved per mole of Fe reacted.

Step-by-Step Solution:

Step 1: Identify the given information.\nMass of Fe = 2.1 g\nHeat evolved = 3.77 kJ

Step 2: Find the number of moles of Fe.\nMolar mass of Fe = 55.85 g/mol\nMoles of Fe = $\frac{2.1}{55.85} = 0.0376$ mol

Step 3: Calculate the heat evolved per mole of Fe.\nHeat evolved per mole = $\frac{3.77}{0.0376} = 100.26$ kJ/mol

Step 4: Since heat is evolved, the heat of formation is negative.\nHeat of formation of FeS = -100.26 kJ/mol $\approx -100.5$ kJ/mol

Why other options are wrong:
- Option A: This is a very small value and doesn't account for the molar mass of Fe. It's likely a calculation error.
- Option C: This is the heat evolved for 2.1g of Fe, not per mole.
- Option D: Since option B is correct, this is incorrect.

Key Concepts:
- Heat of formation
- Stoichiometry

Memory Aids:
- Heat evolved is negative, heat absorbed is positive.

Quick Tips:
- Always convert mass to moles before calculating molar quantities.

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Analyzing +4 | -1

Assertion : Heat absorbed in a reaction at constant temperature and constant volume is

-\Delta G

.

Reason :

\Delta G

should be negative for the reaction to be spontaneous.

View answer and explanation

Correct option: D

If assertion is false but reason is true.

Option D (Assertion is false but reason is true) is correct. Heat absorbed at constant temperature and volume is Δ U, not -Δ G, but a negative Δ G indicates spontaneity.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true. The assertion incorrectly relates heat absorbed to Δ G, while the reason correctly states the spontaneity criterion.

Step-by-Step Solution:

Step 1: Analyze the assertion: Heat absorbed in a reaction at constant temperature and constant volume is $-\Delta G$ .

Step 2: Recall the relationship between heat absorbed and thermodynamic functions: At constant volume, heat absorbed (q_v) is equal to the change in internal energy (Δ U), i.e., $q_v = \Delta U$ . Δ G is related to spontaneity, not directly to heat absorbed at constant volume.

Step 3: Determine the truth value of the assertion: The assertion is false because heat absorbed at constant volume is Δ U, not -Δ G.

Step 4: Analyze the reason: $\Delta G$ should be negative for the reaction to be spontaneous.

Step 5: Recall the Gibbs Free Energy equation: $\Delta G = \Delta H - T\Delta S$ . For a reaction to be spontaneous at a given temperature, Δ G must be negative.

Step 6: Determine the truth value of the reason: The reason is true because a negative Δ G indicates a spontaneous reaction.

Step 7: Determine the relationship between assertion and reason: The assertion is false, and the reason is true. They are not directly related.

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

Key Concepts:
- Gibbs Free Energy (Δ G) and Spontaneity
- Internal Energy (Δ U) and Heat absorbed at constant volume
- Thermodynamic relationships

Memory Aids:
- Δ G < 0: Spontaneous
- Δ G > 0: Non-spontaneous
- Δ G = 0: Equilibrium
- At constant volume: $q_v = \Delta U$

Quick Tips:
- Distinguish between Δ U, Δ H, and Δ G.
- Remember the spontaneity criterion based on Δ G.

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Understanding +4 | -1

Which species have negative value of specific heat?

View answer and explanation

Correct option: D

Saturated vapour

Option D (Saturated vapour) is correct because saturated vapor can have a negative specific heat under certain conditions, meaning adding heat can cause condensation and a decrease in temperature.

Solution approach

Correct Answer + Core Reason: Saturated vapour. Saturated vapor can exhibit negative specific heat under certain conditions.

Step-by-Step Solution:

Step 1: Define specific heat: Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin).

Step 2: Saturated Vapor: Saturated vapor is a vapor that is in equilibrium with its liquid phase. Adding heat to saturated vapor can cause some of the vapor to condense into liquid, which releases heat. If the amount of heat released by condensation is greater than the heat added, the temperature of the system will decrease, resulting in a negative specific heat.

Step 3: Analyze the options:
- A. Ice: Ice has a positive specific heat. Adding heat increases its temperature.
- B. Water: Water has a positive specific heat. Adding heat increases its temperature.
- C. Vapour: Normal vapor (unsaturated) has a positive specific heat. Adding heat increases its temperature.
- D. Saturated vapour: Saturated vapor can have a negative specific heat under specific conditions.

Why other options are wrong:
- Option A: Ice. Ice has a positive specific heat.
- Option B: Water. Water has a positive specific heat.
- Option C: Vapour. Unsaturated vapor typically has a positive specific heat.

Key Concepts:
- Specific Heat
- Saturated Vapor
- Phase Change

Memory Aids:
- Negative specific heat is unusual and typically associated with phase transitions in saturated vapors.

Quick Tips:
- Remember that saturated vapors can exhibit unusual thermodynamic behavior due to phase equilibrium.

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Remembering +4 | -1

An isothermal process is associated with:

View answer and explanation

Correct option: B

Constant temperature

Option B (Constant temperature) is correct because an isothermal process is defined as a process that occurs at a constant temperature.

Solution approach

Correct Answer + Core Reason: Constant temperature. An isothermal process is defined as a process occurring at constant temperature.

Step-by-Step Solution:

Step 1: Define isothermal process: An isothermal process is a thermodynamic process in which the temperature of the system remains constant.

Step 2: Therefore, an isothermal process is associated with constant temperature.

Why other options are wrong:
- Option A: Constant entropy. This is associated with an isentropic process, not an isothermal process.
- Option C: Constant enthalpy. This is associated with an isenthalpic process, not an isothermal process.
- Option D: Large change in heat content. While heat can be exchanged, the temperature remains constant.

Key Concepts:
- Isothermal process
- Thermodynamic processes

Memory Aids:
- Isothermal = same temperature.

Quick Tips:
- Know the definitions of different thermodynamic processes.

58 Medium

58 of 244Question permalink

Application +4 | -1

If

\Delta H_f^\circ

for

{H}_2 {O}_2

and

{H}_2 {O}

are

-188 {KJ} / \mathrm{mol}

and

-286 {KJ} / \mathrm{mol}

, what will be the enthalpy change of the reaction

2 {H}_2 {O}_2({l}) \rightarrow 2 {H}_2 {O}({l})+{O}_2(g)

View answer and explanation

Correct option: A

-196 \,{KJ}

Option A (-196 KJ) is the correct answer because it is calculated using the enthalpy of formation values for the products and reactants.

Solution approach

Correct Answer + Core Reason: -196 KJ. The enthalpy change of the reaction is calculated using the formula: $\Delta H_{reaction} = \sum \Delta H_{f(products)} - \sum \Delta H_{f(reactants)}$ .

Step-by-Step Solution:

Step 1: Identify the given information:
$\Delta H_f^\circ (H_2O_2) = -188 \text{{ KJ/mol}}$
$\Delta H_f^\circ (H_2O) = -286 \text{{ KJ/mol}}$ \nReaction: $2 H_2O_2(l) \rightarrow 2 H_2O(l) + O_2(g)$

Step 2: Write the formula for the enthalpy change of the reaction:
$\Delta H_{reaction} = [2 \times \Delta H_f^\circ (H_2O) + \Delta H_f^\circ (O_2)] - [2 \times \Delta H_f^\circ (H_2O_2)]$

Step 3: Note that the enthalpy of formation of an element in its standard state is zero. Therefore, $\Delta H_f^\circ (O_2) = 0$ .

Step 4: Substitute the given values into the formula:
$\Delta H_{reaction} = [2 \times (-286 \text{{ KJ/mol}}) + 0] - [2 \times (-188 \text{{ KJ/mol}})]$

Step 5: Calculate the enthalpy change:
$\Delta H_{reaction} = -572 \text{{ KJ/mol}} + 376 \text{{ KJ/mol}} = -196 \text{{ KJ/mol}}$

Step 6: Final Answer: -196 KJ

Why other options are wrong:
- Option B (-494 KJ): This might be obtained by incorrectly adding the enthalpies of formation instead of subtracting, or by not multiplying by the stoichiometric coefficients.
- Option C (146 KJ): This is likely due to incorrect signs when subtracting the enthalpies of formation or an error in the arithmetic.
- Option D (-98 KJ): This could result from forgetting to multiply the enthalpies of formation by the stoichiometric coefficients (2 for both $H_2O_2$ and $H_2O$ ).

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

Memory Aids:
- Products minus Reactants (for calculating enthalpy change)

Quick Tips:
- Always remember to multiply the enthalpy of formation by the stoichiometric coefficient in the balanced chemical equation.
- Ensure correct signs when subtracting enthalpies of formation.

59 Medium

59 of 244Question permalink

Remembering +4 | -1

Heat of fusion of a molecular solid is :

View answer and explanation

Correct option: C

Low

Option C (Low) is correct because molecular solids have weak intermolecular forces, requiring less energy to overcome during fusion.

Solution approach

Correct Answer + Core Reason: Low. Molecular solids have weak intermolecular forces.

Step-by-Step Solution:

Step 1: Define heat of fusion.

Step 2: Heat of fusion is the amount of heat required to change a substance from a solid to a liquid at its melting point.

Step 3: Molecular solids are held together by weak intermolecular forces (e.g., van der Waals forces, hydrogen bonds). Therefore, the heat of fusion is low.

Why other options are wrong:
- Option A: Very high is incorrect because molecular solids have weak intermolecular forces.
- Option B: High is incorrect because molecular solids have weak intermolecular forces.
- Option D: None of these is incorrect because option C is the correct answer.

Key Concepts:
- Heat of fusion
- Intermolecular forces
- Molecular solids

Memory Aids:
- Weak forces = Low heat of fusion

Quick Tips:
- Consider the type of bonding and intermolecular forces when determining the heat of fusion.

60 Medium

60 of 244Question permalink

Remembering +4 | -1

\mathrm{C}_{\mathrm{v}}

for monatomic gas is

View answer and explanation

Correct option: B

\frac{3}{2} \mathrm{R}

Option B (

\frac{3}{2} \mathrm{R}

) is correct because for a monatomic gas, the only degrees of freedom are translational, leading to this specific heat at constant volume.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2} \mathrm{R}$ . The specific heat at constant volume for a monatomic gas is directly related to its translational degrees of freedom.

Step-by-Step Solution:

Step 1: Identify the type of gas: Monatomic gas.

Step 2: Recall the degrees of freedom for a monatomic gas. A monatomic gas has 3 translational degrees of freedom.

Step 3: Use the equipartition theorem to find the average energy per degree of freedom: $\frac{1}{2} k_B T$ per molecule, or $\frac{1}{2} R T$ per mole.

Step 4: Calculate the internal energy (U) of one mole of monatomic gas: $U = 3 \times \frac{1}{2} R T = \frac{3}{2} R T$

Step 5: Determine the specific heat at constant volume ( $C_v$ ) using the relation: $C_v = \frac{dU}{dT}$

Step 6: Differentiate the internal energy with respect to temperature: $C_v = \frac{d}{dT} (\frac{3}{2} R T) = \frac{3}{2} R$

Step 7: Final Answer: $C_v = \frac{3}{2} R$

Why other options are wrong:
- Option A: $\frac{7}{2} \mathrm{R}$ is the $C_p$ for a diatomic gas, not a monatomic gas.
- Option C: $\frac{1}{2} \mathrm{R}$ is the energy associated with one degree of freedom, but not the total $C_v$ for a monatomic gas.
- Option D: $\frac{5}{2} \mathrm{R}$ is the $C_v$ for a diatomic gas, not a monatomic gas.

Key Concepts:
- Degrees of freedom
- Equipartition theorem
- Specific heat at constant volume ( $C_v$ )
- Internal energy of an ideal gas

Memory Aids:
- Monatomic: 3 translational degrees of freedom.
- Diatomic: 3 translational + 2 rotational degrees of freedom (at room temperature).

Quick Tips:
- Remember the degrees of freedom for different types of gases.
- Use the equipartition theorem to find the internal energy.
- Differentiate internal energy with respect to temperature to find $C_v$ .

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