Identify the correct statement for change of Gibb’s energy for a system ($\Delta {G_{{\rm{system}}}}$) at constant temperature and pressure:

If $\Delta {G_{{\rm{system}}}}$=0, the system has attained equilibrium

If $\Delta {G_{{\rm{system}}}}$=0, the system is still moving in a particular direction

If $\Delta {G_{{\rm{system}}}}$=-ve, the process is not spontaneous

If $\Delta {G_{{\rm{system}}}}$=+ve, the process is spontaneous

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium, Page 5

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Application +4 | -1

The heat change taking place during the reaction

\;{\mathrm{H}_2}\mathrm{O}\left( l \right)

\to {\mathrm{H}_2}\mathrm{O(g)}

is: [Given,

\Delta {H_f}

of

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

= -57 kcal,

\Delta {H_f} = {\mathrm{H}_2}\mathrm{O}\left( l \right)

= -68.3 kcal]

View answer and explanation

Correct option: A

+ 11.3\mathrm{kcal}

Option A (+11.3 kcal) is correct. The heat change is the enthalpy of vaporization, calculated as the difference between the enthalpy of formation of gaseous water and liquid water.

Solution approach

Correct Answer + Core Reason: $+ 11.3\mathrm{kcal}$ . The heat change is the enthalpy of vaporization, calculated as the difference between the enthalpy of formation of gaseous water and liquid water.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H_f({\mathrm{H}_2}\mathrm{O}(g)) = -57 \text{ kcal}$ , $\Delta H_f({\mathrm{H}_2}\mathrm{O}(l)) = -68.3 \text{ kcal}$ . The reaction is ${\mathrm{H}_2}\mathrm{O}\left( l \right) \to {\mathrm{H}_2}\mathrm{O(g)}$ .

Step 2: Recall that the enthalpy change for the reaction is given by $\Delta H = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants})$ .

Step 3: Apply the formula: $\Delta H = \Delta H_f({\mathrm{H}_2}\mathrm{O}(g)) - \Delta H_f({\mathrm{H}_2}\mathrm{O}(l)) = -57 \text{ kcal} - (-68.3 \text{ kcal})$ .

Step 4: Calculate the enthalpy change: $\Delta H = -57 + 68.3 = 11.3 \text{ kcal}$ .

Why other options are wrong:
- Option B: This option has the wrong sign. The enthalpy of vaporization is positive because energy is required to convert liquid to gas.
- Option C: This option is incorrect. It seems to be adding the enthalpies of formation instead of subtracting.
- Option D: This option is incorrect. It seems to be adding the enthalpies of formation instead of subtracting and also has the wrong sign.

Key Concepts:
- Enthalpy of formation: The enthalpy change when one mole of a substance is formed from its elements in their standard states.
- Enthalpy change of reaction: $\Delta H = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants})$ .

Memory Aids:
- Remember that enthalpy of vaporization is the energy required to convert a liquid to a gas.

Quick Tips:
- Always remember to subtract the enthalpies of formation of reactants from the enthalpies of formation of products.

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Application +4 | -1

How much heat (in kJ) is evolved when 2.7 g of aluminium reacts with a stoichiometric amount of

\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}

?

2\mathrm{A}{\mathrm{l}_2} + \mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} \to 2\mathrm{Fe} + \mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3};\Delta {H^o} = - 852\,\mathrm{kJ}

View answer and explanation

Correct option: C

- 42.6\,\mathrm{kJ}

Option C (-42.6 kJ) is correct because it is calculated based on the stoichiometry of the reaction and the given enthalpy change.

Solution approach

Correct Answer + Core Reason: $- 42.6\,\mathrm{kJ}$ . The heat evolved is calculated based on the moles of aluminum reacting and the given enthalpy change for the reaction.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2\mathrm{Al} + \mathrm{Fe}_2\mathrm{O}_3 \to 2\mathrm{Fe} + \mathrm{Al}_2\mathrm{O}_3;\Delta {H^o} = - 852\,\mathrm{kJ}$

Step 2: Calculate the moles of aluminum: Molar mass of Al = 27 g/mol. Moles of Al = $\frac{2.7 \text{{ g}}}{27 \text{{ g/mol}}} = 0.1 \text{{ mol}}$

Step 3: Determine the heat evolved. From the balanced equation, 2 moles of Al react to evolve 852 kJ of heat. Therefore, 0.1 moles of Al will evolve:
$\frac{0.1 \text{{ mol Al}}}{2 \text{{ mol Al}}} \times (-852 \text{{ kJ}}) = -42.6 \text{{ kJ}}$

Step 4: Final Answer: $- 42.6\,\mathrm{kJ}$

Why other options are wrong:
- Option A (852 kJ): This is the enthalpy change for 2 moles of Al, not 2.7 g.
- Option B (426 kJ): This is the enthalpy change for 1 mole of Al, not 2.7 g.
- Option D (+ 42.6 kJ): The reaction evolves heat, so the sign should be negative.

Key Concepts:
- Stoichiometry
- Enthalpy Change

Memory Aids:
- Remember to use the correct stoichiometric coefficients from the balanced equation.

Quick Tips:
- Always convert mass to moles before using enthalpy changes.
- Pay attention to the sign of the enthalpy change (negative for exothermic, positive for endothermic).

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Application +4 | -1

A solution of 500 mL of 2M KOH is added to 500 mL of 2M HCl and the mixture is well shaken. The rise in temperature

\;{T_1}

is noted. The experiment is then repeated using 250 mL of each solution and rise in temperature

\;{T_2}

is again noted. Assume all heat is taken up by the solution:

View answer and explanation

Correct option: A

\;{T_1}

=

\;{T_2}

Option A (T1=T2) is correct because the heat released per mole of reaction is the same in both cases, and the total volume of solution is also the same, leading to the same temperature rise.

Solution approach

Correct Answer + Core Reason: $\;{T_1}$ = $\;{T_2}$ . The temperature rise depends on the heat released per unit volume, which is the same in both cases.

Step-by-Step Solution:

Step 1: Analyze the reaction:
$KOH + HCl \rightarrow KCl + H_2O$ \nThe reaction is exothermic.

Step 2: Calculate the moles of reactants in the first experiment:\nMoles of KOH = $2 \text{{ M}} \times 0.5 \text{{ L}} = 1 \text{{ mole}}$ \nMoles of HCl = $2 \text{{ M}} \times 0.5 \text{{ L}} = 1 \text{{ mole}}$ \nSince the moles of both reactants are equal, the reaction goes to completion, and 1 mole of heat is released.

Step 3: Calculate the moles of reactants in the second experiment:\nMoles of KOH = $2 \text{{ M}} \times 0.25 \text{{ L}} = 0.5 \text{{ mole}}$ \nMoles of HCl = $2 \text{{ M}} \times 0.25 \text{{ L}} = 0.5 \text{{ mole}}$ \nSince the moles of both reactants are equal, the reaction goes to completion, and 0.5 mole of heat is released.

Step 4: Calculate the total volume of the solution in both experiments:\nExperiment 1: $500 \text{{ mL}} + 500 \text{{ mL}} = 1000 \text{{ mL}} = 1 \text{{ L}}$ \nExperiment 2: $250 \text{{ mL}} + 250 \text{{ mL}} = 500 \text{{ mL}} = 0.5 \text{{ L}}$

Step 5: Calculate the heat released per unit volume:\nExperiment 1: $1 \text{{ mole}} / 1 \text{{ L}} = 1 \text{{ mole/L}}$ \nExperiment 2: $0.5 \text{{ mole}} / 0.5 \text{{ L}} = 1 \text{{ mole/L}}$ \nSince the heat released per unit volume is the same in both experiments, the temperature rise will be the same.

Why other options are wrong:
- Option B: $\;{T_1}$ is 2 times as large as $\;{T_2}$ : Incorrect because the heat released per unit volume is the same.
- Option C: $\;{T_2}$ is twice larger as $\;{T_1}$ : Incorrect because the heat released per unit volume is the same.
- Option D: $\;{T_1}$ is 4 times as large as $\;{T_2}$ : Incorrect because the heat released per unit volume is the same.

Key Concepts:
- Neutralization reaction
- Heat of reaction
- Molarity

Memory Aids:
- Molarity = Moles / Volume

Quick Tips:
- Pay attention to the limiting reactant in neutralization reactions.
- Temperature change is proportional to the heat released per unit volume.

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Application +4 | -1

Heat of combustion ∆H for C(s),

{\mathrm{H}_2}

(g) and

\mathrm{C}{\mathrm{H}_4}

(g) are -94,-68 and -213 kcal/mole then ∆H for

\mathrm{C}\left( s \right) + 2{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)\mathrm{i}s:

View answer and explanation

Correct option: A

-17 kcal

Option A (-17 kcal) is correct because it is calculated using Hess's law by manipulating the given heats of combustion.

Solution approach

Correct Answer + Core Reason: -17 kcal. The enthalpy change is calculated using Hess's Law by summing the heats of formation of the products minus the reactants.

Step-by-Step Solution:

Step 1: Write the target reaction: $\mathrm{C}\left( s \right) + 2{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)$

Step 2: Write the combustion reactions and their enthalpy changes:
- $\mathrm{C}(s) + {O_2}(g) \to {CO_2}(g); \Delta H = -94 \text{{ kcal/mole}}$
- ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l); \Delta H = -68 \text{{ kcal/mole}}$
- ${CH_4}(g) + 2{O_2}(g) \to {CO_2}(g) + 2{H_2}O(l); \Delta H = -213 \text{{ kcal/mole}}$

Step 3: Apply Hess's Law. We need to form the target reaction from the given combustion reactions. Multiply the second equation by 2 and reverse the third equation:
- $\mathrm{C}(s) + {O_2}(g) \to {CO_2}(g); \Delta H = -94 \text{{ kcal/mole}}$
- $2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l); \Delta H = 2 \times (-68) = -136 \text{{ kcal/mole}}$
- ${CO_2}(g) + 2{H_2}O(l) \to {CH_4}(g) + 2{O_2}(g); \Delta H = +213 \text{{ kcal/mole}}$

Step 4: Add the three equations:
$\mathrm{C}(s) + 2{H_2}(g) \to {CH_4}(g); \Delta H = -94 - 136 + 213 = -17 \text{{ kcal/mole}}$

Step 5: Final Answer: -17 kcal

Why other options are wrong:
- Option B (-111 kcal): This is incorrect because it likely involves adding the heats of combustion of C and H2 without subtracting the heat of combustion of CH4 or without proper stoichiometric coefficients.
- Option C (-170 kcal): This is incorrect and doesn't follow Hess's law correctly.
- Option D (-85 kcal): This is incorrect and doesn't follow Hess's law correctly.

Key Concepts:
- Hess's Law
- Heat of Combustion
- Enthalpy Change

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

Quick Tips:
- Always write out the target reaction and the given reactions clearly.
- Manipulate the given reactions to match the target reaction using Hess's Law.
- Remember to reverse the sign of Δ H when reversing a reaction.

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Understanding +4 | -1

Molar heat capacity is given by :

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius, which is mathematically represented as

C = \frac{dQ}{n dT}

, where n is the number of moles.

Solution approach

Correct Answer + Core Reason: None of these. Molar heat capacity is defined as $C = \frac{dQ}{n dT}$ , where n is the number of moles.

Step-by-Step Solution:

Step 1: Define molar heat capacity: Molar heat capacity (C) is the amount of heat (dQ) required to raise the temperature of one mole (n) of a substance by one degree Celsius (dT).

Step 2: Mathematical representation: The formula for molar heat capacity is $C = \frac{dQ}{n dT}$ .

Step 3: Analyze the options:
- A. $\frac{{dQ}}{{dT}}$ : This represents heat capacity, not molar heat capacity. It's missing the division by the number of moles (n).
- B. $dQ \times dT$ : This is incorrect; it's the product of heat and temperature change, which has no physical meaning in this context.
- C. $\sum Q\frac{1}{{dT}}$ : This is also incorrect; it represents a summation of heat divided by temperature change, which is not the definition of molar heat capacity.

Why other options are wrong:
- Option A: $\frac{{dQ}}{{dT}}$ . This is heat capacity, not molar heat capacity.
- Option B: $dQ \times dT$ . This is not a valid representation of molar heat capacity.
- Option C: $\sum Q\frac{1}{{dT}}$ . This is not a valid representation of molar heat capacity.

Key Concepts:
- Molar Heat Capacity
- Heat Capacity

Memory Aids:
- Remember to include the number of moles (n) in the denominator when calculating molar heat capacity.

Quick Tips:
- Pay attention to the units and definitions of heat capacity and molar heat capacity.

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Understanding +4 | -1

The heat of formation is the change in enthalpy accompanying the formation of a substance from its elements at 298 K and 1 atm pressure. Since, the enthalpies of elements in their most stable state are taken to be zero, the heat of formation of compounds is :

View answer and explanation

Correct option: C

Standard heat enthalpy of that compound

Option C (Standard heat enthalpy of that compound) is correct because the heat of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Solution approach

Correct Answer + Core Reason: Standard heat enthalpy of that compound. The heat of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Step-by-Step Solution:

Step 1: Recall the definition of heat of formation.

Step 2: The heat of formation is the enthalpy change ( $\Delta H$ ) when one mole of a compound is formed from its elements in their standard states at 298 K and 1 atm.

Step 3: Since the enthalpies of elements in their most stable state are taken to be zero, the heat of formation of compounds is equal to the standard heat enthalpy of that compound.

Why other options are wrong:
- Option A: Always negative is incorrect because the heat of formation can be positive or negative depending on the compound.
- Option B: Always positive is incorrect because the heat of formation can be positive or negative depending on the compound.
- Option D: Zero is incorrect because the heat of formation is zero only for elements in their standard states.

Key Concepts:
- Heat of Formation
- Enthalpy
- Standard State

Memory Aids:
- Heat of formation = Enthalpy change of formation.

Quick Tips:
- Remember that the heat of formation is zero for elements in their standard states.

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Understanding +4 | -1

To calculate the amount of work done in joules during reversible isothermal expansion of an ideal gas, the volume must be expressed in

View answer and explanation

Correct option: D

Any of these

Option D (Any of these) is correct because the work done calculation depends on the change in volume, and any consistent unit can be used as long as the gas constant is adjusted accordingly.

Solution approach

Correct Answer + Core Reason: Any of these. The volume can be expressed in any unit as long as the gas constant is consistent.

Step-by-Step Solution:

Step 1: The work done during reversible isothermal expansion is given by $w = -nRT \ln{\frac{{V_2}}{{V_1}}}$ .

Step 2: The ratio of volumes $\frac{{V_2}}{{V_1}}$ is dimensionless. Therefore, the units of volume cancel out as long as they are consistent.

Step 3: The gas constant R must be in units consistent with the volume units used.

Step 4: Therefore, ${\mathrm{m}^3}$ , $\mathrm{d}{\mathrm{m}^3}$ , or $\mathrm{c}{\mathrm{m}^3}$ can be used, provided R is adjusted accordingly.

Why other options are wrong:
- Option A: ${\mathrm{m}^3}$ only: This is incorrect because other units can be used if R is adjusted.
- Option B: $\mathrm{d}{\mathrm{m}^3}$ only: This is incorrect because other units can be used if R is adjusted.
- Option C: $\mathrm{c}{\mathrm{m}^3}$ only: This is incorrect because other units can be used if R is adjusted.

Key Concepts:
- Reversible isothermal expansion
- Work done calculation
- Gas constant

Memory Aids:
- Remember the formula for work done in isothermal reversible expansion.

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

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Understanding +4 | -1

Identify the correct statement for change of Gibb’s energy for a system (

\Delta {G_{\mathrm{system}}}

) at constant temperature and pressure:

View answer and explanation

Correct option: D

If

\Delta {G_{\mathrm{system}}}

=0, the system has attained equilibrium

Option D (If

\Delta {G_{\mathrm{system}}}

=0, the system has attained equilibrium) is correct because at equilibrium, the change in Gibbs free energy is zero.

Solution approach

Correct Answer + Core Reason: If $\Delta {G_{\mathrm{system}}}$ =0, the system has attained equilibrium. At equilibrium, the change in Gibbs free energy is zero.

Step-by-Step Solution:

Step 1: Recall the definition of Gibbs free energy and its relation to spontaneity.

Step 2: Remember that at constant temperature and pressure:
- If $\Delta G < 0$ , the process is spontaneous.
- If $\Delta G > 0$ , the process is non-spontaneous.
- If $\Delta G = 0$ , the system is at equilibrium.

Step 3: Evaluate each option:
- Option A: If $\Delta {G_{\mathrm{system}}}$ =0, the system is still moving in a particular direction. Incorrect, it's at equilibrium.
- Option B: If $\Delta {G_{\mathrm{system}}}$ =-ve, the process is not spontaneous. Incorrect, it is spontaneous.
- Option C: If $\Delta {G_{\mathrm{system}}}$ =+ve, the process is spontaneous. Incorrect, it is non-spontaneous.
- Option D: If $\Delta {G_{\mathrm{system}}}$ =0, the system has attained equilibrium. Correct.

Why other options are wrong:
- Option A: At equilibrium, there is no net change, so the system is not moving in a particular direction.
- Option B: A negative change in Gibbs free energy indicates a spontaneous process.
- Option C: A positive change in Gibbs free energy indicates a non-spontaneous process.
- Option D: Correct, as explained above.

Key Concepts:
- Gibbs free energy, Spontaneity, Equilibrium

Memory Aids:
- Remember the sign convention for Gibbs free energy: negative for spontaneous, positive for non-spontaneous, and zero for equilibrium.

Quick Tips:
- Always consider the sign of $\Delta G$ to determine the spontaneity of a process at constant temperature and pressure.

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Understanding +4 | -1

Heat of neutralisation of which acid-base reaction is - 57.32 kJ for?

View answer and explanation

Correct option: D

\mathrm{HN}{\mathrm{O}_3} + \mathrm{LiOH}

Option D (

\mathrm{HNO_3} + \mathrm{LiOH}

) is correct because it represents the neutralization of a strong acid (

\mathrm{HNO_3}

) with a strong base (

\mathrm{LiOH}

), which results in a heat of neutralization of -57.32 kJ.

Solution approach

Correct Answer + Core Reason: $\mathrm{HN}{\mathrm{O}_3} + \mathrm{LiOH}$ . Strong acid-strong base neutralization releases approximately -57.32 kJ/mol.

Step-by-Step Solution:

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

Step 2: Identify strong acids and strong bases:
- Strong acids: HCl, H2SO4, HNO3, etc.
- Strong bases: NaOH, KOH, LiOH, etc.

Step 3: Analyze the given options:
- A: CH3COOH (weak acid) + NaOH (strong base)
- B: HCl (strong acid) + NH4OH (weak base)
- C: HCOOH (weak acid) + KOH (strong base)
- D: HNO3 (strong acid) + LiOH (strong base)

Step 4: Conclusion: Only option D involves the reaction of a strong acid and a strong base.

Why other options are wrong:
- Option A: CH3COOH is a weak acid, so the heat of neutralization will be less negative than -57.32 kJ/mol due to the energy required to ionize the weak acid.
- Option B: NH4OH is a weak base, so the heat of neutralization will be less negative than -57.32 kJ/mol due to the energy required to ionize the weak base.
- Option C: HCOOH is a weak acid, so the heat of neutralization will be less negative than -57.32 kJ/mol due to the energy required to ionize the weak acid.

Key Concepts:
- Heat of neutralization
- Strong acids and bases
- Weak acids and bases

Memory Aids:
- Strong acid + Strong base \u2192 -57.32 kJ/mol

Quick Tips:
- Identify strong acids and bases to quickly determine the correct answer.

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Analyzing +4 | -1

Given that

\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2: \Delta \mathrm{H}^{\circ}=-\mathrm{x} \mathrm{kJ}

2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2: \Delta \mathrm{H}^{\circ}=-\mathrm{y} \mathrm{kJ}

the enthalpy of formation of carbon monoxide will be

View answer and explanation

Correct option: B

\frac{y-2 x}{2}

Option B (

\frac{{y-2x}}{{2}}

) is correct because it is derived from Hess's Law by manipulating the given equations to obtain the formation equation for CO.

Solution approach

Correct Answer + Core Reason: $\frac{{y-2 x}}{{2}}$ . The enthalpy of formation of carbon monoxide is calculated using Hess's Law by manipulating the given reactions.

Step-by-Step Solution:

Step 1: Write the target equation for the formation of carbon monoxide:
$\mathrm{{C(s) + \frac{{1}}{{2}}O_2(g) \rightarrow CO(g) : \Delta H_f^\circ = ?}}$

Step 2: Write the given equations:
(1) $\mathrm{{C(s) + O_2(g) \rightarrow CO_2(g) : \Delta H^\circ = -x kJ}}$
(2) $\mathrm{{2CO(g) + O_2(g) \rightarrow 2CO_2(g) : \Delta H^\circ = -y kJ}}$

Step 3: Reverse equation (2) and divide by 2:
$\mathrm{{CO_2(g) \rightarrow CO(g) + \frac{{1}}{{2}}O_2(g) : \Delta H^\circ = \frac{{y}}{{2}} kJ}}$

Step 4: Add equation (1) and the modified equation (2):
$\mathrm{{C(s) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{{1}}{{2}}O_2(g)}}$

Step 5: Simplify the equation:
$\mathrm{{C(s) + \frac{{1}}{{2}}O_2(g) \rightarrow CO(g)}}$

Step 6: Calculate the enthalpy change for the formation of CO:
$\Delta H_f^\circ = -x + \frac{{y}}{{2}} = \frac{{y - 2x}}{{2}}$

Why other options are wrong:
- Option A: $\frac{{2x-y}}{{2}}$ . This is incorrect because the equations are not manipulated correctly. The sign of y is incorrect.
- Option C: $2x-y$ . This is incorrect because the equations are not manipulated correctly, and the division by 2 is missing.
- Option D: $y-2x$ . This is incorrect because the division by 2 is missing.

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

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

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

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Understanding +4 | -1

Absorption of gasses on solid surface is generally exothermic because

View answer and explanation

Correct option: B

Entropy decreases

Option B (Entropy decreases) is correct because when gases are adsorbed onto a solid surface, their freedom of movement is restricted, leading to a decrease in entropy.

Solution approach

Correct Answer + Core Reason: Entropy decreases. Adsorption restricts the movement of gas molecules, leading to a decrease in entropy.

Step-by-Step Solution:

Step 1: Adsorption: Adsorption is the adhesion of atoms, ions, or molecules from a gas, liquid, or dissolved solid to a surface.

Step 2: Entropy of Gases: Gases have high entropy due to the large freedom of movement of their molecules.

Step 3: Entropy Change during Adsorption: When a gas is adsorbed onto a solid surface, the gas molecules lose some of their freedom of movement. They are now confined to the surface, which is a more ordered state compared to the gaseous state. Therefore, the entropy decreases during adsorption.

Step 4: Exothermic Nature: Adsorption is generally exothermic because the decrease in entropy is compensated by the release of heat, making the Gibbs Free Energy change negative ( $\Delta G = \Delta H - T\Delta S$ ).

Step 5: Conclusion: Therefore, absorption of gases on a solid surface is generally exothermic because entropy decreases.

Why other options are wrong:
- Option A: Enthalpy is positive: Adsorption is generally exothermic, meaning enthalpy is negative, not positive.
- Option C: Entropy increases: Entropy decreases during adsorption because the gas molecules become more ordered on the surface.
- Option D: Free energy increases: For a spontaneous process like adsorption, the free energy must decrease, not increase.

Key Concepts:
- Adsorption
- Entropy (S)
- Enthalpy (H)
- Gibbs Free Energy (G)

Memory Aids:
- Adsorption = Order = Decreased Entropy

Quick Tips:
- Remember that adsorption is generally exothermic and involves a decrease in entropy.

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Understanding +4 | -1

The work done during the process when 1 mole of gas is allowed to expand freely into vacuum is:

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because when a gas expands freely into a vacuum, the external pressure is zero, resulting in no work being done.

Solution approach

Correct Answer + Core Reason: Zero. When a gas expands freely into a vacuum, no work is done.

Step-by-Step Solution:

Step 1: Define Work Done: Work done (W) is given by $W = -P_{ext} \Delta V$ , where $P_{ext}$ is the external pressure and $\Delta V$ is the change in volume.

Step 2: Free Expansion into Vacuum: In free expansion into a vacuum, the external pressure $P_{ext} = 0$ .

Step 3: Calculate Work Done: $W = -0 \times \Delta V = 0$

Step 4: Conclusion: The work done is zero.

Why other options are wrong:
- Option B: +ve work would imply compression, which is not happening in free expansion into a vacuum.
- Option C: -ve work would imply expansion against an external pressure, which is not the case here.
- Option D: Either of these is incorrect because the work done is definitively zero.

Key Concepts:
- Work Done
- Free Expansion
- Vacuum

Memory Aids:
- Free expansion into vacuum: W = 0

Quick Tips:
- Always remember that work done is zero in free expansion, regardless of the gas or volume change.

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Remembering +4 | -1

Which one among the following is the correct expression for isothermal reversible change?

View answer and explanation

Correct option: B

w = - 2.303\,\,nRT\,\,\log \frac{{{V_f}}}{{{V_i}}}

Option B is correct because the correct expression for isothermal reversible work is w = -2.303nRTlog(Vf/Vi).

Solution approach

Correct Answer + Core Reason: $w = - 2.303\,\,nRT\,\,\log \frac{{{V_f}}}{{{V_i}}}$ . This is the correct formula for isothermal reversible work.

Step-by-Step Solution:

Step 1: Isothermal Process: A process that occurs at constant temperature.

Step 2: Reversible Process: A process that can be reversed without any net change in the system or surroundings.

Step 3: Work done in an isothermal reversible process: $w = - \int_{V_i}^{V_f} P dV$

Step 4: Ideal Gas Law: $PV = nRT$ , so $P = \frac{nRT}{V}$

Step 5: Substitute into the work equation: $w = - \int_{V_i}^{V_f} \frac{nRT}{V} dV = -nRT \int_{V_i}^{V_f} \frac{dV}{V}$

Step 6: Integrate: $w = -nRT \ln \frac{V_f}{V_i}$

Step 7: Convert natural logarithm to base-10 logarithm: $w = -2.303 nRT \log \frac{V_f}{V_i}$

Step 8: Conclusion: The correct expression for isothermal reversible change is $w = - 2.303\,\,nRT\,\,\log \frac{{{V_f}}}{{{V_i}}}$ .

Why other options are wrong:
- Option A: Incorrect; the temperature T is missing.
- Option C: Incorrect; the volume ratio is inverted.
- Option D: Incorrect; the factor 2.303 is missing.

Key Concepts:
- Isothermal Process
- Reversible Process
- Work Done

Memory Aids:
- Isothermal: Constant Temperature
- $w = -nRT \ln \frac{V_f}{V_i}$ (remember the formula)

Quick Tips:
- Remember to convert between natural and base-10 logarithms when necessary.

134 Medium

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Analyzing +4 | -1

\Delta {S^ \circ }

will be highest for the reaction:

View answer and explanation

Correct option: B

CaC{O_3}(s) \to CaO(s) + C{O_2}(g)

Option B (

CaC{O_3}(s) \to CaO(s) + C{O_2}(g)

) is correct because it involves the decomposition of a solid into a solid and a gas, leading to a significant increase in entropy.

Solution approach

Correct Answer + Core Reason: $CaC{O_3}(s) \to CaO(s) + C{O_2}(g)$ . This reaction has the highest $\Delta S^\circ$ because it involves the decomposition of a solid into a solid and a gas, resulting in a significant increase in disorder.

Step-by-Step Solution:
Step 1: Analyze each reaction:
A. $Ca(s) + \frac{1}{2}{O_2}(g) \to CaO(s)$ : Decrease in the number of moles of gas, so $\Delta S$ is negative.
B. $CaC{O_3}(s) \to CaO(s) + C{O_2}(g)$ : Formation of a gas from a solid, so $\Delta S$ is positive and large.
C. $C(s) + {O_2}(g) \to C{O_2}(g)$ : Number of moles of gas remains the same, so $\Delta S$ is small.
D. ${N_2}(g) + {O_2}(g) \to 2NO(g)\,$ : Number of moles of gas remains the same, so $\Delta S$ is small.
Step 2: Compare the entropy changes: The reaction that produces a gas from a solid will have the largest positive entropy change.

Why other options are wrong:
- Option A: $Ca(s) + \frac{1}{2}{O_2}(g) \to CaO(s)$ . This is incorrect because the number of moles of gas decreases, leading to a negative entropy change.
- Option C: $C(s) + {O_2}(g) \to C{O_2}(g)$ . This is incorrect because the number of moles of gas remains the same, so the entropy change is small.
- Option D: ${N_2}(g) + {O_2}(g) \to 2NO(g)\,$ . This is incorrect because the number of moles of gas remains the same, so the entropy change is small.

Key Concepts:
- Entropy Change
- Phase Transitions
- Number of Moles

Memory Aids:
- Remember that the formation of a gas increases entropy, while the disappearance of a gas decreases entropy.

Quick Tips:
- Look for reactions that involve a change in the number of moles of gas.
- Reactions that produce gases from solids or liquids tend to have large positive entropy changes.

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Remembering +4 | -1

Entropy change of fusion at constant pressure is given by:

View answer and explanation

Correct option: A

\Delta {S_{\left( f \right)}} = \frac{{\Delta {H_f}}}{T}

Option A (

\Delta {S_{\left( f \right)}} = \frac{{\Delta {H_f}}}{T}

) is correct because the entropy change of fusion at constant pressure is defined as the enthalpy of fusion divided by the temperature at the melting point.

Solution approach

Correct Answer + Core Reason: $\Delta {S_{\left( f \right)}} = \frac{{\Delta {H_f}}}{T}$ . This formula directly relates entropy change during fusion to the enthalpy of fusion and temperature.

Step-by-Step Solution:

Step 1: Recall the definition of entropy change during a phase transition.

Step 2: At constant pressure, the entropy change ( $\Delta S$ ) is related to the enthalpy change ( $\Delta H$ ) and the temperature (T) by the equation: $\Delta S = \frac{{\Delta H}}{T}$

Step 3: For fusion (melting), the entropy change is given by: $\Delta {S_{\left( f \right)}} = \frac{{\Delta {H_f}}}{T}$

Why other options are wrong:
- Option B: This is incorrect because it uses Gibbs free energy change instead of enthalpy change.
- Option C: This is incorrect because it divides by the change in temperature, not the temperature itself.
- Option D: This is incorrect because option A is the correct formula.

Key Concepts:
- Entropy change during phase transition
- Enthalpy of fusion

Memory Aids:
- $\Delta S = \frac{{\Delta H}}{T}$ (for phase transitions)

Quick Tips:
- Remember that this formula applies specifically to phase transitions at constant pressure.

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Remembering +4 | -1

The enthalpy change accompanying a reaction
is called

View answer and explanation

Correct option: A

reaction enthalpy

Option A (reaction enthalpy) is correct because the enthalpy change accompanying a reaction is specifically termed the reaction enthalpy.

Solution approach

Correct Answer + Core Reason: reaction enthalpy. The enthalpy change accompanying a reaction is called reaction enthalpy.

Step-by-Step Solution:

Step 1: Define enthalpy change. Enthalpy change is the heat absorbed or released during a chemical reaction at constant pressure.

Step 2: Define reaction enthalpy. Reaction enthalpy is the specific term for the enthalpy change accompanying a reaction.

Step 3: Conclusion: Therefore, the enthalpy change accompanying a reaction is called reaction enthalpy.

Why other options are wrong:
- Option B: standard enthalpy: Standard enthalpy refers to the enthalpy change under standard conditions (298 K and 1 atm).
- Option C: thermochemical equation: A thermochemical equation is a chemical equation that includes the enthalpy change.
- Option D: enthalpy of atomization: Enthalpy of atomization is the enthalpy change when one mole of a substance is completely converted into gaseous atoms.

Key Concepts:
- Enthalpy change
- Reaction enthalpy

Memory Aids:
- Reaction enthalpy is specific to a chemical reaction.

Quick Tips:
- Understand the different types of enthalpy changes and their definitions.

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Application +4 | -1

The enthalpy and entropy change for the reaction,

\mathrm{B}{\mathrm{r}_2}\left( l \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{BrCl}\left( \mathrm{g} \right)

Are 30 kJ

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

and

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

respectively. The temperature at which the reaction will be in equilibrium is:

View answer and explanation

Correct option: C

285.7 K

Option C (285.7 K) is correct because the equilibrium temperature is calculated using T = ∆H/∆S.

Solution approach

Correct Answer + Core Reason: 285.7 K. Equilibrium temperature calculation.

Step-by-Step Solution:

Step 1: Given: $\Delta H = 30 \text{{ kJ/mol}} = 30000 \text{{ J/mol}}$ and $\Delta S = 105 \text{{ J/K/mol}}$ .

Step 2: At equilibrium, $\Delta G = 0$ , so $\Delta H = T\Delta S$ .

Step 3: Solving for T, we get $T = \frac{{\Delta H}}{{\Delta S}} = \frac{{30000}}{{105}} \approx 285.7 \text{{ K}}$ .

Why other options are wrong:
- Option A: 450 K: This is incorrect due to a calculation error.
- Option B: 300 K: This is incorrect due to a calculation error.
- Option D: 273 K: This is incorrect due to a calculation error.

Key Concepts:
- Gibbs free energy
- Enthalpy
- Entropy
- Equilibrium

Memory Aids:
- At equilibrium, $\Delta G = 0$ , so $\Delta H = T\Delta S$ .

Quick Tips:
- Remember the condition for equilibrium.

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Application +4 | -1

The value of

\Delta S

for freezing of 10g of

{H_2}O\left( \ell \right)

(enthalpy of fusion is 80 cal/g) at

0^\circ C

and 1 atm pressure is:

View answer and explanation

Correct option: D

-12.25 J/K

Option D (-12.25 J/K) is correct because the change in entropy during freezing is calculated using the formula

\Delta S = \frac{-Q}{T}

, where Q is the heat released during freezing and T is the temperature in Kelvin.

Solution approach

Correct Answer + Core Reason: -12.25 J/K. The change in entropy is calculated using $\Delta S = \frac{-Q}{T}$ .

Step-by-Step Solution:

Step 1: Calculate the heat released during freezing: The enthalpy of fusion is the heat absorbed during melting. The heat released during freezing is the negative of the enthalpy of fusion. So, $Q = -m \times \Delta H_{fusion}$ , where $m$ is the mass and $\Delta H_{fusion}$ is the enthalpy of fusion.

Step 2: Given values: $m = 10 \text{{ g}}$ and $\Delta H_{fusion} = 80 \text{{ cal/g}}$ .

Step 3: Calculate $Q$ : $Q = -10 \text{{ g}} \times 80 \text{{ cal/g}} = -800 \text{{ cal}}$

Step 4: Convert calories to joules: $Q = -800 \text{{ cal}} \times 4.184 \text{{ J/cal}} = -3347.2 \text{{ J}}$

Step 5: Convert temperature to Kelvin: $T = 0^\circ C + 273.15 = 273.15 \text{{ K}}$

Step 6: Calculate the change in entropy: $\Delta S = \frac{-Q}{T} = \frac{-(-3347.2 \text{{ J}})}{273.15 \text{{ K}}} = \frac{3347.2 \text{{ J}}}{273.15 \text{{ K}}} = -12.25 \text{{ J/K}}$ (Since it's freezing, entropy decreases, so we use the negative sign)

Why other options are wrong:
- Option A: 12.25 J/K. This is the magnitude of the entropy change but has the wrong sign. Freezing decreases entropy.
- Option B: -0.244J/K. This is incorrect; it likely results from a miscalculation or using incorrect units.
- Option C: -2.93 J/K. This is incorrect; it likely results from a miscalculation or using incorrect units.

Key Concepts:
- Entropy Change
- Enthalpy of Fusion
- Freezing Process

Memory Aids:
- $\Delta S = \frac{Q}{T}$
- Freezing is an exothermic process (heat is released), so Q is negative, and entropy decreases.

Quick Tips:
- Pay attention to the sign of the heat change (positive for endothermic, negative for exothermic).
- Always convert temperature to Kelvin for thermodynamic calculations.

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Application +4 | -1

The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ

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

. The heat released when 0.5 mole of

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

solution is mixed with 0.2 mole of KOH is

View answer and explanation

Correct option: B

11.4 kJ

Option B (11.4 kJ) is correct because the heat released is determined by the limiting reactant (KOH) 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 mixed with 0.2 mole of KOH.

Step-by-Step Solution:

Step 1: Identify the given information.
- Heat of neutralization of strong acid and strong base: 57.0 kJ/mol
- Moles of $HNO_3$ : 0.5 mol
- Moles of $KOH$ : 0.2 mol

Step 2: Determine the limiting reactant.
Since both $HNO_3$ and $KOH$ are strong, the reaction is:
$HNO_3(aq) + KOH(aq) \rightarrow KNO_3(aq) + H_2O(l)$
The stoichiometry is 1:1. Since we have 0.5 mol of $HNO_3$ and 0.2 mol of $KOH$ , $KOH$ 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 = 0.2 mol x 57.0 kJ/mol = 11.4 kJ

Step 4: Final Answer: 11.4 kJ

Why other options are wrong:
- Option A: 57.0 kJ is incorrect. It's the heat of neutralization per mole, but not the total heat released.
- Option C: 28.5 kJ is incorrect. It's likely a result of using 0.5 moles (of HNO3) instead of using the limiting reactant (0.2 moles of KOH) or some other calculation error.
- Option D: 34.9 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.

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Understanding +4 | -1

In an irreversible process taking place at constant T and P and in which only pressure volume work is being done, the change in Gibbs free energy (dG) and change in entropy

View answer and explanation

Correct option: B

{\left( {dS} \right)_{V,E}} > 0{\left( {dG} \right)_{T,P}} < 0

Option B (

{\left( {dS} \right)_{V,E}} > 0{\left( {dG} \right)_{T,P}} < 0

) is correct because, in an irreversible process at constant T and P, entropy increases and Gibbs free energy decreases.

Solution approach

Correct Answer + Core Reason: ${\left( {dS} \right)_{V,E}} > 0{\left( {dG} \right)_{T,P}} < 0$ . In an irreversible process at constant T and P, entropy increases and Gibbs free energy decreases.

Step-by-Step Solution:

Step 1: Define Irreversible Process: An irreversible process is a process that cannot be reversed without external work.

Step 2: Entropy Change: For an irreversible process, the entropy of the system increases, i.e., ${\left( {dS} \right)_{V,E}} > 0$ .

Step 3: Gibbs Free Energy Change: At constant T and P, the Gibbs free energy change for a spontaneous (irreversible) process is negative, i.e., ${\left( {dG} \right)_{T,P}} < 0$ .

Step 4: Conclusion: Therefore, ${\left( {dS} \right)_{V,E}} > 0$ and ${\left( {dG} \right)_{T,P}} < 0$ .

Why other options are wrong:
- Option A: ${\left( {dS} \right)_{V,E}} < 0{\left( {dG} \right)_{T,P}} < 0$ is incorrect because entropy increases in an irreversible process.
- Option C: ${\left( {dS} \right)_{V,E}} > 0{\left( {dG} \right)_{T,P}} = 0$ is incorrect because dG = 0 only for reversible processes at equilibrium.
- Option D: ${\left( {dS} \right)_{V,E}} = 0{\left( {dG} \right)_{T,P}} > 0$ is incorrect because entropy increases and Gibbs free energy decreases in an irreversible process.

Key Concepts:
- Irreversible Process
- Entropy
- Gibbs Free Energy
- Spontaneity

Memory Aids:
- Irreversible process: $\Delta S > 0$ , $\Delta G < 0$ (at constant T and P)

Quick Tips:
- Remember that for irreversible processes, entropy always increases, and Gibbs free energy decreases at constant T and P.

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Understanding +4 | -1

The process that is not endothermic in nature is

View answer and explanation

Correct option: C

H(g) + {e^ - } \to {H^ - }(g)

Option C (

H(g) + {e^ - } \to {H^ - }(g)

) is correct because the formation of

H^-

from

H

is exothermic due to high hydration energy.

Solution approach

Correct Answer + Core Reason: $H(g) + {e^ - } \to {H^ - }(g)$ . This process is exothermic because of the relatively high electron affinity of hydrogen.

Step-by-Step Solution:

Step 1: Define endothermic process: A process that requires energy input (absorbs heat).

Step 2: Define exothermic process: A process that releases energy (releases heat).

Step 3: Analyze the given options:
- A: ${O^ - }(g) + {e^ - } \to {O^{2 - }}(g)$ : Adding an electron to a negatively charged ion is generally endothermic due to electron-electron repulsion.
- B: $Na(g) \to N{a^ + }(g) + {e^ - }$ : Ionization energy is always endothermic.
- C: $H(g) + {e^ - } \to {H^ - }(g)$ : Electron affinity of hydrogen is exothermic.
- D: $Ar(g) + {e^ - } \to A{r^ - }(g)$ : Adding an electron to a noble gas is generally endothermic because they have stable electronic configurations.

Step 4: Conclusion: The only exothermic process is the formation of $H^-$ from $H$ .

Why other options are wrong:
- Option A: ${O^ - }(g) + {e^ - } \to {O^{2 - }}(g)$ : This is endothermic due to electron-electron repulsion.
- Option B: $Na(g) \to N{a^ + }(g) + {e^ - }$ : This is endothermic because it requires ionization energy.
- Option D: $Ar(g) + {e^ - } \to A{r^ - }(g)$ : This is endothermic because noble gases have stable configurations.

Key Concepts:
- Endothermic and Exothermic Processes
- Ionization Energy
- Electron Affinity
- Electronic Configuration

Memory Aids:
- Ionization is always endothermic.
- Adding electrons to negative ions is usually endothermic.

Quick Tips:
- Consider the electronic configurations and the charges of the species involved.

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Application +4 | -1

Calculate the free energy change of

2\mathrm{CuO}\left( s \right) \to \mathrm{C}{\mathrm{u}_2}\mathrm{O}\left( s \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right)

Given, ∆H=145.6kJ per mol ∆S=116 J per mol per K

View answer and explanation

Correct option: A

113.8 kJ per mol

Option A (113.8 kJ per mol) is correct because it is the result of calculating the Gibbs free energy change using the given enthalpy and entropy changes.

Solution approach

Correct Answer + Core Reason: 113.8 kJ per mol. The Gibbs free energy change is calculated using the formula $\Delta G = \Delta H - T\Delta S$ .

Step-by-Step Solution:

Step 1: Identify the given values: $\Delta H = 145.6 \text{ kJ/mol}$ , $\Delta S = 116 \text{ J/mol K}$ . Assume standard conditions, so $T = 298 \text{ K}$ .

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

Step 3: Convert $\Delta S$ to kJ/mol K: $\Delta S = 116 \text{ J/mol K} = 0.116 \text{ kJ/mol K}$ .

Step 4: Substitute the values into the equation: $\Delta G = 145.6 \text{ kJ/mol} - (298 \text{ K})(0.116 \text{ kJ/mol K})$ .

Step 5: Calculate: $\Delta G = 145.6 - 34.568 = 111.032 \text{ kJ/mol}$ .

Step 6: Round to one decimal place: $\Delta G \approx 111.0 \text{ kJ/mol}$ . Note: The provided correct answer is 113.8 kJ/mol, there might be an error in the provided answer or the question's data. Assuming the correct answer should be around 111.0 kJ/mol.

Step 7: Final Answer: 111.0 kJ/mol (approximately). Note: The provided correct answer is 113.8 kJ/mol, there might be an error in the provided answer or the question's data.

Why other options are wrong:
- Option B (221.5 kJ per mol): This is incorrect as it likely involves adding the enthalpy and entropy terms instead of subtracting them, or an error in calculation.
- Option C (55.4 kJ per mol): This is incorrect as it likely involves subtracting the enthalpy and entropy terms in the wrong order, or an error in calculation.
- Option D (145.6 kJ per mol): This is incorrect as it only considers the enthalpy change and neglects the entropy change.

Key Concepts:
- Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$
- Standard Conditions: $T = 298 \text{ K}$
- Enthalpy Change, Entropy Change

Memory Aids:
- G = H - TS (Give Help To Students)

Quick Tips:
- Always convert entropy to the same units as enthalpy before calculating Gibbs free energy.
- Remember to use the correct temperature (usually standard temperature 298 K unless otherwise specified).

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Analyzing +4 | -1

Consider the following reaction:

{C_6}{H_6}\left( l \right) + \frac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( g \right)

Sign of

\Delta H,\,\,\,\Delta G\,\,\,\& \,\,\,\Delta S

for the above reaction will be

View answer and explanation

Correct option: B

- ,\,\, - ,\,\, +

Option B ( - , - , + ) is correct because combustion reactions are exothermic (negative Δ H), spontaneous (negative Δ G), and increase disorder (positive Δ S).

Solution approach

Correct Answer + Core Reason: $- ,\,\, - ,\,\, +$ . Combustion reactions are exothermic, spontaneous, and increase disorder.

Step-by-Step Solution:
Step 1: Analyze the reaction: ${C_6}{H_6}\left( l \right) + \frac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( g \right)$ .
Step 2: Determine the sign of $\Delta H$ : Combustion reactions are exothermic, so $\Delta H$ is negative.
Step 3: Determine the sign of $\Delta G$ : Combustion reactions are spontaneous, so $\Delta G$ is negative.
Step 4: Determine the sign of $\Delta S$ : The number of moles of gas increases (15/2 to 6+3=9), so the disorder increases, and $\Delta S$ is positive.

Why other options are wrong:
- Option A: $+ ,\,\, - ,\,\, +$ . This is incorrect because $\Delta H$ is negative for combustion reactions.
- Option C: $- ,\,\, + ,\,\, +$ . This is incorrect because $\Delta G$ is negative for spontaneous reactions.
- Option D: $+ ,\,\, + ,\,\, -$ . This is incorrect because $\Delta H$ is negative and $\Delta S$ is positive.

Key Concepts:
- Enthalpy Change
- Gibbs Free Energy Change
- Entropy Change
- Combustion Reactions

Memory Aids:
- Remember that combustion reactions are exothermic and spontaneous.

Quick Tips:
- Know the sign conventions for $\Delta H$ , $\Delta G$ , and $\Delta S$ for different types of reactions.

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Remembering +4 | -1

The calorific value of fat is :

View answer and explanation

Correct option: D

More than that of carbohydrate and protein

Option D (More than that of carbohydrate and protein) is correct because fats have a higher calorific value due to their chemical structure and higher proportion of carbon-hydrogen bonds.

Solution approach

Correct Answer + Core Reason: More than that of carbohydrate and protein. Fats provide more energy per unit mass compared to carbohydrates and proteins.

Step-by-Step Solution:

Step 1: Recall the definition of calorific value.

Step 2: Calorific value is the amount of heat produced when one gram of a substance is completely burned.

Step 3: Fats have a higher proportion of carbon-hydrogen bonds, which release more energy upon oxidation compared to carbohydrates and proteins.

Step 4: Therefore, the calorific value of fat is higher than that of carbohydrate and protein.

Why other options are wrong:
- Option A: This is incorrect because fats have a higher calorific value.
- Option B: This is incorrect because fats have a higher calorific value than both protein and carbohydrates.
- Option C: This is incorrect because fats have a higher calorific value.

Key Concepts:
- Calorific value
- Energy content of biomolecules

Memory Aids:
- Fats = High energy

Quick Tips:
- Remember that fats are more energy-dense than carbohydrates and proteins.

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Understanding +4 | -1

Tea placed in thermos flask is an example of

View answer and explanation

Correct option: C

Isolated system

Option C (Isolated system) is correct because a thermos flask is designed to prevent the exchange of both heat and matter with the surroundings, approximating an isolated system.

Solution approach

Correct Answer + Core Reason: Isolated system. A thermos flask minimizes heat and matter exchange.

Step-by-Step Solution:

Step 1: Define an isolated system as one that exchanges neither energy nor matter with its surroundings.

Step 2: Consider the properties of a thermos flask: it is designed to minimize heat transfer (through insulation and vacuum) and prevent matter exchange (through a tight seal).

Step 3: Recognize that while not perfect, a thermos flask approximates an isolated system.

Why other options are wrong:
- Option A: Open system is incorrect because an open system exchanges both energy and matter.
- Option B: Close system is incorrect because a closed system exchanges energy but not matter.
- Option D: It can't act as system is incorrect because the tea in the thermos flask can be considered a system.

Key Concepts:
- Isolated system
- Open system
- Closed system

Memory Aids:
- Isolated = no exchange.

Quick Tips:
- Consider the definitions of open, closed, and isolated systems in real-world examples.

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Understanding +4 | -1

Which of the following is correct for an ideal gas :

View answer and explanation

Correct option: B

{\left( {\frac{{\partial E}}{{\partial P}}} \right)_T}

= 0

Option B (

{\left( {\frac{{\partial E}}{{\partial P}}} \right)_T}

= 0) is correct because for an ideal gas, internal energy (E) depends only on temperature (T), not on pressure (P).

Solution approach

Correct Answer + Core Reason: ${\left( {\frac{{\partial E}}{{\partial P}}} \right)_T}$ = 0. For an ideal gas, internal energy depends only on temperature.

Step-by-Step Solution:

Step 1: Define internal energy (E) for an ideal gas: The internal energy of an ideal gas is a function of temperature only. This means that at a constant temperature, the internal energy remains constant, regardless of changes in pressure or volume.

Step 2: Partial derivative representation: The partial derivative ${\left( {\frac{{\partial E}}{{\partial P}}} \right)_T}$ represents the change in internal energy (E) with respect to a change in pressure (P) at constant temperature (T).

Step 3: Apply to ideal gas: Since the internal energy of an ideal gas depends only on temperature, changing the pressure at constant temperature will not change the internal energy. Therefore, ${\left( {\frac{{\partial E}}{{\partial P}}} \right)_T} = 0$ .

Why other options are wrong:
- Option A: ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_V}$ = 0. This is incorrect. The internal energy of an ideal gas *does* change with temperature. ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_V} = nC_v$ , where $n$ is the number of moles and $C_v$ is the molar heat capacity at constant volume.
- Option C: $\;{\left( {\frac{{\partial E}}{{\partial T}}} \right)_P}$ = 0. This is also incorrect. Similar to option A, internal energy changes with temperature. ${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = nC_p$ , where $C_p$ is the molar heat capacity at constant pressure.
- Option D: All of these. Only option B is correct.

Key Concepts:
- Internal Energy of an Ideal Gas
- Partial Derivatives
- Ideal Gas Law

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

Quick Tips:
- Understand the relationship between internal energy, temperature, pressure, and volume for ideal gases.

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Analyzing +4 | -1

A: All reactions which are exothermic are spontaneous.
R: All reactions in which entropy increases are spontaneous.

View answer and explanation

Correct option: D

If both A and R are incorrect

Option D is correct because neither exothermic reactions are always spontaneous, nor reactions with increased entropy are always spontaneous; spontaneity depends on both enthalpy and entropy.

Solution approach

Correct Answer + Core Reason: If both A and R are incorrect. Spontaneity depends on both enthalpy and entropy, as described by Gibbs Free Energy.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): All reactions which are exothermic are spontaneous. This statement is incorrect. While exothermic reactions (negative Δ H) tend to be spontaneous, spontaneity also depends on the entropy change (Δ S) and temperature (T). The Gibbs Free Energy equation, Δ G = Δ H - TΔ S, determines spontaneity. A reaction is spontaneous only if Δ G < 0.

Step 2: Analyze Reason (R): All reactions in which entropy increases are spontaneous. This statement is also incorrect. While an increase in entropy (positive Δ S) favors spontaneity, the enthalpy change (Δ H) and temperature (T) also play a role. Again, the Gibbs Free Energy equation, Δ G = Δ H - TΔ S, determines spontaneity.

Step 3: Conclusion: Both the assertion and the reason are incorrect.

Why other options are wrong:
- Option A: Both A and R are not correct.
- Option B: Both A and R are not correct.
- Option C: A is not correct.

Key Concepts:
- Gibbs Free Energy (Δ G)
- Enthalpy (Δ H)
- Entropy (Δ S)
- Spontaneity

Memory Aids:
- Δ G = Δ H - TΔ S (Gibbs Free Energy equation).
- Spontaneous reactions have Δ G < 0.

Quick Tips:
- Remember that spontaneity depends on both enthalpy and entropy changes, not just one of them.

148 Medium

148 of 244Question permalink

Understanding +4 | -1

The heat evolved during neutralisation is maximum in the reaction of :

View answer and explanation

Correct option: D

\mathrm{NaOH}\,\,\mathrm{and}\,\,\mathrm{HCl}

Option D (

\mathrm{NaOH}\,\mathrm{and}\,\mathrm{HCl}

) is correct because the reaction between a strong acid (HCl) and a strong base (NaOH) releases the maximum amount of heat during neutralization.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaOH}\,\mathrm{and}\,\mathrm{HCl}$ . Strong acid-strong base reactions release the maximum heat during neutralization.

Step-by-Step Solution:

Step 1: Recall that the heat evolved during neutralization is maximum for strong acid-strong base reactions.

Step 2: Identify strong acids and bases:
- Strong acids: HCl, H2SO4, HNO3, etc.
- Strong bases: NaOH, KOH, LiOH, etc.

Step 3: Analyze the given options:
- A: NH4OH (weak base) and CH3COOH (weak acid)
- B: NH4OH (weak base) and HCl (strong acid)
- C: NaOH (strong base) and CH3COOH (weak acid)
- D: NaOH (strong base) and HCl (strong acid)

Step 4: Conclusion: Only option D involves the reaction of a strong acid and a strong base.

Why other options are wrong:
- Option A: Both NH4OH and CH3COOH are weak, so the heat evolved will be less due to incomplete ionization.
- Option B: NH4OH is weak, so the heat evolved will be less due to incomplete ionization.
- Option C: CH3COOH is weak, so the heat evolved will be less due to incomplete ionization.

Key Concepts:
- Heat of neutralization
- Strong acids and bases
- Weak acids and bases

Memory Aids:
- Strong + Strong = Maximum heat evolved

Quick Tips:
- Identify strong acids and bases to quickly determine the correct answer.

149 Medium

149 of 244Question permalink

Understanding +4 | -1

The numerical value of

{c_p} - {c_v}

is equal to:

View answer and explanation

Correct option: B

R/M

Option B (

R/M

) is correct because the difference between molar heat capacity at constant pressure and constant volume (

c_p - c_v

) for an ideal gas is equal to

R/M

, where R is the gas constant and M is the molar mass.

Solution approach

Correct Answer + Core Reason: $R/M$ . The difference between molar heat capacity at constant pressure and constant volume is equal to $R/M$ .

Step-by-Step Solution:

Step 1: Define molar heat capacities: $c_p$ is the molar heat capacity at constant pressure, and $c_v$ is the molar heat capacity at constant volume.

Step 2: Relationship between $c_p$ and $c_v$ : For an ideal gas, the relationship between $c_p$ and $c_v$ is given by $c_p - c_v = \frac{R}{M}$ , where $R$ is the gas constant and $M$ is the molar mass.

Why other options are wrong:
- Option A: $R$ . This is the difference between *molar* heat capacities, not specific heat capacities.
- Option C: $M/R$ . This is the inverse of the correct relationship.
- Option D: None of these. Option B is correct.

Key Concepts:
- Molar Heat Capacity at Constant Pressure ( $c_p$ )
- Molar Heat Capacity at Constant Volume ( $c_v$ )
- Ideal Gas Law

Memory Aids:
- Remember that $c_p$ is always greater than $c_v$ for gases.

Quick Tips:
- Pay attention to whether the question is asking about molar or specific heat capacities.

150 Medium

150 of 244Question permalink

Understanding +4 | -1

A reaction accompanied with the absorption of energy is :

View answer and explanation

Correct option: C

Electrolysis of water

The correct answer is C (Electrolysis of water) because electrolysis requires an external energy source to drive the non-spontaneous decomposition of water.

Solution approach

Correct Answer + Core Reason: Electrolysis of water. Electrolysis requires energy input to break chemical bonds.

Step-by-Step Solution:

Step 1: Define absorption of energy: A reaction accompanied by the absorption of energy is an endothermic reaction.

Step 2: Analyze each option:
- A. Burning of a candle: This is combustion, which releases energy (exothermic).
- B. Rusting of iron: This is oxidation, which releases energy (exothermic).
- C. Electrolysis of water: This requires electrical energy to break down water into hydrogen and oxygen (endothermic).
- D. Digestion of food: This is primarily breaking down complex molecules, releasing energy (exothermic).

Step 3: Identify the endothermic reaction: Electrolysis of water is the only endothermic reaction among the options.

Why other options are wrong:
- Option A: Burning of a candle: This is an exothermic process.
- Option B: Rusting of iron: This is an exothermic process.
- Option D: Digestion of food: This is primarily an exothermic process.

Key Concepts:
- Endothermic reaction
- Exothermic reaction
- Electrolysis

Memory Aids:
- Electrolysis needs electricity (energy input).

Quick Tips:
- Remember that electrolysis is always an endothermic process.

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