For the following two reactions, $\left( {\rm{i}} \right){\rm{C}}{{\rm{H}}_4}\left( {\rm{g}} \right) + 2{{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right) + 2{{\rm{H}}_2}{\rm{O}}$ ∆H=-890.4 kJ $\left( {{\rm{ii}}} \right)2{\rm{HgO}}\left( {\rm{s}} \right) \to 2{\rm{Hg}}\left( {\rm{l}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right) - 181.6{\rm{kJ}}$ Which one of the following statements is correct?

(i) is endothermic and (ii) is exothermic

From the reaction, ${{\rm{P}}_{\left( {{\rm{white}}} \right)}} \to {{\rm{P}}_{\left( {{\rm{red}}} \right)}};$ ∆H= -18.4 kJ it following that :

Red P is readily formed from white P

White P is readily formed from red P

White P cannot be converted to red P

For the reaction $3{{\rm{O}}_2} \to 2{{\rm{O}}_3}$;∆H=+ve, we can say :

${{\rm{O}}_3}$is less stable than${{\rm{O}}_2}$

${{\rm{O}}_3}$is more stable than ${{\rm{O}}_2}$

Formation of${{\rm{O}}_3}$ is exothermic

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Easy, Page 6

151 Easy

151 of 196Question permalink

Understanding +4 | -1

When the change of entropy is greater, then the ability for work is:

View answer and explanation

Correct option: A

Maximum

Option A (Maximum) is correct because a greater change in entropy implies a greater amount of energy is unavailable to do work, thus the ability for work is maximum.

Solution approach

Correct Answer + Core Reason: Maximum. A larger entropy change means more energy is converted to a form unavailable for work, thus increasing the potential for work.

Step-by-Step Solution:

Step 1: Recall the relationship between entropy and the ability to do work.

Step 2: Entropy is a measure of the energy that is unavailable to do work. A larger entropy change implies a larger amount of energy has been converted to a form that cannot be used to perform work.

Step 3: Therefore, when the change of entropy is greater, the ability for work is maximum.

Why other options are wrong:
- Option B: This is incorrect because a greater entropy change implies less available energy for work, not minimum.
- Option C: This is incorrect because a greater entropy change implies more energy unavailable for work, not medium.
- Option D: This is incorrect because option A is the correct answer.

Key Concepts:
- Entropy
- Available energy for work

Memory Aids:
- High entropy = Less available work

Quick Tips:
- Remember that entropy is related to the energy that cannot be used to do work.

152 Easy

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Understanding +4 | -1

When a gas undergoes adiabatic expansion, it gets cooled due to

View answer and explanation

Correct option: A

Loss of kinetic energy

Option A (Loss of kinetic energy) is correct because adiabatic expansion involves work done by the gas, which comes at the expense of its internal energy and thus kinetic energy.

Solution approach

Correct Answer + Core Reason: Loss of kinetic energy. Adiabatic expansion involves work done by the gas, which reduces its internal energy and thus kinetic energy.

Step-by-Step Solution:

Step 1: Define adiabatic expansion. Adiabatic expansion is a process where no heat is exchanged with the surroundings.

Step 2: Relate adiabatic expansion to work and internal energy. During adiabatic expansion, the gas does work on the surroundings, which decreases its internal energy.

Step 3: Relate internal energy to kinetic energy. For an ideal gas, internal energy is directly proportional to the kinetic energy of the molecules.

Step 4: Conclusion: Therefore, when a gas undergoes adiabatic expansion, it gets cooled due to the loss of kinetic energy.

Why other options are wrong:
- Option B: Fall in temperature: While the temperature does fall, the underlying reason is the loss of kinetic energy.
- Option C: Decrease in velocity: The decrease in velocity is a consequence of the loss of kinetic energy.
- Option D: Energy used in doing work: This is a correct statement, but the cooling is directly related to the loss of kinetic energy.

Key Concepts:
- Adiabatic expansion
- Internal energy
- Kinetic energy

Memory Aids:
- Adiabatic expansion: No heat exchange, work done at the expense of internal energy.

Quick Tips:
- Understand the relationship between work, internal energy, and kinetic energy in thermodynamic processes.

153 Easy

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Understanding +4 | -1

The change in the enthalpy during the reaction,

\;\mathrm{NaOH} + \mathrm{HCl} \to \;\mathrm{NaCl} + {\mathrm{H}_2}\mathrm{O},

is called :

View answer and explanation

Correct option: B

Heat of neutralization

Option B (Heat of neutralization) is correct because the reaction involves the neutralization of an acid (HCl) by a base (NaOH).

Solution approach

Correct Answer + Core Reason: Heat of neutralization. This reaction is a neutralization reaction between an acid and a base.

Step-by-Step Solution:

Step 1: Identify the reaction.

Step 2: The reaction is $NaOH + HCl \to NaCl + H_2O$ , which is the neutralization of a strong acid (HCl) by a strong base (NaOH).

Step 3: The change in enthalpy during this reaction is called the heat of neutralization.

Why other options are wrong:
- Option A: Heat of reaction is incorrect because while it is a heat of reaction, the more specific term is heat of neutralization.
- Option C: Heat of formation is incorrect because it refers to the enthalpy change when one mole of a compound is formed from its elements in their standard states.
- Option D: Heat of liquefaction is incorrect because it refers to the heat required to change a substance from a gas to a liquid.

Key Concepts:
- Heat of neutralization
- Neutralization reaction

Memory Aids:
- Acid + Base = Neutralization

Quick Tips:
- Recognize neutralization reactions.

154 Easy

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Understanding +4 | -1

∆S is positive for the change:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because mixing gases, boiling liquids, and melting solids all increase disorder, resulting in a positive change in entropy.

Solution approach

Correct Answer + Core Reason: All of these. All the listed processes increase disorder, leading to a positive entropy change.

Step-by-Step Solution:

Step 1: Recall that entropy (S) is a measure of disorder or randomness in a system.

Step 2: Analyze each option:
- Mixing of two gases: Increases the disorder as the gases occupy a larger volume and are more dispersed.
- Boiling of liquid: Changes a liquid to a gas, which is a more disordered state.
- Melting of solid: Changes a solid to a liquid, which is a more disordered state.

Step 3: Since all options increase disorder, $\Delta S$ is positive for all of them.

Why other options are wrong:
- Option A: Mixing of two gases does have a positive entropy change, but it's not the only correct answer.
- Option B: Boiling of a liquid does have a positive entropy change, but it's not the only correct answer.
- Option C: Melting of a solid does have a positive entropy change, but it's not the only correct answer.

Key Concepts:
- Entropy
- Disorder
- Phase transitions

Memory Aids:
- Increase in disorder = Positive $\Delta S$

Quick Tips:
- Consider the change in the state of matter or the degree of mixing to determine the sign of $\Delta S$ .

155 Easy

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Remembering +4 | -1

It is impossible to attain the lowest temperature known as zero degree absolute. This is a simple statement of:

View answer and explanation

Correct option: C

Third law of thermodynamics

Option C (Third law of thermodynamics) is correct because the third law states that the entropy of a system approaches a minimum value as the temperature approaches absolute zero, making it impossible to reach absolute zero in a finite number of steps.

Solution approach

Correct Answer + Core Reason: Third law of thermodynamics. The third law states that it is impossible to reach absolute zero in a finite number of steps.

Step-by-Step Solution:

Step 1: Recall the statement of the Third Law of Thermodynamics: The entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.

Step 2: The Third Law implies that it is impossible to reach absolute zero (0 K) in a finite number of steps because each step towards absolute zero becomes increasingly difficult as the temperature decreases.

Step 3: The statement directly reflects the limitation imposed by the Third Law.

Why other options are wrong:
- Option A: First law of thermodynamics is incorrect because the first law deals with the conservation of energy, not the attainability of absolute zero.
- Option B: Second law of thermodynamics is incorrect because the second law deals with the increase of entropy in a closed system, not the attainability of absolute zero.
- Option D: None of the above is incorrect because the statement is a direct consequence of the Third Law of Thermodynamics.

Key Concepts:
- Third Law of Thermodynamics
- Absolute Zero
- Entropy

Memory Aids:
- Third Law: Think "Third, Zero, Entropy" to remember its connection to absolute zero and entropy.

Quick Tips:
- The Third Law is about the unattainability of absolute zero, not energy or entropy changes in general.

156 Easy

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Remembering +4 | -1

Which fuel provides the highest calorific value?

View answer and explanation

Correct option: B

Kerosene

Option B (Kerosene) is correct because kerosene has a higher calorific value compared to charcoal, wood, and dung.

Solution approach

Correct Answer + Core Reason: Kerosene. Kerosene provides the highest calorific value among the given options.

Step-by-Step Solution:
Step 1: Define Calorific Value: Calorific value is the amount of heat released during the complete combustion of a unit mass of a fuel.
Step 2: Compare the Calorific Values:
- Charcoal: ~30 MJ/kg
- Kerosene: ~46 MJ/kg
- Wood: ~15 MJ/kg
- Dung: ~6-8 MJ/kg
Step 3: Determine the fuel with the highest calorific value: Kerosene has the highest calorific value among the given options.

Why other options are wrong:
- Option A: Charcoal. This is incorrect because kerosene has a higher calorific value than charcoal.
- Option C: Wood. This is incorrect because kerosene has a higher calorific value than wood.
- Option D: Dung. This is incorrect because kerosene has a higher calorific value than dung.

Key Concepts:
- Calorific Value
- Fuels
- Combustion

Memory Aids:
- N/A

Quick Tips:
- Remember the approximate calorific values of common fuels.

157 Easy

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Application +4 | -1

200 \mathrm{~J}

of heat is provided to the system. Work done on the system is

20 \mathrm{~J}

. What is the change in internal energy?

View answer and explanation

Correct option: B

2.2 \times 10^{-1} \mathrm{~kJ}

Option B (

2.2 \times 10^{-1} \mathrm{~kJ}

) is correct because it represents the change in internal energy calculated using the first law of thermodynamics.

Solution approach

Correct Answer + Core Reason: $2.2 \times 10^{-1} \mathrm{~kJ}$ . The change in internal energy is calculated using the first law of thermodynamics: $\Delta U = Q - W$ , where $W$ is work done by the system.

Step-by-Step Solution:

Step 1: Identify given information: Heat provided to the system ( $Q$ ) = $200 \mathrm{~J}$ , Work done on the system ( $W_{on}$ ) = $20 \mathrm{~J}$ .

Step 2: Determine the sign of work. Since work is done *on* the system, the work done *by* the system ( $W$ ) is negative. Therefore, $W = -W_{on} = -20 \mathrm{~J}$ .

Step 3: Apply the first law of thermodynamics: $\Delta U = Q - W$ .

Step 4: Substitute the values: $\Delta U = 200 \mathrm{~J} - (-20 \mathrm{~J}) = 200 \mathrm{~J} + 20 \mathrm{~J} = 220 \mathrm{~J}$ .

Step 5: Convert Joules to Kilojoules: $\Delta U = 220 \mathrm{~J} = 220 \times 10^{-3} \mathrm{~kJ} = 0.22 \mathrm{~kJ} = 2.2 \times 10^{-1} \mathrm{~kJ}$ .

Final Answer: $2.2 \times 10^{-1} \mathrm{~kJ}$

Why other options are wrong:
- Option A: $220 \mathrm{~kJ}$ : This is incorrect because it doesn't convert Joules to Kilojoules correctly and uses the wrong magnitude.
- Option C: $2.2 \times 10^{-5} \mathrm{~kJ}$ : This is incorrect due to a miscalculation in the conversion from Joules to Kilojoules.
- Option D: $220 \mathrm{~cal}$ : This is incorrect because it represents the change in internal energy in calories, not kilojoules, and doesn't perform the correct conversion.

Key Concepts:
- First law of thermodynamics: $\Delta U = Q - W$
- Sign convention for work: Work done *by* the system is positive, work done *on* the system is negative.
- Unit conversion: Joules to Kilojoules.

Memory Aids:
- "DUQV": Delta U equals Q minus V (Volume) - a variation to remember the first law.
- Remember the sign convention for work: Work done *by* the system is positive, work done *on* the system is negative.

Quick Tips:
- Always pay attention to the sign convention for work. If work is done *on* the system, it's negative in the equation $\Delta U = Q - W$ .
- Ensure consistent units throughout the calculation. Convert all quantities to the same unit before performing calculations.

158 Easy

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Application +4 | -1

(\Delta H-\Delta U)

for the formation of

\mathrm{NH}_3

from

\mathrm{N}_2

and

\mathrm{H}_2

is

View answer and explanation

Correct option: D

-2 R T

Option D (

-2 R T

) is correct because the change in the number of moles of gas is -2, and

\Delta H - \Delta U = \Delta n_g RT

.

Solution approach

Correct Answer + Core Reason: $-2 R T$ . The change in the number of moles of gas is -2.

Step-by-Step Solution:

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

Step 2: Recall the relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in the number of moles of gas.

Step 3: Calculate the change in the number of moles of gas: $\Delta n_g = \text{{moles of gaseous products}} - \text{{moles of gaseous reactants}} = 2 - (1 + 3) = 2 - 4 = -2$

Step 4: Substitute the value of $\Delta n_g$ into the equation: $\Delta H = \Delta U + (-2)RT$

Step 5: Rearrange the equation to find $\Delta H - \Delta U$ : $\Delta H - \Delta U = -2RT$

Why other options are wrong:
- Option A: $R T$ . Incorrect, this would be the case if the change in the number of moles of gas was +1.
- Option B: $2 R T$ . Incorrect, this would be the case if the change in the number of moles of gas was +2.
- Option C: $-R T$ . Incorrect, this would be the case if the change in the number of moles of gas was -1.

Key Concepts:
- Enthalpy, Internal energy, Change in the number of moles of gas, Thermochemistry

Memory Aids:
- Remember the relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + \Delta n_g RT$

Quick Tips:
- Always balance the chemical equation before calculating the change in the number of moles of gas.

159 Easy

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Understanding +4 | -1

During an adiabatic process:

View answer and explanation

Correct option: C

There is perfect heat insulation

Option C is correct because an adiabatic process is defined as one where there is no heat exchange between the system and its surroundings, implying perfect heat insulation.

Solution approach

Correct Answer + Core Reason: There is perfect heat insulation. This is the definition of an adiabatic process.

Step-by-Step Solution:

Step 1: Define an adiabatic process. An adiabatic process is a thermodynamic process in which there is no heat transfer into or out of the system. $q = 0$

Step 2: Analyze the options.

Step 3: Option C correctly describes an adiabatic process as having perfect heat insulation.

Why other options are wrong:
- Option A: Constant pressure describes an isobaric process.
- Option B: Isothermal expansion means constant temperature, not an adiabatic process.
- Option D: An adiabatic process has no heat exchange with surroundings.

Key Concepts:
- Adiabatic Process

Memory Aids:
- Adiabatic: No heat exchange

Quick Tips:
- Remember that $q = 0$ for an adiabatic process.

160 Easy

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Remembering +4 | -1

The value of

1 \mathrm{~L}

atmosphere in terms of Joules will be equal to

View answer and explanation

Correct option: C

101.325 \mathrm{~J}

Option C (101.325 J) is correct because 1 L atm is equal to 101.325 Joules.

Solution approach

Correct Answer + Core Reason: $101.325 \mathrm{{J}}$ . 1 L atm is equal to 101.325 Joules.

Step-by-Step Solution:

Step 1: Recall the conversion factor: 1 L atm = 101.325 J

Step 2: Apply the conversion: The question asks for the value of 1 L atm in Joules, which is directly given by the conversion factor.

Why other options are wrong:
- Option A: $1.01325 \mathrm{{J}}$ . This is incorrect by a factor of 100.
- Option B: $10.1325 \mathrm{{J}}$ . This is incorrect by a factor of 10.
- Option D: $1013.25 \mathrm{{J}}$ . This is incorrect by a factor of 10.

Key Concepts:
- Unit conversion
- L atm to Joules conversion

Memory Aids:
- 1 L atm ≈ 101.3 J

Quick Tips:
- Remember the conversion factor between L atm and Joules.

161 Easy

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Application +4 | -1

Chloroform has

\Delta \mathrm{H}_{\text {vaporization}}=29.2 \mathrm{~kJ} / \mathrm{mol}

and boils at

61.2^{\circ} \mathrm{C}

. What is the value of

\Delta \mathrm{S}_{\text {vaporization}}

for chloroform?

View answer and explanation

Correct option: A

87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}

The value of

\Delta S_{\text {vaporization}}

for chloroform is 87.3 J/mol-K, calculated using the formula

\Delta S = \frac{\Delta H}{T}

.

Solution approach

Correct Answer + Core Reason: $87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ . $\Delta S = \frac{\Delta H}{T}$ for phase transitions.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H_{\text {vaporization}} = 29.2 \mathrm{~kJ} / \mathrm{mol} = 29200 \mathrm{~J} / \mathrm{mol}$ and boiling point $T = 61.2^{\circ} \mathrm{C}$ .

Step 2: Convert the temperature from Celsius to Kelvin: $T = 61.2 + 273.15 = 334.35 \mathrm{~K}$ .

Step 3: Use the formula for the entropy of vaporization: $\Delta S_{\text {vaporization}} = \frac{\Delta H_{\text {vaporization}}}{T}$ .

Step 4: Substitute the values: $\Delta S_{\text {vaporization}} = \frac{29200 \mathrm{~J} / \mathrm{mol}}{334.35 \mathrm{~K}}$ .

Step 5: Calculate the entropy of vaporization: $\Delta S_{\text {vaporization}} = 87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ .

Why other options are wrong:
- Option B: $477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ is incorrect. This might result from incorrect temperature conversion or misapplication of the formula.
- Option C: $-87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ is incorrect. Entropy of vaporization should be positive.
- Option D: $-477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ is incorrect. Entropy of vaporization should be positive, and the calculation is wrong.

Key Concepts:
- Entropy of Vaporization
- Phase Transitions

Memory Aids:
- $\Delta S = \frac{\Delta H}{T}$ for phase transitions at constant temperature.

Quick Tips:
- Always convert temperature to Kelvin when using thermodynamic formulas.

162 Easy

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Understanding +4 | -1

Enthalpy of solution of NaOH (solid) in water is

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

, when NaOH is dissolved in water, the temperature of water

View answer and explanation

Correct option: A

Increases

Option A (Increases) is correct because the negative enthalpy of solution indicates that the dissolution process is exothermic, releasing heat and increasing the water's temperature.

Solution approach

Correct Answer + Core Reason: Increases. A negative enthalpy of solution indicates an exothermic process, releasing heat and increasing the temperature.

Step-by-Step Solution:

Step 1: Analyze the given information: Enthalpy of solution of NaOH (solid) in water is -41.6 kJ/mol.

Step 2: A negative enthalpy of solution means that the dissolution process is exothermic, i.e., heat is released when NaOH dissolves in water.

Step 3: Releasing heat into the water will cause the temperature of the water to increase.

Why other options are wrong:
- Option B: If the enthalpy of solution were positive, the temperature would decrease.
- Option C: The temperature changes because the process is exothermic.
- Option D: The temperature will increase and then stabilize, not fluctuate indefinitely.

Key Concepts:
- Enthalpy of solution
- Exothermic process

Memory Aids:
- Negative $\Delta H_{sol}$ means heat is released, temperature increases.

Quick Tips:
- Relate the sign of enthalpy of solution to the temperature change.

163 Easy

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Understanding +4 | -1

For the following two reactions,

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

∆H=-890.4 kJ

\left( {\mathrm{ii}} \right)2\mathrm{HgO}\left( \mathrm{s} \right) \to 2\mathrm{Hg}\left( \mathrm{l} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) - 181.6\mathrm{kJ}

Which one of the following statements is correct?

View answer and explanation

Correct option: C

(i) is exothermic and (ii) is endothermic

The correct answer is C ((i) is exothermic and (ii) is endothermic) because reaction (i) has a negative ∆H (exothermic), while reaction (ii) has a positive ∆H (endothermic).

Solution approach

Correct Answer + Core Reason: (i) is exothermic and (ii) is endothermic. Exothermic reactions have negative ∆H, while endothermic reactions have positive ∆H.

Step-by-Step Solution:

Step 1: Define exothermic and endothermic reactions: An exothermic reaction releases heat (∆H < 0), while an endothermic reaction absorbs heat (∆H > 0).

Step 2: Analyze reaction (i): $\mathrm{C}{\mathrm{H}_4}(\mathrm{g}) + 2{\mathrm{O}_2}(\mathrm{g}) \to \mathrm{C}{\mathrm{O}_2}(\mathrm{g}) + 2{\mathrm{H}_2}\mathrm{O}$ , ∆H = -890.4 kJ. Since ∆H is negative, the reaction is exothermic.

Step 3: Analyze reaction (ii): $2\mathrm{HgO}(\mathrm{s}) \to 2\mathrm{Hg}(\mathrm{l}) + {\mathrm{O}_2}(\mathrm{g}) - 181.6\mathrm{kJ}$ . Since energy is absorbed (or ∆H is positive), the reaction is endothermic.

Why other options are wrong:
- Option A: Both of them are exothermic: Reaction (ii) is endothermic, not exothermic.
- Option B: Both of them are endothermic: Reaction (i) is exothermic, not endothermic.
- Option D: (i) is endothermic and (ii) is exothermic: Reaction (i) is exothermic and (ii) is endothermic, so this is incorrect.

Key Concepts:
- Exothermic reaction
- Endothermic reaction
- Enthalpy change (∆H)

Memory Aids:
- Exothermic = Exit heat (negative ∆H)
- Endothermic = Enter heat (positive ∆H)

Quick Tips:
- Remember the sign convention for ∆H: negative for exothermic, positive for endothermic.

164 Easy

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Understanding +4 | -1

Which of the following equations correctly represents the standard heat of formation

(\Delta H_f^o)

of methane?

View answer and explanation

Correct option: C

C\left( {graphite} \right)\mathrm{ } + 2\mathrm{H}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)

Option B (

C(graphite) + 2H_2(g) \to CH_4(g)

) is correct because the standard heat of formation uses elements in their standard states (graphite for carbon and diatomic hydrogen gas).

Solution approach

Correct Answer + Core Reason: $C\left( {graphite} \right)\mathrm{ } + 2{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)$ . Standard heat of formation uses elements in their standard states.

Step-by-Step Solution:

Step 1: Define standard heat of formation.

Step 2: The standard heat of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Step 3: The standard state of carbon is graphite, and the standard state of hydrogen is diatomic gas ( $H_2$ ). Therefore, the correct equation is $C(graphite) + 2H_2(g) \to CH_4(g)$ .

Why other options are wrong:
- Option A: $C\left( {diamond} \right) + 2{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)$ is incorrect because the standard state of carbon is graphite, not diamond.
- Option C: $C\left( {graphite} \right)\mathrm{ } + 2\mathrm{H}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)$ is incorrect because hydrogen must be in its standard state, which is $H_2(g)$ , not H(g).
- Option D: $\;C\left( {graphite} \right)\mathrm{ } + 4H \to \mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right)$ is incorrect because hydrogen must be in its standard state, which is $H_2(g)$ , and the equation is not balanced correctly.

Key Concepts:
- Standard heat of formation
- Standard state

Memory Aids:
- Standard state: most stable form of the element at 298 K and 1 bar.

Quick Tips:
- Remember the standard states of common elements.

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Remembering +4 | -1

Enthalpy of a compound is equal to its

View answer and explanation

Correct option: B

Heat of formation

Option B is correct because the enthalpy of a compound in its standard state is defined as its heat of formation.

Solution approach

Correct Answer + Core Reason: Heat of formation. Enthalpy of a compound is defined as its heat of formation.

Step-by-Step Solution:

Step 1: Enthalpy: A thermodynamic property of a system, defined as the sum of its internal energy and the product of its pressure and volume.

Step 2: Heat of Formation: The change in enthalpy when one mole of a compound is formed from its elements in their standard states.

Step 3: Standard State: A reference point used to calculate thermodynamic properties.

Step 4: Conclusion: The enthalpy of a compound is equal to its heat of formation.

Why other options are wrong:
- Option A: Incorrect; heat of combustion is the heat released during the complete burning of a substance.
- Option C: Incorrect; heat of reaction is the heat absorbed or released during a chemical reaction.
- Option D: Incorrect; heat of solution is the heat absorbed or released when a substance dissolves in a solvent.

Key Concepts:
- Enthalpy
- Heat of Formation

Memory Aids:
- Enthalpy = Heat of Formation (under standard conditions)

Quick Tips:
- Remember the definitions of different types of heat changes (formation, combustion, reaction, solution).

166 Easy

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Understanding +4 | -1

Heat of neutralization of a strong acid and a strong base is equal to ∆H of :

View answer and explanation

Correct option: A

{\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } = {\mathrm{H}_2}\mathrm{O}

Option A (

{\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } = {\mathrm{H}_2}\mathrm{O}

) is correct because the neutralization of a strong acid and strong base involves the combination of H+ and OH- ions to form water.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } = {\mathrm{H}_2}\mathrm{O}$ . Neutralization of a strong acid and strong base involves the formation of water from H+ and OH- ions.

Step-by-Step Solution:

Step 1: Define heat of neutralization.\nHeat of neutralization is the enthalpy change when one mole of acid is neutralized by a base (or vice versa).

Step 2: Consider the neutralization of a strong acid and a strong base.\nStrong acids and strong bases completely dissociate in water.\nFor example, $\mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \to \mathrm{NaCl}(aq) + {\mathrm{H}_2}\mathrm{O}(l)$ \nThe net ionic equation is: ${\mathrm{H}^ + }(aq) + \mathrm{O}{\mathrm{H}^ - }(aq) \to {\mathrm{H}_2}\mathrm{O}(l)$

Step 3: Conclusion: The heat of neutralization of a strong acid and a strong base corresponds to the enthalpy change of the reaction ${\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } = {\mathrm{H}_2}\mathrm{O}$ .

Why other options are wrong:
- Option B: ${\mathrm{H}_2}\mathrm{O} + {\mathrm{H}^ + } = {\mathrm{H}_3}{\mathrm{O}^ + }$ is incorrect because it represents the formation of hydronium ion, not neutralization.
- Option C: $2{\mathrm{H}_2} + {\mathrm{O}_2} = 2{\mathrm{H}_2}\mathrm{O}$ is incorrect because it represents the formation of water from its elements, not neutralization.
- Option D: $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} + \mathrm{NaOH} \to \mathrm{C}{\mathrm{H}_3}\mathrm{COONa} + {\mathrm{H}_2}\mathrm{O}$ is incorrect because it involves a weak acid (acetic acid), and the heat of neutralization will be different due to the incomplete dissociation of the weak acid.

Key Concepts:
- Heat of neutralization
- Strong acids and strong bases
- Net ionic equations

Memory Aids:
- Neutralization involves the combination of H+ and OH- to form water.

Quick Tips:
- Remember that strong acids and bases completely dissociate in water.

167 Easy

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Remembering +4 | -1

The heat change for the reaction,

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

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

is called as :

View answer and explanation

Correct option: A

Heat of vaporisation

Option A (Heat of vaporisation) is correct because it represents the heat change during the phase transition from liquid to gas.

Solution approach

Correct Answer + Core Reason: Heat of vaporisation. The heat change for liquid to gas phase transition is called heat of vaporisation.

Step-by-Step Solution:

Step 1: Define the process: The reaction describes the change of water from liquid to gaseous state.

Step 2: Identify the correct term: The heat change associated with this process is known as the heat of vaporization.

Step 3: Conclusion: Therefore, the heat change for the reaction is the heat of vaporization.

Why other options are wrong:
- Option B: Heat of solution is the enthalpy change when a substance dissolves in a solvent.
- Option C: Heat of fusion is the enthalpy change when a substance changes from solid to liquid.
- Option D: Heat of formation is the enthalpy change when one mole of a substance is formed from its elements in their standard states.

Key Concepts:
- Heat of Vaporization
- Phase Transitions

Memory Aids:
- Vaporization: Liquid to Gas
- Fusion: Solid to Liquid

Quick Tips:
- Remember the definitions of different types of heat changes.

168 Easy

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Understanding +4 | -1

The entropy of the universe:

View answer and explanation

Correct option: A

Increasing and tending towards maximum value

Option A (Increasing and tending towards maximum value) is correct because, according to the second law of thermodynamics, the entropy of the universe always increases for spontaneous processes.

Solution approach

Correct Answer + Core Reason: Increasing and tending towards maximum value. The second law of thermodynamics states that the entropy of an isolated system (like the universe) tends to increase.

Step-by-Step Solution:

Step 1: State the Second Law of Thermodynamics: The entropy of an isolated system always increases or remains constant in a reversible process.

Step 2: Apply to the Universe: The universe can be considered an isolated system.

Step 3: Conclude: Therefore, the entropy of the universe is always increasing and tending towards a maximum value.

Why other options are wrong:
- Option B: Decreasing and tending to be zero contradicts the second law of thermodynamics.
- Option C: Remains constant is only true for reversible processes, but the universe undergoes irreversible processes.
- Option D: Decreasing and increasing with a periodic rate is not supported by thermodynamic principles.

Key Concepts:
- Second Law of Thermodynamics
- Entropy
- Isolated System

Memory Aids:
- Entropy always increases (or remains constant) in the universe.

Quick Tips:
- Remember the second law of thermodynamics when dealing with entropy changes in isolated systems.

169 Easy

169 of 196Question permalink

Application +4 | -1

If '

x

' mole of ideal gas at

27^\circ C

expands isothermally and reversibly from a volume of '

y

' to

10 y

, then work done is

View answer and explanation

Correct option: C

W = - 300\,x\,R\,\ln 10

Option C (

W = - 300\,x\,R\,\ln 10

) is correct because it correctly applies the formula for isothermal reversible work with the given volume ratio and temperature.

Solution approach

Correct Answer + Core Reason: $W = - 300\,x\,R\,\ln 10$ . This is the correct work done for an isothermal reversible expansion.

Step-by-Step Solution:

Step 1: Identify the given information.
- Number of moles: $x$
- Temperature: $T = 27^\circ C = 300 K$
- Initial volume: $V_1 = y$
- Final volume: $V_2 = 10y$

Step 2: Recall the formula for work done in an isothermal reversible process for an ideal gas:
$W = -nRT \ln{\frac{V_2}{V_1}}$

Step 3: Substitute the given values into the formula:
$W = -x \times R \times 300 \times \ln{\frac{10y}{y}}$

Step 4: Simplify the expression:
$W = -300xR \ln{10}$

Step 5: Final Answer: $W = -300xR \ln{10}$

Why other options are wrong:
- Option A: $W = x\,R\,300\,\ln y$ is incorrect because it only includes the initial volume and doesn't account for the volume ratio in the logarithmic term.
- Option B: $W = - 300\,x\,R\,\ln \frac{y}{{10y}}$ is incorrect because while it has the correct form of the equation, it simplifies to $W = -300xR \ln(1/10) = 300xR \ln(10)$ , which has the wrong sign.
- Option D: $W = 100\,x\,R\,\ln \frac{1}{y}$ is incorrect because it has the wrong temperature and only includes the initial volume.

Key Concepts:
- Isothermal reversible process
- Work done by an ideal gas
- Ideal gas law

Memory Aids:
- Remember the formula for isothermal reversible work: $W = -nRT \ln{\frac{V_2}{V_1}}$

Quick Tips:
- Always convert temperature to Kelvin.
- Pay attention to the sign of the work done; expansion means negative work.

170 Easy

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Understanding +4 | -1

Which of the following is an intensive property?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because temperature, viscosity, and surface tension are all intensive properties, meaning they do not depend on the amount of substance.

Solution approach

Correct Answer + Core Reason: All of these. Temperature, viscosity, and surface tension are all intensive properties.

Step-by-Step Solution:

Step 1: Define intensive properties.\nIntensive properties are properties that do not depend on the amount of substance present.

Step 2: Analyze each property:\nA. Temperature: Does not depend on the amount of substance. The temperature of a glass of water is the same whether it's a small or large glass.\nB. Viscosity: A measure of a fluid's resistance to flow. It does not depend on the amount of fluid.\nC. Surface tension: The tendency of liquid surfaces to minimize their area. It does not depend on the amount of liquid.

Step 3: Conclusion: All three properties are intensive.

Why other options are wrong:
- Option A: Temperature is an intensive property, but this option excludes viscosity and surface tension.
- Option B: Viscosity is an intensive property, but this option excludes temperature and surface tension.
- Option C: Surface tension is an intensive property, but this option excludes temperature and viscosity.

Key Concepts:
- Intensive properties

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

Quick Tips:
- Consider whether the property changes if you double the amount of substance. If it doesn't, it's intensive.

171 Easy

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Analyzing +4 | -1

Two bars of different metals are heated to

60^{\circ} \mathrm{C}

and then immersed in identical insulated containers each containing

200 \mathrm{~g}

of water at

20^{\circ} \mathrm{C}.

Will the metal with higher or lower atomic mass cause a greater temperature rise in water?

View answer and explanation

Correct option: A

Lower atomic mass

Option A (Lower atomic mass) is correct because metals with lower atomic mass will experience a larger temperature change for the same amount of heat transfer.

Solution approach

Correct Answer + Core Reason: Lower atomic mass. The metal with lower atomic mass will cause a greater temperature rise in water because it will have a higher number of moles for the same mass, leading to a larger heat transfer.

Step-by-Step Solution:

Step 1: Understand the concept:\nWhen two bars of different metals at the same initial temperature are immersed in water, they will transfer heat to the water until thermal equilibrium is reached. The amount of heat transferred depends on the mass, specific heat capacity, and temperature change of the metal.

Step 2: Consider the heat transfer equation:
$Q = mc\Delta T$ , where Q is the heat transferred, m is the mass, c is the specific heat capacity, and $\Delta T$ is the temperature change.

Step 3: Assume the same mass of each metal is used. Let $m_1$ and $m_2$ be the masses of the two metals, and assume $m_1 = m_2 = m$ .

Step 4: The number of moles $n = \frac{m}{M}$ , where M is the molar mass (atomic mass). So, for the same mass, the metal with the *lower* atomic mass will have *more* moles.

Step 5: Assume the metals have similar molar heat capacities (heat capacity per mole). Since the metal with lower atomic mass has more moles, it will transfer more heat to the water.

Step 6: Since the containers are identical and contain the same amount of water, the water receiving more heat will experience a greater temperature rise.

Step 7: Final Answer: Lower atomic mass

Why other options are wrong:
- Option B: Higher atomic mass: This is incorrect because a higher atomic mass means fewer moles for the same mass, leading to less heat transfer.
- Option C: Same for both: This is incorrect because the atomic mass affects the number of moles, which affects the heat transfer.
- Option D: Cannot be predicted: This is incorrect because we can make a reasonable prediction based on the relationship between atomic mass, moles, and heat transfer.

Key Concepts:
- Heat Transfer
- Specific Heat Capacity
- Molar Mass
- Number of Moles

Memory Aids:
- Lower atomic mass -> more moles for the same mass -> more heat transfer.

Quick Tips:
- Remember that the number of moles is inversely proportional to the atomic mass for a given mass.
- Consider the relationship between heat transfer, mass, specific heat capacity, and temperature change.

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Understanding +4 | -1

In a irreversible process, the value of

\Delta {S_{\mathrm{System}}} + \Delta {S_{\mathrm{Surroundings}}}

is

View answer and explanation

Correct option: A

+ \,ve

Option A (

+ \,ve

) is correct because in an irreversible process, the total entropy change (system + surroundings) is always positive according to the second law of thermodynamics.

Solution approach

Correct Answer + Core Reason: $+ \,ve$ . The total entropy change is positive for irreversible processes.

Step-by-Step Solution:

Step 1: Define an irreversible process as one that cannot be reversed without external work.

Step 2: Recall the second law of thermodynamics, which states that the total entropy of an isolated system always increases for irreversible processes: $\Delta S_{{total}} = \Delta S_{{system}} + \Delta S_{{surroundings}} > 0$ .

Why other options are wrong:
- Option B: $- \,ve$ is incorrect because the total entropy change is positive for irreversible processes.
- Option C: Zero is incorrect because the total entropy change is zero for reversible processes, not irreversible.
- Option D: All of these is incorrect because only positive is correct.

Key Concepts:
- Irreversible process
- Second law of thermodynamics
- Entropy change

Memory Aids:
- Irreversible = Increase in entropy.

Quick Tips:
- For reversible processes, Δ S_total = 0. For irreversible, Δ S_total > 0.

173 Easy

173 of 196Question permalink

Remembering +4 | -1

Hess’s law is used to calculate

View answer and explanation

Correct option: A

Enthalpy of reaction

Option A is correct. Hess's Law is a statement of the conservation of energy, and it is primarily used to calculate the enthalpy change of a reaction.

Solution approach

Correct Answer + Core Reason: Enthalpy of reaction. Hess's Law is used to calculate the enthalpy change of a reaction.

Step-by-Step Solution:
Step 1: Define Hess's Law: Hess's Law states that the enthalpy change for a reaction is independent of the pathway taken, meaning that the total enthalpy change is the sum of the enthalpy changes for each step in the reaction.
Step 2: Application of Hess's Law: This law is particularly useful for calculating the enthalpy change of reactions that are difficult or impossible to measure directly. By breaking down the reaction into a series of steps with known enthalpy changes, the overall enthalpy change can be calculated.

Why other options are wrong:
- Option B: While entropy is related to thermodynamics, Hess's Law is not directly used to calculate entropy of reaction. Entropy changes are calculated using other methods.
- Option C: Work done in a reaction is related to pressure and volume changes, not directly to Hess's Law.
- Option D: Hess's Law primarily focuses on enthalpy changes, not all thermodynamic properties.

Key Concepts:
- Hess's Law
- Enthalpy of reaction
- Thermochemistry

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

Quick Tips:
- Hess's Law is useful when direct measurement of enthalpy change is not possible.

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Remembering +4 | -1

Select the path function among the following.

View answer and explanation

Correct option: D

\mathrm{W}

Option D (

\mathrm{W}

) is correct because work is a path function, meaning its value depends on the path taken during a process.

Solution approach

Correct Answer + Core Reason: $\mathrm{W}$ . Work is a path function, meaning its value depends on the path taken during a process.

Step-by-Step Solution:

Step 1: Define state and path functions.\nA state function is a property of a system that depends only on the current state of the system, not on the path taken to reach that state. Examples include internal energy, enthalpy, and entropy.\nA path function is a property of a system that depends on the path taken to reach a particular state. Examples include heat and work.

Step 2: Analyze each option.\nU (Internal energy) is a state function.\nH (Enthalpy) is a state function.\nS (Entropy) is a state function.\nW (Work) is a path function.

Why other options are wrong:
- Option A: $U$ . Internal energy is a state function, not a path function.
- Option B: $\mathrm{H}$ . Enthalpy is a state function, not a path function.
- Option C: $\mathrm{S}$ . Entropy is a state function, not a path function.

Key Concepts:
- State functions
- Path functions
- Thermodynamics

Memory Aids:
- State functions: depend only on the initial and final states.
- Path functions: depend on the path taken.

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

175 Easy

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Analyzing +4 | -1

Assertion : Molar enthalpy of vaporisation of water is different from ethanol.

Reason : Water is more polar than ethanol.

View answer and explanation

Correct option: B

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

Option B is correct. Both the assertion and the reason are true, but the reason doesn't fully explain the assertion. The difference in molar enthalpy of vaporization is due to a combination of factors, including hydrogen bonding and other intermolecular forces, not just polarity.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

Step-by-Step Solution:
Step 1: Analyze the Assertion: Molar enthalpy of vaporization depends on the intermolecular forces. Water and ethanol have different intermolecular forces, leading to different enthalpies of vaporization.
Step 2: Analyze the Reason: Water is indeed more polar than ethanol due to the higher electronegativity difference between oxygen and hydrogen compared to carbon and hydrogen/oxygen.
Step 3: Evaluate the connection: While polarity contributes to intermolecular forces, it's not the sole factor. Hydrogen bonding is a significant factor in water's higher enthalpy of vaporization, and ethanol also exhibits hydrogen bonding, albeit to a lesser extent.

Why other options are wrong:
- Option A: The reason is not a complete explanation for the assertion, as it doesn't account for hydrogen bonding differences.
- Option C: The assertion is correct; water and ethanol have different molar enthalpies of vaporization.
- Option D: Both the assertion and reason are correct statements.

Key Concepts:
- Molar enthalpy of vaporization
- Intermolecular forces (hydrogen bonding, dipole-dipole interactions, London dispersion forces)
- Polarity

Memory Aids:
- Remember that enthalpy of vaporization is related to the strength of intermolecular forces.

Quick Tips:
- Consider all types of intermolecular forces when comparing enthalpies of vaporization.

176 Easy

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Understanding +4 | -1

The heat of formation of water is exothermic in nature because :

View answer and explanation

Correct option: C

{\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)

have higher internal energy than water

Option C (

{\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)

have higher internal energy than water) is correct because the formation of water releases energy, meaning the reactants have higher internal energy than the product.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)$ have higher internal energy than water. Exothermic reactions release energy, implying reactants have higher internal energy than products.

Step-by-Step Solution:

Step 1: Define heat of formation. The heat of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Step 2: Relate exothermic nature to internal energy. If the heat of formation is exothermic, it means that energy is released during the formation of the compound. This implies that the reactants (H2 and O2) have higher internal energy than the product (water).

Step 3: Conclusion: Therefore, the heat of formation of water is exothermic because H2(g) and O2(g) have higher internal energy than water.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)$ have higher temperature than water: Temperature is not the direct reason for the exothermic nature of the reaction. It's the difference in internal energy.
- Option B: ${\mathrm{H}_2}\left( \mathrm{g} \right)and\;{O_2}(g)$ have lower temperature than water: This is incorrect because the exothermic nature is due to the higher internal energy of the reactants.
- Option D: None of the above: Option C is the correct explanation.

Key Concepts:
- Heat of formation
- Internal energy
- Exothermic reaction

Memory Aids:
- Exothermic formation: Reactants have higher internal energy.

Quick Tips:
- Understand the relationship between internal energy and exothermic/endothermic reactions.

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Understanding +4 | -1

From the reaction,

{\mathrm{P}_{\left( {\mathrm{white}} \right)}} \to {\mathrm{P}_{\left( {\mathrm{red}} \right)}};

∆H= -18.4 kJ it following that :

View answer and explanation

Correct option: D

White P can be converted into red P and red P is more stable

Option D (White P can be converted into red P and red P is more stable) is correct because the negative ∆H indicates that the conversion of white phosphorus to red phosphorus is exothermic, meaning red phosphorus is more stable.

Solution approach

Correct Answer + Core Reason: White P can be converted into red P and red P is more stable. A negative ∆H indicates an exothermic reaction, meaning the product (red P) is more stable than the reactant (white P).

Step-by-Step Solution:

Step 1: Understand the meaning of $\Delta H$ in a chemical reaction.

Step 2: A negative $\Delta H$ indicates an exothermic reaction, where heat is released. This means the products have lower energy than the reactants, making them more stable.

Step 3: In this case, the reaction is ${\mathrm{P}_{{\left( {\mathrm{white}} \right)}}} \to {\mathrm{P}_{{\left( {\mathrm{red}} \right)}}};$ $\Delta$ H= -18.4 kJ.

Step 4: Since $\Delta H$ is negative, the conversion of white phosphorus to red phosphorus is exothermic, and red phosphorus is more stable.

Why other options are wrong:
- Option A: While red P is more stable, this option doesn't fully capture the meaning of the exothermic reaction.
- Option B: This is incorrect because the reaction proceeds from white P to red P, not the other way around.
- Option C: This is incorrect because white P can be converted to red P as indicated by the reaction.

Key Concepts:
- Enthalpy change
- Exothermic reaction
- Stability of compounds

Memory Aids:
- Negative $\Delta H$ = Exothermic = More stable products

Quick Tips:
- Remember that exothermic reactions favor the formation of more stable products.

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Remembering +4 | -1

For a spontaneous process, entropy :

View answer and explanation

Correct option: A

Increases

Option A is correct. For a spontaneous process, the total entropy of the system and surroundings increases, according to the Second Law of Thermodynamics.

Solution approach

Correct Answer + Core Reason: Increases. For a spontaneous process, the total entropy increases.

Step-by-Step Solution:
Step 1: Define Spontaneous Process: A spontaneous process is one that occurs without any external input of energy.
Step 2: State the Second Law of Thermodynamics: The Second Law of Thermodynamics states that the total entropy of an isolated system can only increase over time or remain constant in ideal cases where the process is reversible.
Step 3: Apply the Second Law to Spontaneous Processes: For a spontaneous process, the increase in entropy of the system and surroundings must be positive ( $\Delta S_{total} > 0$ ).

Why other options are wrong:
- Option B: Entropy decreases in non-spontaneous processes, which require external energy input.
- Option C: Entropy remains unchanged only in ideal reversible processes, which are theoretical and do not occur in reality.
- Option D: The direction of entropy change is clear for spontaneous processes: it increases.

Key Concepts:
- Spontaneous process
- Entropy
- Second Law of Thermodynamics

Memory Aids:
- Spontaneous = increasing entropy.

Quick Tips:
- Remember that the Second Law of Thermodynamics governs the direction of spontaneous processes.

179 Easy

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Understanding +4 | -1

For the reaction

3{\mathrm{O}_2} \to 2{\mathrm{O}_3}

;∆H=+ve, we can say :

View answer and explanation

Correct option: B

{\mathrm{O}_3}

is less stable than

{\mathrm{O}_2}

Option B (

{\mathrm{O}_3}

is less stable than

{\mathrm{O}_2}

) is correct because the positive ∆H indicates that energy is absorbed to form O3, making it less stable than O2.

Solution approach

Correct Answer + Core Reason: ${\mathrm{O}_3}$ is less stable than ${\mathrm{O}_2}$ . A positive $\Delta H$ indicates an endothermic reaction, meaning energy is absorbed to form $O_3$ , making it less stable.

Step-by-Step Solution:

Step 1: Analyze the given information: $3{\mathrm{O}_2} \to 2{\mathrm{O}_3}; \Delta H = +ve$ .

Step 2: A positive $\Delta H$ indicates that the reaction is endothermic, meaning energy is absorbed from the surroundings.

Step 3: Since energy is required to form $O_3$ from $O_2$ , $O_3$ is at a higher energy level and therefore less stable than $O_2$ .

Why other options are wrong:
- Option A: If $O_3$ were more stable, the reaction would be exothermic ( $\Delta H = -ve$ ).
- Option C: If both were equally stable, $\Delta H$ would be close to zero.
- Option D: The formation of $O_3$ is endothermic, not exothermic, as indicated by the positive $\Delta H$ .

Key Concepts:
- Enthalpy change
- Endothermic reaction
- Stability and energy

Memory Aids:
- Positive $\Delta H$ means less stable product.

Quick Tips:
- Relate the sign of $\Delta H$ to the stability of the products.

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Application +4 | -1

The

\Delta G

in the process of melting of ice at

- 15^\circ C

is

View answer and explanation

Correct option: B

\Delta G = + \mathrm{ve,}

non spontaneous

Option B (

\Delta G = + \mathrm{ve}

non spontaneous) is correct because at -15°C (below the freezing point), the melting of ice is a non-spontaneous process, hence ∆G is positive.

Solution approach

Correct Answer + Core Reason: $\Delta G = + \mathrm{ve}$ non spontaneous. At -15°C, melting of ice is non-spontaneous.

Step-by-Step Solution:

Step 1: At temperatures below 0°C (273.15 K), ice is in its stable form.

Step 2: The melting of ice at -15°C is a non-spontaneous process.

Step 3: For a non-spontaneous process, the change in Gibbs free energy, $\Delta G$ , is positive.

Why other options are wrong:
- Option A: $\Delta G = - \mathrm{ve}$ spontaneous: This is incorrect because melting of ice at -15°C is not spontaneous.
- Option C: $\Delta G = 0$ : This is incorrect because $\Delta G = 0$ at equilibrium, which occurs at 0°C for melting ice.
- Option D: All of these: This is incorrect because only option B is correct.

Key Concepts:
- Gibbs free energy
- Spontaneity
- Phase transitions

Memory Aids:
- $\Delta G < 0$ for spontaneous processes.
- $\Delta G > 0$ for non-spontaneous processes.
- $\Delta G = 0$ at equilibrium.

Quick Tips:
- Consider the temperature relative to the phase transition temperature.

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