THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Easy, Page 4

91 Easy

Remembering +4 | -1

Heat of formation of compound is defined as :

View answer and explanation

Correct option: A

Heat evolved to form one mole of the compound from its elements

Option A is correct because the heat of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.

Solution approach

Correct Answer + Core Reason: Heat evolved to form one mole of the compound from its elements. This is the standard definition of heat of formation.

Step-by-Step Solution:

Step 1: Define heat of formation. Heat of formation is the change in enthalpy during the formation of one mole of a substance from its constituent elements, with all reactants and products in their standard states.

Step 2: Relate the definition to the options.\nOption A directly matches the definition.

Step 3: Conclusion: Option A is the correct answer.

Why other options are wrong:
- Option B: Heat required to form one molecule of a compound: This is incorrect because heat of formation is defined for one *mole*, not one molecule, of the compound.
- Option C: Change in heat content of the system when one molecule of a compound is formed: Similar to option B, this is incorrect because heat of formation is defined for one *mole*, not one molecule, of the compound. Also, it should be formed from its elements.
- Option D: None of the above: This is incorrect because option A is the correct answer.

Key Concepts:
- Heat of Formation
- Enthalpy Change
- Standard States

Memory Aids:
- Formation = From Elements
- One Mole is Key

Quick Tips:
- Remember that heat of formation refers to the formation of one mole of a substance from its elements in their standard states.

92 Easy

92 of 196Question permalink

Remembering +4 | -1

Mathematical from of 1st Law

View answer and explanation

Correct option: C

\Delta u = Q + W

Option C (

\Delta u = Q + W

) is correct because it accurately represents the mathematical formulation of the first law of thermodynamics.

Solution approach

Correct Answer + Core Reason: $\Delta u = Q + W$ . This equation represents the mathematical form of the first law of thermodynamics, stating that the change in internal energy of a system is equal to the heat added to the system plus the work done on the system.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: The first law of thermodynamics states that energy cannot be created or destroyed, only converted from one form to another. Mathematically, it is expressed as: Δ U = Q + W, where Δ U is the change in internal energy, Q is the heat added to the system, and W is the work done *on* the system. (Note: some conventions define W as work done *by* the system, in which case the equation becomes Δ U = Q - W).

Step 2: Analyze the given options:
- Option A: ${C_V} = a{T^3}$ . This is the Debye model for the heat capacity of solids at low temperatures, not the first law of thermodynamics.
- Option B: $\Delta {S_{\left( {total} \right)}} = \Delta {S_{Sys}} + \Delta {S_{svr}}$ . This represents the total entropy change, which is related to the second law of thermodynamics.
- Option C: $\Delta u = Q + W$ . This is the correct mathematical form of the first law of thermodynamics.
- Option D: $\Delta {S_{sys}} = \frac{{{q_{rev}}}}{T}$ . This represents the entropy change for a reversible process.

Why other options are wrong:
- Option A: This is related to heat capacity, not the first law.
- Option B: This is related to the second law of thermodynamics (entropy).
- Option D: This is related to entropy change in a reversible process, not the first law.

Key Concepts:
- First law of thermodynamics
- Internal energy
- Heat
- Work

Memory Aids:
- Δ U = Q + W (First Law)

Quick Tips:
- Remember the sign conventions for heat and work: Q is positive when heat is added to the system, and W is positive when work is done *on* the system.

93 Easy

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Analyzing +4 | -1

Which of the following gas will liberate maximum heat on combustion per gram?

View answer and explanation

Correct option: A

Methane

Option A (Methane) is correct because it has the highest hydrogen-to-carbon ratio, leading to the greatest heat release per gram upon combustion.

Solution approach

Correct Answer + Core Reason: Methane. Methane has the highest hydrogen-to-carbon ratio, which leads to the maximum heat release per gram upon combustion.

Step-by-Step Solution:

Step 1: Write the combustion reactions for each gas:\nMethane ( $CH_4$ ): $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ \nEthane ( $C_2H_6$ ): $C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$ \nAcetylene ( $C_2H_2$ ): $C_2H_2 + \frac{5}{2}O_2 \rightarrow 2CO_2 + H_2O$ \nEthylene ( $C_2H_4$ ): $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Step 2: Consider the heat released per mole of each gas (approximate values):\nMethane: -890 kJ/mol\nEthane: -1560 kJ/mol\nAcetylene: -1300 kJ/mol\nEthylene: -1410 kJ/mol

Step 3: Calculate the molar mass of each gas:\nMethane: 16 g/mol\nEthane: 30 g/mol\nAcetylene: 26 g/mol\nEthylene: 28 g/mol

Step 4: Calculate the heat released per gram for each gas:\nMethane: $\frac{890}{16} \approx 55.6$ kJ/g\nEthane: $\frac{1560}{30} = 52$ kJ/g\nAcetylene: $\frac{1300}{26} = 50$ kJ/g\nEthylene: $\frac{1410}{28} \approx 50.4$ kJ/g

Step 5: Compare the values and determine which gas releases the most heat per gram.\nMethane releases the most heat per gram (approximately 55.6 kJ/g).

Step 6: Final Answer: Methane

Why other options are wrong:
- Option B: Ethane: Ethane releases less heat per gram than methane.
- Option C: Acetylene: Acetylene releases less heat per gram than methane.
- Option D: Ethylene: Ethylene releases less heat per gram than methane.

Key Concepts:
- Heat of Combustion
- Molar Mass
- Stoichiometry

Memory Aids:
- Higher hydrogen-to-carbon ratio generally leads to more heat released per gram.

Quick Tips:
- Calculate the heat released per gram by dividing the heat of combustion per mole by the molar mass.
- Compare the values to determine which gas releases the most heat per gram.

94 Easy

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Understanding +4 | -1

In which case, a spontaneous reaction is possible at any temperature?

View answer and explanation

Correct option: A

∆H-ve,∆S+ve

Option A (∆H-ve,∆S+ve) is correct because a negative enthalpy change (exothermic) and a positive entropy change always result in a negative Gibbs free energy change, making the reaction spontaneous at all temperatures.

Solution approach

Correct Answer + Core Reason: ∆H-ve,∆S+ve. This combination always results in a negative Gibbs Free Energy change, making the reaction spontaneous at all temperatures.

Step-by-Step Solution:

Step 1: Gibbs Free Energy Equation: The Gibbs Free Energy change ( $\Delta G$ ) is given by the equation: $\Delta G = \Delta H - T\Delta S$ , where $\Delta H$ is the enthalpy change, T is the temperature, and $\Delta S$ is the entropy change.

Step 2: Spontaneity Criterion: A reaction is spontaneous if $\Delta G < 0$ .

Step 3: Analyzing the Options:
- Option A: $\Delta H < 0$ and $\Delta S > 0$ . In this case, $\Delta G = (-) - T(+)$ , which is always negative, regardless of the temperature.
- Option B: $\Delta H < 0$ and $\Delta S < 0$ . In this case, $\Delta G = (-) - T(-)$ , which can be negative or positive depending on the temperature. Spontaneous at low temperatures.
- Option C: $\Delta H > 0$ and $\Delta S > 0$ . In this case, $\Delta G = (+) - T(+)$ , which can be negative or positive depending on the temperature. Spontaneous at high temperatures.

Step 4: Conclusion: Therefore, a spontaneous reaction is possible at any temperature when $\Delta H$ is negative and $\Delta S$ is positive.

Why other options are wrong:
- Option B: ∆H-ve,∆S-ve: This combination is spontaneous only at low temperatures.
- Option C: ∆H+ve,∆S+ve: This combination is spontaneous only at high temperatures.
- Option D: In none of the cases: This is incorrect because option A provides the correct conditions.

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

Memory Aids:
- Negative H, Positive S = Always Spontaneous

Quick Tips:
- Remember the Gibbs Free Energy equation and how the signs of $\Delta H$ and $\Delta S$ affect spontaneity at different temperatures.

95 Easy

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Understanding +4 | -1

For an ideal gas, the relation between the enthalpy change and internal energy change at constant temperature is given by :

View answer and explanation

Correct option: B

∆H= ∆U+ ∆nRT

Option B (∆H= ∆U+ ∆nRT) is correct because it represents the relationship between enthalpy change and internal energy change for an ideal gas, where ∆n is the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: ∆H= ∆U+ ∆nRT. This equation directly relates enthalpy and internal energy changes for ideal gases.

Step-by-Step Solution:

Step 1: Recall the definition of enthalpy (H) and its relationship to internal energy (U).

Step 2: Enthalpy is defined as: $H = U + PV$

Step 3: For a change at constant temperature, $\Delta H = \Delta U + \Delta (PV)$

Step 4: For an ideal gas, $PV = nRT$ , so $\Delta (PV) = \Delta (nRT) = RT \Delta n$ , where $\Delta n$ is the change in the number of moles of gas.

Step 5: Therefore, $\Delta H = \Delta U + \Delta nRT$

Why other options are wrong:
- Option A: This is incorrect because it represents the relationship between enthalpy, internal energy, pressure, and volume at a specific state, not the change in these quantities.
- Option C: This is incorrect because it rearranges the equation incorrectly.
- Option D: This is incorrect because it relates enthalpy change to Gibbs free energy change and entropy change, which is a different thermodynamic relationship.

Key Concepts:
- Enthalpy
- Internal energy
- Ideal gas law

Memory Aids:
- H = U + PV

Quick Tips:
- Remember that this formula applies specifically to ideal gases.

96 Easy

96 of 196Question permalink

Application +4 | -1

Equal volume of

{C_2}{H_2}

and

{H_2}

are combusted under identical condition. The ratio of their heat of combustion is:

{H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\Delta H = - 241.8\,\,\,kJ

{C_2}{H_2}(g) + 2\frac{1}{2}{O_2}(g) \to 2C{O_2}(g) + {H_2}O(g);\,\,\Delta H = - 1300\,\,\,kJ

View answer and explanation

Correct option: A

5.37/1

Option A (5.37/1) is correct because the ratio of the heat of combustion is calculated by dividing the heat of combustion of C2H2 by the heat of combustion of H2, considering equal volumes imply equal moles.

Solution approach

Correct Answer + Core Reason: 5.37/1. The ratio is calculated by dividing the heat of combustion of acetylene by that of hydrogen.

Step-by-Step Solution:

Step 1: Identify the given heats of combustion:
- For ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\Delta H = - 241.8\,\,\,kJ$
- For ${C_2}{H_2}(g) + 2\frac{1}{2}{O_2}(g) \to 2C{O_2}(g) + {H_2}O(g);\,\,\Delta H = - 1300\,\,\,kJ$

Step 2: Since equal volumes are combusted under identical conditions, it implies equal number of moles.

Step 3: Calculate the ratio of their heat of combustion:
Ratio = $\frac{{\Delta H_{C_2H_2}}}{{\Delta H_{H_2}}} = \frac{{1300}}{{241.8}} \approx 5.37$

Step 4: Express the ratio as a fraction: 5.37/1

Why other options are wrong:
- Option B: 1/5.37 is the inverse of the correct ratio.
- Option C: 44562 is not a reasonable answer based on the given data. It's likely a calculation error.
- Option D: None of these is incorrect because option A is correct.

Key Concepts:
- Heat of combustion
- Stoichiometry
- Molar volume

Memory Aids:
- Remember that equal volumes at the same conditions imply equal moles (Avogadro's Law).

Quick Tips:
- Always pay attention to the stoichiometry of the reactions.
- Ensure the units are consistent before calculating the ratio.

97 Easy

97 of 196Question permalink

Application +4 | -1

The change in internal energy, if

60 {~J}

of heat is given to the system at constant pressure and

30 {~J}

of work is done by the system, will be:

View answer and explanation

Correct option: D

+30 {~J}

Option D (+30 J) is correct because the change in internal energy is the difference between the heat added to the system and the work done by the system.

Solution approach

Correct Answer + Core Reason: $+30 {~J}$ . The change in internal energy is the difference between heat added and work done by the system.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta U = q - w$

Step 2: Identify the given values: $q = +60 \text{{ J}}$ (heat given to the system), $w = +30 \text{{ J}}$ (work done by the system).

Step 3: Substitute the values into the equation: $\Delta U = 60 \text{{ J}} - 30 \text{{ J}}$

Step 4: Calculate the change in internal energy: $\Delta U = 30 \text{{ J}}$

Why other options are wrong:
- Option A: $-90 {~J}$ . Incorrect, this would be the case if the work was done *on* the system instead of *by* the system, and the heat was leaving the system.
- Option B: $+90 {~J}$ . Incorrect, this would be the case if the work was done *on* the system instead of *by* the system.
- Option C: $-30 {~J}$ . Incorrect, this would be the case if the work was done *on* the system instead of *by* the system, or if the heat was leaving the system.

Key Concepts:
- First law of thermodynamics, Internal energy, Heat, Work

Memory Aids:
- Remember the sign convention: heat added to the system is positive, heat removed is negative; work done by the system is positive, work done on the system is negative.

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

98 Easy

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Understanding +4 | -1

Equilibrium constant of a reaction is related to:

View answer and explanation

Correct option: A

Standard Gibbs energy change ∆G°

Option A (Standard Gibbs energy change ∆G°) is correct because the equilibrium constant is directly related to the standard Gibbs free energy change by the equation \

∆G&deg; = -RTlnK

.

Solution approach

Correct Answer + Core Reason: Standard Gibbs energy change ∆G°. The equilibrium constant is related to the standard Gibbs free energy change through the equation $\Delta G^{\circ} = -RT \ln K$ .

Step-by-Step Solution:

Step 1: Recall the relationship between the equilibrium constant (K) and the Gibbs free energy change.

Step 2: The standard Gibbs free energy change is related to the equilibrium constant by the equation: $\Delta G^{\circ} = -RT \ln K$ .

Step 3: This equation shows that K is directly related to $\Delta G^{\circ}$ .

Why other options are wrong:
- Option B: Gibbs energy change ∆G is incorrect because it is the Gibbs free energy change under non-standard conditions, not directly related to K without further information.
- Option C: Heat enthalpy is incorrect because while enthalpy changes can influence equilibrium, the direct relationship is with Gibbs free energy.
- Option D: None of the above is incorrect because option A is the correct answer.

Key Concepts:
- Equilibrium Constant
- Gibbs Free Energy

Memory Aids:
- Remember the equation: $\Delta G^{\circ} = -RT \ln K$ .

Quick Tips:
- Always use the standard Gibbs free energy change when relating it to the equilibrium constant.

99 Easy

99 of 196Question permalink

Application +4 | -1

For a gas

\left( {R/{C_v}} \right) = 0.67

the gas is made up of molecules which are:

View answer and explanation

Correct option: A

Monoatomic

Option A (Monoatomic) is correct because for a monoatomic gas, the ratio R/Cv = 2/3 = 0.67.

Solution approach

Correct Answer + Core Reason: Monoatomic. For a monoatomic gas, R/Cv = 0.67.

Step-by-Step Solution:

Step 1: Recall the relationship between $C_p$ , $C_v$ , and R: $C_p - C_v = R$

Step 2: Given $\frac{R}{{{C_v}}} = 0.67$ , we need to determine the type of gas.

Step 3: Consider different types of gases:
- Monoatomic: $C_v = \frac{3}{2}R$ , so $\frac{R}{{{C_v}}} = \frac{R}{{\frac{3}{2}R}} = \frac{2}{3} \approx 0.67$
- Diatomic: $C_v = \frac{5}{2}R$ , so $\frac{R}{{{C_v}}} = \frac{R}{{\frac{5}{2}R}} = \frac{2}{5} = 0.4$
- Polyatomic: $C_v > \frac{5}{2}R$ , so $\frac{R}{{{C_v}}} < 0.4$

Step 4: Conclusion: Since $\frac{R}{{{C_v}}} = 0.67$ , the gas is monoatomic.

Why other options are wrong:
- Option B: Diatomic gases have R/Cv = 0.4, not 0.67.
- Option C: Polyatomic gases have R/Cv < 0.4, not 0.67.
- Option D: A mixture of gases would have a different R/Cv ratio depending on the composition, but it's not the most direct answer given the information.

Key Concepts:
- Molar heat capacity at constant volume ( $C_v$ )
- Ideal gas constant (R)
- Degrees of freedom for different types of gases

Memory Aids:
- Monoatomic: $C_v = \frac{3}{2}R$
- Diatomic: $C_v = \frac{5}{2}R$

Quick Tips:
- Remember the formulas for $C_v$ for different types of gases.

100 Easy

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Analyzing +4 | -1

Statement I Due to atomisation of gaseous molecule,
entropy increases.
Statement II Atoms are more ordered than the molecules.

View answer and explanation

Correct option: C

Statement I is correct but Statement II is incorrect

Statement I is correct, but Statement II is incorrect. Atomization increases entropy because atoms have more freedom than molecules.

Solution approach

Correct Answer + Core Reason: Statement I is correct but Statement II is incorrect. Atomization increases entropy, and atoms are less ordered than molecules.

Step-by-Step Solution:

Step 1: Analyze Statement I: \"Due to atomisation of gaseous molecule, entropy increases.\"\nAtomization is the process of breaking a molecule into its constituent atoms. When a molecule breaks into individual atoms, the atoms have more freedom of movement and occupy a larger volume. This leads to an increase in disorder and, therefore, an increase in entropy. So, Statement I is correct.

Step 2: Analyze Statement II: \"Atoms are more ordered than the molecules.\"\nAtoms are individual particles, while molecules are formed by bonding atoms together. Molecules have specific structures and arrangements, which impose constraints on their movement. Atoms, on the other hand, have more freedom of movement and are less constrained. Therefore, atoms are less ordered than molecules. So, Statement II is incorrect.

Why other options are wrong:
- Option A: Both statements are not correct.
- Option B: Both statements are not correct.
- Option D: Statement II is not correct.

Key Concepts:
- Entropy
- Atomization
- Molecular Structure

Memory Aids:
- Entropy is a measure of disorder.
- Atomization increases disorder.

Quick Tips:
- Remember that entropy increases with increasing disorder and freedom of movement.

101 Easy

101 of 196Question permalink

Remembering +4 | -1

In which state matter have highest entropy?

View answer and explanation

Correct option: C

Gas

Option C is correct. Gases have the highest entropy because their molecules have the greatest freedom of movement and occupy the largest volume, leading to the highest degree of disorder.

Solution approach

Correct Answer + Core Reason: Gas. Gases have the highest entropy due to the greatest molecular freedom.

Step-by-Step Solution:
Step 1: Define Entropy: Entropy is a measure of disorder or randomness in a system.
Step 2: Compare Molecular Arrangement in Different States of Matter:
- Solids: Molecules are tightly packed in a fixed arrangement, resulting in low entropy.
- Liquids: Molecules are more loosely packed than in solids, allowing for some movement and disorder, resulting in higher entropy than solids.
- Gases: Molecules are widely dispersed and move randomly, resulting in the highest entropy.
Step 3: Conclusion: Gases have the highest entropy because their molecules have the greatest freedom of movement and occupy the largest volume, leading to the highest degree of disorder.

Why other options are wrong:
- Option A: Solids have the lowest entropy due to their ordered structure.
- Option B: Liquids have intermediate entropy, higher than solids but lower than gases.
- Option D: Entropy is not equal in all states; it varies based on the degree of disorder.

Key Concepts:
- Entropy
- States of matter (solid, liquid, gas)
- Molecular arrangement and disorder

Memory Aids:
- Gas = Greatest disorder = Highest entropy.

Quick Tips:
- Remember the order of increasing entropy: Solid < Liquid < Gas.

102 Easy

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Remembering +4 | -1

The work done by a system in an expansion against a constant external pressure is:

View answer and explanation

Correct option: B

- P.\Delta V

Option B is correct because the work done by a system during expansion against a constant external pressure is given by -PΔV.

Solution approach

Correct Answer + Core Reason: $- P.\Delta V$ . This is the formula for work done against constant external pressure.

Step-by-Step Solution:

Step 1: Definition of Work: Work is defined as force times distance.

Step 2: Work done by a gas: When a gas expands against an external pressure, it does work.

Step 3: Formula for work done at constant pressure: $w = -P_{ext} \Delta V$ , where $P_{ext}$ is the external pressure and $\Delta V$ is the change in volume.

Step 4: Conclusion: The work done by a system in an expansion against a constant external pressure is $- P.\Delta V$ .

Why other options are wrong:
- Option A: Incorrect; $\Delta P.\Delta V$ is not the correct expression for work done at constant external pressure.
- Option C: Incorrect; Q represents heat, not work.
- Option D: Incorrect; $V.\Delta P$ is not the correct expression for work done at constant external pressure.

Key Concepts:
- Work done by a gas
- Constant external pressure

Memory Aids:
- Work = -PΔV (at constant pressure)

Quick Tips:
- Remember the sign convention for work: work done *by* the system is negative.

103 Easy

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Remembering +4 | -1

A thermodynamic state function is:

View answer and explanation

Correct option: C

A quantity whose value is independent of the path

Option C is correct because a thermodynamic state function's value depends only on the initial and final states, not the path taken.

Solution approach

Correct Answer + Core Reason: A quantity whose value is independent of the path. This is the defining characteristic of a state function.

Step-by-Step Solution:

Step 1: Define a thermodynamic state function. A thermodynamic state function is a property of a system that depends only on the current state of the system, not on how the system reached that state.

Step 2: Analyze the options.

Step 3: Option C correctly describes a state function as a quantity whose value is independent of the path.

Why other options are wrong:
- Option A: While state functions are used in thermodynamics, obeying the laws of thermodynamics is not the defining characteristic of a state function.
- Option B: State functions can be used to measure thermal changes, but this is not their defining characteristic.
- Option D: State functions can be used to express pressure-volume work, but this is not their defining characteristic.

Key Concepts:
- State Function

Memory Aids:
- State functions: Value depends only on the state, not the path.

Quick Tips:
- Remember the key difference between state and path functions.

104 Easy

104 of 196Question permalink

Understanding +4 | -1

For the reaction,

{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right),

which is true?

View answer and explanation

Correct option: B

∆H< ∆U

Option B, ∆H< ∆U, is correct because for the given reaction, the number of moles of gas decreases, leading to a negative value for

\Delta n_g

, and thus ∆H< ∆U.

Solution approach

Correct Answer + Core Reason: ∆H< ∆U. For this reaction, the change in the number of moles of gas is negative, making ∆H less than ∆U.

Step-by-Step Solution:

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

Step 2: Calculate $\Delta n_g$ for the reaction: ${\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)$ .
$\Delta n_g = n_{products} - n_{reactants} = 2 - (1 + 3) = 2 - 4 = -2$

Step 3: Since $\Delta n_g$ is negative, $\Delta n_g RT$ is also negative.

Step 4: Therefore, $\Delta H = \Delta U + (-2)RT = \Delta U - 2RT$

Step 5: This means that $\Delta H < \Delta U$

Step 6: Final Answer: ∆H< ∆U

Why other options are wrong:
- Option A: ∆H= ∆U is incorrect because $\Delta n_g$ is not zero.
- Option C: ∆H> ∆U is incorrect because $\Delta n_g$ is negative, making $\Delta H$ less than $\Delta U$ .
- Option D: None of these is incorrect because option B is correct.

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

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

Quick Tips:
- Pay attention to the change in the number of moles of gas when relating $\Delta H$ and $\Delta U$ .

105 Easy

105 of 196Question permalink

Application +4 | -1

For the melting of

NaCl

heat required is

7.26\,K\,cal\,mo{l^{ - 1}}

and

\Delta S

increases by

6.73\,cal\,mo{l^{ - 1}}.{K^{ - 1}}.

The melting point of the salt is

View answer and explanation

Correct option: A

805.75^\circ C

Option A (

805.75^\circ C

) is correct because the melting point is calculated using the formula

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

and then converted from Kelvin to Celsius.

Solution approach

Correct Answer + Core Reason: $805.75^\circ C$ . The melting point is calculated using $T = \frac{{\Delta H}}{{\Delta S}}$ and converted to Celsius.

Step-by-Step Solution:

Step 1: Use the formula for melting point: $T = \frac{{\Delta H}}{{\Delta S}}$

Step 2: Substitute the given values: $T = \frac{{7.26 \times 10^3 \,cal/mol}}{{6.73 \,cal/mol.K}} = 1078.75 \,K$

Step 3: Convert Kelvin to Celsius: $T(^\circ C) = T(K) - 273 = 1078.75 - 273 = 805.75 ^\circ C$

Why other options are wrong:
- Option B (500 K): This is incorrect because it doesn't use the correct formula to calculate the melting point.
- Option C (1.77 K): This is incorrect because it seems to be a result of dividing $\Delta S$ by $\Delta H$ instead of the other way around.
- **Option D ( $1.77^\circ C$ ):** This is incorrect because it's the result of dividing $\Delta S$ by $\Delta H$ and then converting to Celsius, which is not the correct approach.

Key Concepts:
- Melting point calculation: $T = \frac{{\Delta H}}{{\Delta S}}$
- Conversion between Kelvin and Celsius: $T(^\circ C) = T(K) - 273$

Memory Aids:
- Remember the formula for melting point and the conversion between Kelvin and Celsius.

Quick Tips:
- Ensure the units are consistent before performing the calculation.
- Always convert the temperature to the required unit (Celsius or Kelvin) at the end.

106 Easy

106 of 196Question permalink

Remembering +4 | -1

Which of the following is an example of extensive property?

View answer and explanation

Correct option: C

Mass

Option C (Mass) is correct because mass is an extensive property, meaning it depends on the amount of substance, while temperature, density, and pressure are intensive properties.

Solution approach

Correct Answer + Core Reason: Mass. Mass is an extensive property because it depends on the amount of substance.

Step-by-Step Solution:

Step 1: Define an extensive property. An extensive property is a property that depends on the amount of matter in the system (e.g., mass, volume, enthalpy).

Step 2: Define an intensive property. An intensive property is a property that does not depend on the amount of matter in the system (e.g., temperature, pressure, density).

Step 3: Analyze the options.

Step 4: Mass is an extensive property.

Why other options are wrong:
- Option A: Temperature is an intensive property.
- Option B: Density is an intensive property.
- Option D: Pressure is an intensive property.

Key Concepts:
- Extensive Property
- Intensive Property

Memory Aids:
- Extensive: Depends on amount
- Intensive: Independent of amount

Quick Tips:
- Think about whether doubling the amount of substance doubles the property. If yes, it's extensive.

107 Easy

107 of 196Question permalink

Understanding +4 | -1

A gas is allowed to expand under reversible adiabatic conditions then :-

View answer and explanation

Correct option: C

\Delta {S}=0

Option C (

\Delta {S}=0

) is correct because in a reversible adiabatic process, the change in entropy is zero.

Solution approach

Correct Answer + Core Reason: $\Delta {S}=0$ . In a reversible adiabatic process, the change in entropy is zero because there is no heat transfer.

Step-by-Step Solution:

Step 1: Define reversible adiabatic conditions: A reversible adiabatic process is one that is both adiabatic (no heat exchange with the surroundings, $q = 0$ ) and reversible (occurs infinitesimally slowly).

Step 2: Entropy change formula: The change in entropy ( $\Delta S$ ) is given by $\Delta S = \frac{q_{rev}}{T}$ , where $q_{rev}$ is the heat exchanged in a reversible process and $T$ is the temperature.

Step 3: Apply to adiabatic process: Since $q = 0$ in an adiabatic process, $\Delta S = \frac{0}{T} = 0$ .

Why other options are wrong:
- Option A: $\Delta {G}=0$ . Gibbs free energy change is not necessarily zero in an adiabatic expansion. It depends on the specific conditions and the nature of the gas.
- Option B: $\Delta {T}=0$ . Temperature changes in an adiabatic expansion. As the gas expands, it does work, which causes the internal energy and thus the temperature to decrease.
- Option D: None of these. Option C is correct.

Key Concepts:
- Adiabatic process
- Reversible process
- Entropy

Memory Aids:
- Adiabatic: No heat exchange.
- Reversible: Infinitesimally slow.

Quick Tips:
- Remember that adiabatic processes imply no heat transfer, which directly relates to entropy change.

108 Easy

108 of 196Question permalink

Application +4 | -1

What is the entropy change in

\mathrm{JK}^{-1}

during the melting of

27.3 \mathrm{~g}

of ice at

0^{\circ} \mathrm{C}

? (Latent heat of fusion of ice

=330 \mathrm{Jg}^{-1}

)

View answer and explanation

Correct option: D

33

Option D (33) is correct because the entropy change is calculated by dividing the heat absorbed during melting by the temperature in Kelvin.

Solution approach

Correct Answer + Core Reason: 33. Entropy change is calculated using $\Delta S = \frac{Q}{T}$ .

Step-by-Step Solution:

Step 1: Identify the given information:
- Mass of ice, $m = 27.3\, \mathrm{g}$
- Temperature, $T = 0^{\circ} \mathrm{C} = 273\, \mathrm{K}$
- Latent heat of fusion, $L_f = 330\, \mathrm{J/g}$

Step 2: Calculate the total heat absorbed during melting:
$Q = m \times L_f = 27.3\, \mathrm{g} \times 330\, \mathrm{J/g} = 9009\, \mathrm{J}$

Step 3: Calculate the entropy change:
$\Delta S = \frac{Q}{T} = \frac{9009\, \mathrm{J}}{273\, \mathrm{K}} = 33\, \mathrm{J/K}$

Why other options are wrong:
- Option A: 330 is the latent heat of fusion, not the entropy change.
- Option B: 12.1 is likely an incorrect calculation.
- Option C: 3.3 is likely an incorrect calculation.

Key Concepts:
- Entropy change
- Latent heat of fusion

Memory Aids:
- Entropy change: $\Delta S = \frac{Q}{T}$

Quick Tips:
- Always convert temperature to Kelvin when calculating entropy change.
- Ensure the units are consistent.

109 Easy

109 of 196Question permalink

Remembering +4 | -1

Hess’s law is related to:

View answer and explanation

Correct option: A

Change in heat during a reaction

Option A is correct because Hess's law states that the change in enthalpy during a reaction depends only on the initial and final states, not the path taken.

Solution approach

Correct Answer + Core Reason: Change in heat during a reaction. Hess's Law states that the enthalpy change of a reaction is independent of the path taken, meaning it is related to the change in heat during the reaction.

Step-by-Step Solution:

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

Step 2: Relate Hess's Law to the given options:
- Option A: Change in heat during a reaction. This is directly related to Hess's Law, as it deals with the enthalpy change (heat change at constant pressure) of a reaction.
- Option B: Rates of reaction. Hess's Law does not directly relate to the rates of reaction, which are studied in chemical kinetics.
- Option C: Equilibrium constant. Hess's Law can be used to calculate the enthalpy change, which can then be used to determine the temperature dependence of the equilibrium constant (using the van't Hoff equation), but it doesn't directly relate to the equilibrium constant itself.
- Option D: Influence of pressure on volume of a gas. This is related to gas laws (e.g., Boyle's Law) and not directly to Hess's Law.

Why other options are wrong:
- Option B: Hess's Law does not deal with reaction rates.
- Option C: Hess's Law is indirectly related to the equilibrium constant through thermodynamics, but it doesn't directly determine it.
- Option D: Hess's Law does not describe the relationship between pressure and volume of a gas.

Key Concepts:
- Hess's Law
- Enthalpy change

Memory Aids:
- Hess's Law: The enthalpy change is a state function.

Quick Tips:
- Remember that Hess's Law is a consequence of enthalpy being a state function.

110 Easy

110 of 196Question permalink

Understanding +4 | -1

Which is not a spontaneous process?

View answer and explanation

Correct option: C

Heat flowing from colder body to a hotter body

Option C (Heat flowing from colder body to a hotter body) is correct because heat naturally flows from a hotter body to a colder body, not the other way around. This violates the second law of thermodynamics.

Solution approach

Correct Answer + Core Reason: Heat flowing from colder body to a hotter body. This violates the second law of thermodynamics.

Step-by-Step Solution:

Step 1: Define a spontaneous process as one that occurs without external intervention.

Step 2: Consider the second law of thermodynamics, which states that entropy tends to increase in a closed system.

Step 3: Recognize that heat flowing from a colder body to a hotter body would decrease entropy, requiring external work and thus not being spontaneous.

Why other options are wrong:
- Option A: Expansion of a gas into a vacuum is a spontaneous process as it increases entropy.
- Option B: Water flowing downhill is a spontaneous process due to gravity.
- Option D: Evaporation of water from clothes during drying is a spontaneous process as it increases entropy.

Key Concepts:
- Spontaneous process
- Second law of thermodynamics
- Entropy

Memory Aids:
- Spontaneous processes increase entropy.

Quick Tips:
- Consider the entropy change for each process.

111 Easy

111 of 196Question permalink

Understanding +4 | -1

For the precipitation of AgCl by

\mathrm{A}{\mathrm{g}^ + }

ions and HCl:

View answer and explanation

Correct option: C

∆G=-ve

Option C (∆G=-ve) is correct because the precipitation of AgCl is a spontaneous process under normal conditions, and spontaneous processes have a negative change in Gibbs free energy.

Solution approach

Correct Answer + Core Reason: ∆G=-ve. Precipitation reactions are generally spontaneous, and spontaneous reactions have a negative Gibbs free energy change.

Step-by-Step Solution:

Step 1: Recall that a spontaneous process has a negative change in Gibbs free energy ( $\Delta G < 0$ ).

Step 2: The precipitation of AgCl from $\mathrm{A}{\mathrm{g}^ + }$ ions and HCl is a spontaneous process under normal conditions.

Step 3: Therefore, $\Delta G$ for this reaction is negative.

Why other options are wrong:
- Option A: ∆H=0 is incorrect because the enthalpy change is not necessarily zero for precipitation reactions.
- Option B: ∆G=0 is incorrect because $\Delta G = 0$ indicates equilibrium, not a spontaneous precipitation.
- Option D: ∆H=∆G is incorrect because there is no general relationship stating that enthalpy change equals Gibbs free energy change.

Key Concepts:
- Gibbs Free Energy
- Spontaneity
- Precipitation Reactions

Memory Aids:
- Spontaneous reactions have negative ∆G.

Quick Tips:
- Remember that precipitation reactions are typically spontaneous.

112 Easy

112 of 196Question permalink

Remembering +4 | -1

Calorie is equivalent to:

View answer and explanation

Correct option: B

4.184 Joule

Option B (4.184 Joule) is correct because 1 calorie is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius, which is equivalent to 4.184 Joules.

Solution approach

Correct Answer + Core Reason: 4.184 Joule. 1 calorie is equivalent to 4.184 Joules.

Step-by-Step Solution:

Step 1: Recall the definition of a calorie.\nA calorie is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.

Step 2: Recall the conversion factor between calories and Joules.
1 calorie = 4.184 Joules

Why other options are wrong:
- Option A: 0.4184 Joule. Incorrect conversion factor.
- Option C: 41.84 Joule. Incorrect conversion factor.
- Option D: 418.4 Joule. Incorrect conversion factor.

Key Concepts:
- Calorie
- Joule
- Energy conversion

Memory Aids:
- 1 calorie = 4.184 Joules

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

113 Easy

113 of 196Question permalink

Understanding +4 | -1

The value of

\Delta E

for an open and isolated system respectively is

View answer and explanation

Correct option: C

Non-zero; zero

Option C (Non-zero; zero) is correct because an open system can exchange both energy and matter with the surroundings, so its internal energy can change, while an isolated system cannot exchange either, so its internal energy remains constant.

Solution approach

Correct Answer + Core Reason: Non-zero; zero. Open systems can exchange energy, isolated systems cannot.

Step-by-Step Solution:

Step 1: Define an open system as one that can exchange both energy and matter with its surroundings.

Step 2: Define an isolated system as one that can exchange neither energy nor matter with its surroundings.

Step 3: Recognize that if a system can exchange energy, its internal energy can change (non-zero Δ E). If a system cannot exchange energy, its internal energy remains constant (zero Δ E).

Why other options are wrong:
- Option A: Zero; non-zero is incorrect because the open system can have a non-zero Δ E and the isolated system has a zero Δ E.
- Option B: Zero is incorrect because it only describes the isolated system.
- Option D: Non-zero is incorrect because it only describes the open system.

Key Concepts:
- Open system
- Isolated system
- Internal energy (E)

Memory Aids:
- Open = exchange, Isolated = no exchange.

Quick Tips:
- Consider the definitions of open, closed, and isolated systems.

114 Easy

114 of 196Question permalink

Application +4 | -1

If

{\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } \to {\mathrm{H}_2}\mathrm{O} + 13.7

kcal then the heat of neutralization for complete neutralization of one mole of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

by a base will be :

View answer and explanation

Correct option: B

- 27.4\mathrm{kcal}

Option B (-27.4 kcal) is correct because the heat of neutralization for a strong acid and strong base is additive, and sulfuric acid has two acidic protons.

Solution approach

Correct Answer + Core Reason: $- 27.4\mathrm{kcal}$ . The heat of neutralization for two moles of ${\mathrm{H}^ + }$ is twice that of one mole.

Step-by-Step Solution:
Step 1: Understand that the heat of neutralization for ${\mathrm{H}^ + } + \mathrm{O}{\mathrm{H}^ - } \to {\mathrm{H}_2}\mathrm{O}$ is -13.7 kcal.
Step 2: Recognize that ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ is a diprotic acid, meaning it has two ionizable protons (2 ${\mathrm{H}^ + }$ ).
Step 3: For complete neutralization of one mole of ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ , two moles of ${\mathrm{H}^ + }$ will react with two moles of ${\mathrm{O}{\mathrm{H}^ - }}$ .
Step 4: The total heat released will be twice the heat released for one mole of ${\mathrm{H}^ + }$ :
$2 \times (-13.7 \text{ kcal}) = -27.4 \text{ kcal}$

Why other options are wrong:
- Option A: This is the heat of neutralization for only one mole of ${\mathrm{H}^ + }$ , not for the complete neutralization of ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ .
- Option C: This is half the heat of neutralization for one mole of ${\mathrm{H}^ + }$ , which is incorrect.
- Option D: This is one-fourth the heat of neutralization for one mole of ${\mathrm{H}^ + }$ , which is incorrect.

Key Concepts:
- Heat of neutralization
- Strong acid-strong base neutralization
- Diprotic acids

Memory Aids:
- Remember that the heat of neutralization is additive for multiple moles of ${\mathrm{H}^ + }$ .

Quick Tips:
- Identify the number of ionizable protons in the acid to determine the total heat released during neutralization.

115 Easy

115 of 196Question permalink

Understanding +4 | -1

All the naturally occurring processes, i.e., spontaneous proceed spontaneously in a direction which leads to:

View answer and explanation

Correct option: A

Decrease of free energy

Option A (Decrease of free energy) is correct because spontaneous processes always proceed in a direction that lowers the Gibbs free energy of the system.

Solution approach

Correct Answer + Core Reason: Decrease of free energy. Spontaneous processes occur in the direction of decreasing Gibbs Free Energy.

Step-by-Step Solution:

Step 1: Spontaneous Process: A spontaneous process is a process that occurs without any external intervention.

Step 2: Gibbs Free Energy and Spontaneity: The Gibbs Free Energy (G) is a thermodynamic potential that can be used to predict the spontaneity of a process at constant temperature and pressure.

Step 3: Criterion for Spontaneity: For a spontaneous process, the change in Gibbs Free Energy ( $\Delta G$ ) must be negative ( $\Delta G < 0$ ). This means that the process proceeds in a direction that leads to a decrease in free energy.

Step 4: Conclusion: Therefore, naturally occurring processes proceed spontaneously in a direction that leads to a decrease of free energy.

Why other options are wrong:
- Option B: Increase of free energy: An increase in free energy indicates a non-spontaneous process.
- Option C: Decrease of entropy: While entropy often increases in spontaneous processes, it is not a universal requirement. The Gibbs Free Energy considers both enthalpy and entropy.
- Option D: Increase of enthalpy: An increase in enthalpy indicates an endothermic process, which may or may not be spontaneous depending on the entropy change and temperature.

Key Concepts:
- Gibbs Free Energy (G)
- Spontaneity
- Thermodynamics

Memory Aids:
- Spontaneous = Negative Delta G

Quick Tips:
- Remember that the Gibbs Free Energy equation is $\Delta G = \Delta H - T\Delta S$ . Spontaneity depends on the balance between enthalpy and entropy changes.

116 Easy

116 of 196Question permalink

Understanding +4 | -1

At constant pressure and temperature, the direction of any chemical reaction is one where, the … decrease.

View answer and explanation

Correct option: C

Gibbs energy

Option C (Gibbs energy) is correct because at constant pressure and temperature, the direction of a spontaneous chemical reaction is towards a decrease in Gibbs free energy.

Solution approach

Correct Answer + Core Reason: Gibbs energy. Spontaneous reactions at constant pressure and temperature proceed towards decreasing Gibbs energy.

Step-by-Step Solution:

Step 1: Gibbs Free Energy: The Gibbs Free Energy (G) is a thermodynamic potential that combines enthalpy (H) and entropy (S) to determine the spontaneity of a process at constant temperature and pressure.

Step 2: Spontaneity Criterion: At constant pressure and temperature, a chemical reaction is spontaneous if the change in Gibbs Free Energy ( $\Delta G$ ) is negative ( $\Delta G < 0$ ).

Step 3: Direction of Reaction: The direction of a spontaneous reaction is the one that leads to a decrease in Gibbs Free Energy.

Step 4: Conclusion: Therefore, at constant pressure and temperature, the direction of any chemical reaction is one where the Gibbs energy decreases.

Why other options are wrong:
- Option A: Entropy: While entropy often increases in spontaneous processes, it is not the sole determining factor at constant pressure and temperature. Gibbs Free Energy considers both enthalpy and entropy.
- Option B: Enthalpy: Enthalpy changes are important, but the Gibbs Free Energy, which includes both enthalpy and entropy, determines spontaneity at constant pressure and temperature.
- Option D: None of these: This is incorrect because Gibbs energy is the correct answer.

Key Concepts:
- Gibbs Free Energy (G)
- Spontaneity
- Thermodynamics

Memory Aids:
- Spontaneous = Negative Delta G (at constant P and T)

Quick Tips:
- Remember the Gibbs Free Energy equation: $\Delta G = \Delta H - T\Delta S$ . Spontaneity depends on the balance between enthalpy and entropy changes at constant temperature and pressure.

117 Easy

117 of 196Question permalink

Application +4 | -1

Heat of formation of

CO

and

C{O_2}

are -26.4, -94.0 Kcal/mole respectively. What is the heat of combustion of CO in K cals?

View answer and explanation

Correct option: C

- 67.6

Option C (

- 67.6

) is correct because the heat of combustion of CO is calculated using the heats of formation of CO and CO2 and applying Hess's Law.

Solution approach

Correct Answer + Core Reason: $- 67.6$ . The heat of combustion is calculated using Hess's Law.

Step-by-Step Solution:

Step 1: Write the reaction for the combustion of CO: $CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)$

Step 2: Use Hess's Law: $\Delta H_{combustion} = \Delta H_{f(CO_2)} - \Delta H_{f(CO)}$

Step 3: Substitute the given values: $\Delta H_{combustion} = -94.0 - (-26.4) = -94.0 + 26.4 = -67.6$ Kcal/mole

Why other options are wrong:
- Option A: 26.4 is the heat of formation of CO with the wrong sign. It doesn't account for the formation of CO2.
- Option B: 120.6 is incorrect. It's likely a result of adding the heats of formation instead of subtracting.
- Option D: 135.2 is incorrect. It's likely a result of adding the absolute values of the heats of formation and making a sign error.

Key Concepts:
- Hess's Law
- Heat of combustion
- Heat of formation

Memory Aids:
- $\Delta H_{reaction} = \sum \Delta H_{f(products)} - \sum \Delta H_{f(reactants)}$

Quick Tips:
- Remember to subtract the heats of formation of reactants from the heats of formation of products.

118 Easy

118 of 196Question permalink

Application +4 | -1

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at

25^{\circ} \mathrm{C}

is

View answer and explanation

Correct option: B

298 \times 10^7 \times 8.31 \times 2.303 \log 2

Option B (

298 \times 10^7 \times 8.31 \times 2.303 \log 2

) is correct because it represents the work done for a reversible expansion, calculated using the formula

W = -2.303nRT\log(V_2/V_1)

with R in ergs.

Solution approach

Correct Answer + Core Reason: $298 \times 10^7 \times 8.31 \times 2.303 \log 2$ . The work done for reversible expansion is calculated using $W = -2.303nRT\log(\frac{V_2}{V_1})$ .

Step-by-Step Solution:

Step 1: Identify the given information.\nMoles of gas, n = 1 mol\nInitial volume, $V_1$ = 10 L\nFinal volume, $V_2$ = 20 L\nTemperature, T = 25°C = 298 K\nR = 8.314 J/mol.K = $8.314 \times 10^7$ erg/mol.K

Step 2: Apply the formula for work done in a reversible expansion.\nW = $-2.303nRT\log(\frac{V_2}{V_1})$

Step 3: Substitute the given values.\nW = $-2.303 \times 1 \times 8.314 \times 10^7 \times 298 \times \log(\frac{20}{10})$ \nW = $-2.303 \times 8.314 \times 10^7 \times 298 \times \log(2)$ \nSince the question asks for the magnitude of work done, we can ignore the negative sign.\nW = $2.303 \times 8.314 \times 10^7 \times 298 \times \log(2)$ ergs

Why other options are wrong:
- Option A: This uses R in L atm, not ergs. Also, it's missing the $10^7$ factor for converting J to ergs.
- Option C: This uses $\log(0.5)$ instead of $\log(2)$ , and also uses R in L atm.
- Option D: This is a combination of incorrect values and operations.

Key Concepts:
- Reversible expansion
- Work done in reversible process
- Ideal gas constant in different units

Memory Aids:
- $W = -2.303nRT\log(\frac{V_2}{V_1})$

Quick Tips:
- Pay attention to the units of R and ensure consistency.

119 Easy

119 of 196Question permalink

Understanding +4 | -1

Standard state Gibbs free energy changes for the isomerization reaction,cis-2-pentene⇌trans-2-pentene is -3.67 kJ/mol at 400K. If more trans-2-pentene is added to the reaction vessel, then:

View answer and explanation

Correct option: C

More cis-2-pentene is formed

Option C (More cis-2-pentene is formed) is correct because adding more trans-2-pentene will shift the equilibrium towards the reactants (cis-2-pentene) to counteract the change, according to Le Chatelier's principle.

Solution approach

Correct Answer + Core Reason: More cis-2-pentene is formed. According to Le Chatelier's principle, adding more product will shift the equilibrium towards the reactants.

Step-by-Step Solution:

Step 1: State the reaction: cis-2-pentene ⇌ trans-2-pentene

Step 2: Apply Le Chatelier's principle. Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Step 3: Analyze the effect of adding trans-2-pentene. Adding more trans-2-pentene (product) will shift the equilibrium to the left, favoring the formation of cis-2-pentene (reactant).

Step 4: Conclusion: Therefore, more cis-2-pentene is formed.

Why other options are wrong:
- Option A: Equilibrium remains unaffected: This is incorrect because adding more product will shift the equilibrium.
- Option B: Additional trans-2-pentene is formed: This is incorrect because adding more trans-2-pentene will shift the equilibrium towards the reactants.
- Option D: Equilibrium is shifted in forward direction: This is incorrect because adding more trans-2-pentene will shift the equilibrium in the reverse direction.

Key Concepts:
- Le Chatelier's principle
- Chemical equilibrium

Memory Aids:
- Le Chatelier's principle: System shifts to relieve stress.

Quick Tips:
- Remember Le Chatelier's principle and how it applies to changes in concentration, pressure, and temperature.

120 Easy

120 of 196Question permalink

Application +4 | -1

What is the melting point of benzene if

\Delta H_{\text {fusion }}=9.95 \mathrm{~kJ} / \mathrm{mol}

and

\Delta S_{\text {fusion }}=35.7 \mathrm{~J} / \mathrm{K}-\mathrm{mol}

?

View answer and explanation

Correct option: B

278.7 \mathrm{~K}

Option B (

278.7 \mathrm{~K}

) is correct because the melting point is calculated using the formula

T = \frac{\Delta H_{\text {fusion }}}{\Delta S_{\text {fusion }}}

.

Solution approach

Correct Answer + Core Reason: $278.7 \mathrm{~K}$ . The melting point is calculated using $T = \frac{\Delta H_{\text {fusion }}}{\Delta S_{\text {fusion }}}$ .

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H_{\text {fusion }} = 9.95 \mathrm{~kJ} / \mathrm{mol} = 9950 \mathrm{~J} / \mathrm{mol}$ and $\Delta S_{\text {fusion }} = 35.7 \mathrm{~J} / (\mathrm{K} \cdot \mathrm{mol})$ .

Step 2: Apply the formula for melting point: $T = \frac{\Delta H_{\text {fusion }}}{\Delta S_{\text {fusion }}}$ .

Step 3: Substitute the values: $T = \frac{9950 \mathrm{~J} / \mathrm{mol}}{35.7 \mathrm{~J} / (\mathrm{K} \cdot \mathrm{mol})}$ .

Step 4: Calculate the melting point: $T = 278.71 \mathrm{~K}$ .

Step 5: Round to the correct number of significant figures: $T \approx 278.7 \mathrm{~K}$ .

Final Answer: $278.7 \mathrm{~K}$

Why other options are wrong:
- Option A: $278.7^{\circ} \mathrm{C}$ is incorrect. This is the value in Kelvin converted to Celsius, but the question asks for the melting point, which is directly calculated in Kelvin.
- Option C: $300 \mathrm{~K}$ is incorrect. This is a rounded value and not the accurate calculation.
- Option D: $298 \mathrm{~K}$ is incorrect. This is a common reference temperature (room temperature) but not the melting point calculated from the given data.

Key Concepts:
- Melting point calculation: $T = \frac{\Delta H_{\text {fusion }}}{\Delta S_{\text {fusion }}}$

Memory Aids:
- Remember that melting point is the temperature at which solid and liquid phases are in equilibrium.

Quick Tips:
- Ensure that the units of $\Delta H$ and $\Delta S$ are consistent before performing the calculation.

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