Heat energy change during the chemical reaction, ${\rm{CO}} + \frac{1}{2}{{\rm{O}}_2} \to {\rm{C}}{{\rm{O}}_2}$is known as :

Latent heat of ${\rm{C}}{{\rm{O}}_2}$

Heat of formation of ${\rm{C}}{{\rm{O}}_2}$

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium, Page 4

91 Medium

Analyzing +4 | -1

The normal boiling point and

\Delta {H}_{\text {vap }}

of a liquid '

{X}

' are

400 {~K}

and

40 {~kJ} {~mol}^{-1}

respectively. Assuming

\Delta {H}_{\text {vap }}

to be constant, which of the following is correct?
(I)

\Delta {S}_{\text {vap }}>100 {JK}^{-1} {~mol}^{-1}

at

400 {~K}

and

0.5 {~atm}

(II)

\Delta {S}_{\text {vap }} < 100 {JK}^{-1} {~mol}^{-1}

at

400 {~K}

and

1 {~atm}

(III)

\Delta {S}_{\text {vap }} < 100 {JK}^{-1} {~mol}^{-1}

at

400 {~K}

and

2 {~atm}

(IV)

\Delta {S}_{\text {vap }}=100 {JK}^{-1} {~mol}^{-1}

at

400 {~K}

and

1 {~atm}

View answer and explanation

Correct option: D

I, III and IV

Option D (I, III and IV) is correct. At the normal boiling point and 1 atm, ΔSvap = ΔHvap/T = 100 J/Kmol. Increasing pressure decreases ΔSvap.

Solution approach

Correct Answer + Core Reason: I, III and IV. At normal boiling point and 1 atm, $\Delta S_{vap} = \Delta H_{vap} / T$ . Increasing pressure decreases $\Delta S_{vap}$ .

Step-by-Step Solution:

Step 1: Calculate $\Delta S_{vap}$ at the normal boiling point (400 K and 1 atm):
$\Delta S_{vap} = \frac{{\Delta H_{vap}}}{{T}} = \frac{{40000 \text{{ J/mol}}}}{{400 \text{{ K}}}} = 100 \text{{ J/K mol}}$

Step 2: Analyze statement IV: $\Delta S_{vap} = 100 \text{{ J/K mol}}$ at 400 K and 1 atm. This is correct based on our calculation.

Step 3: Analyze statement II: $\Delta S_{vap} < 100 \text{{ J/K mol}}$ at 400 K and 1 atm. This is incorrect because we calculated it to be equal to 100 J/K mol.

Step 4: Consider the effect of pressure on entropy of vaporization. Increasing the pressure will decrease the entropy of the gas phase, thus decreasing $\Delta S_{vap}$ .

Step 5: Analyze statement I: $\Delta S_{vap} > 100 \text{{ J/K mol}}$ at 400 K and 0.5 atm. Since decreasing pressure increases entropy, this statement is correct.

Step 6: Analyze statement III: $\Delta S_{vap} < 100 \text{{ J/K mol}}$ at 400 K and 2 atm. Since increasing pressure decreases entropy, this statement is correct.

Step 7: Conclusion: Statements I, III, and IV are correct.

Why other options are wrong:
- Option A: II and IV only - Statement II is incorrect.
- Option B: II only - Statement II is incorrect.
- Option C: I and III only - Statement IV is also correct.

Key Concepts:
- Entropy of vaporization
- Relationship between $\Delta G$ , $\Delta H$ , and $\Delta S$
- Effect of pressure on entropy

Memory Aids:
- $\Delta G = \Delta H - T\Delta S$
- $\Delta S = \frac{{\Delta H}}{{T}}$ at equilibrium

Quick Tips:
- Remember the relationship between $\Delta H$ , $\Delta S$ , and T at phase transitions.
- Consider the effect of pressure on entropy.

92 Medium

92 of 244Question permalink

Remembering +4 | -1

Specific heat is defined as:

View answer and explanation

Correct option: A

Heat capacity/g

Option A (Heat capacity/g) is correct because specific heat is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.

Solution approach

Correct Answer + Core Reason: Heat capacity/g. Specific heat is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.

Step-by-Step Solution:

Step 1: Define Heat Capacity: Heat capacity is the amount of heat required to raise the temperature of a substance by one degree Celsius (or one Kelvin).

Step 2: Define Specific Heat: Specific heat is the amount of heat required to raise the temperature of *one gram* of a substance by one degree Celsius (or one Kelvin). Therefore, it is heat capacity divided by mass (in grams).

Step 3: Define Molar Heat Capacity: Molar heat capacity is the amount of heat required to raise the temperature of *one mole* of a substance by one degree Celsius (or one Kelvin). Therefore, it is heat capacity divided by the number of moles.

Why other options are wrong:
- Option B: Heat capacity/mol is the definition of molar heat capacity, not specific heat.
- Option C: Heat capacity at constant pressure is a type of heat capacity, but not the definition of specific heat.
- Option D: Heat capacity at constant volume is a type of heat capacity, but not the definition of specific heat.

Key Concepts:
- Heat Capacity
- Specific Heat
- Molar Heat Capacity

Memory Aids:
- Specific = per gram
- Molar = per mole

Quick Tips:
- Distinguish between heat capacity, specific heat, and molar heat capacity.

93 Medium

93 of 244Question permalink

Application +4 | -1

{H_2}

(g)+Cl(g)=2HCl(g); ∆H(298 K)=22.06 kcal. For this reaction, ∆U is equal to:

View answer and explanation

Correct option: D

- 22.06\mathrm{kcal}

Option D (-22.06 kcal) is correct because for the given reaction, the change in the number of moles of gas is zero, therefore ∆U = ∆H.

Solution approach

Correct Answer + Core Reason: $- 22.06\mathrm{kcal}$ . Since the change in the number of moles of gas is zero, ∆U = ∆H.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H = -22.06 \text{ kcal}$ at $298 \text{ K}$ . The reaction is ${H_2} (g)+Cl(g)=2HCl(g)$ .

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 gaseous products and reactants.

Step 3: Calculate $\Delta n_g$ : $\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - (1+1) = 0$ .

Step 4: Substitute $\Delta n_g = 0$ into the equation: $\Delta H = \Delta U + (0)RT \Rightarrow \Delta H = \Delta U$ .

Step 5: Therefore, $\Delta U = -22.06 \text{ kcal}$ .

Why other options are wrong:
- Option A: This option incorrectly adds a term involving R and T, but since $\Delta n_g = 0$ , this term should be zero.
- Option B: This option also incorrectly adds a term involving R and T, but since $\Delta n_g = 0$ , this term should be zero.
- Option C: This option incorrectly subtracts a term involving R and T, but since $\Delta n_g = 0$ , this term should be zero.

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

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

Quick Tips:
- Always check if there are any gaseous reactants or products. If not, $\Delta n_g = 0$ and $\Delta H = \Delta U$ .

94 Medium

94 of 244Question permalink

Analyzing +4 | -1

48 g of C (diamond) on complete combustion evolves 1584 kJ of heat. The standard heat of formation of gaseous carbon is 725 kJ/mol. The energies required for the process
(i)C (graphite) → C(gas)
(ii)C (diamond) → C(gas)are:

View answer and explanation

Correct option: C

725, 723

Option C (725, 723) is correct because it calculates the energy required for sublimation of graphite and diamond using the heat of combustion and heat of formation.

Solution approach

Correct Answer + Core Reason: 725, 723. The energy required for C(graphite) -> C(gas) is the standard heat of formation, and for C(diamond) -> C(gas), it's calculated using the heat of combustion.

Step-by-Step Solution:

Step 1: Define the given information.
- Heat evolved during combustion of 48g C(diamond) = 1584 kJ
- Standard heat of formation of gaseous carbon = 725 kJ/mol

Step 2: Calculate the heat of combustion per mole of C(diamond).
- Molar mass of C = 12 g/mol
- Moles of C = 48g / 12g/mol = 4 mol
- Heat evolved per mole = 1584 kJ / 4 mol = 396 kJ/mol
- Since heat is evolved, Δ H_{combustion} = -396 kJ/mol

Step 3: Apply Hess's Law to find the energy required for C(diamond) -> C(gas).
- C(diamond) + O_2(g) -> CO_2(g) Δ H = -396 kJ/mol
- C(graphite) + O_2(g) -> CO_2(g) Δ H = ?
- C(graphite) -> C(gas) Δ H = 725 kJ/mol
- C(diamond) -> C(gas) Δ H = x
- C(graphite) -> C(diamond) Δ H = 725 - x

Step 4: Use the relationship between Δ H_{combustion} of graphite and diamond.
- Δ H_{combustion}(graphite) = Δ H_{formation}(CO_2) = -393.5 kJ/mol (approximate value)
- C(graphite) + O_2(g) -> CO_2(g) Δ H = -393.5 kJ/mol
- C(diamond) + O_2(g) -> CO_2(g) Δ H = -396 kJ/mol
- C(graphite) -> C(diamond) Δ H = -393.5 - (-396) = 2.5 kJ/mol

Step 5: Calculate the energy required for C(diamond) -> C(gas).
- C(graphite) -> C(diamond) Δ H = 2.5 kJ/mol
- C(graphite) -> C(gas) Δ H = 725 kJ/mol
- C(diamond) -> C(gas) Δ H = 725 - 2.5 = 722.5 kJ/mol ≈ 723 kJ/mol

Step 6: Final Answer: C(graphite) -> C(gas) = 725 kJ/mol, C(diamond) -> C(gas) = 723 kJ/mol

Why other options are wrong:
- Option A: Incorrect calculation of energy required for C(diamond) -> C(gas).
- Option B: Incorrect calculation of energy required for C(diamond) -> C(gas).
- Option D: The correct answer is among the given options.

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

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

Quick Tips:
- Always balance the chemical equations before applying Hess's Law.
- Pay attention to the signs of enthalpy changes (positive for endothermic, negative for exothermic).

95 Medium

95 of 244Question permalink

Remembering +4 | -1

The word standard in molar enthalpy change implies

View answer and explanation

Correct option: C

Any temperature and pressure 1 bar

Option C (Any temperature and pressure 1 bar) is correct because the standard condition for molar enthalpy change is defined at 1 bar pressure, while temperature is not standardized.

Solution approach

Correct Answer + Core Reason: Any temperature and pressure 1 bar. Standard conditions for molar enthalpy changes are defined at 1 bar pressure.

Step-by-Step Solution:

Step 1: Define standard conditions. Standard conditions are a reference point for thermodynamic measurements.

Step 2: Identify the standard pressure. The standard pressure is 1 bar (100 kPa).

Step 3: Consider the temperature. The temperature is not specified in the definition of standard conditions for molar enthalpy change; it can be any temperature.

Step 4: Conclusion. Therefore, the word \"standard\" implies any temperature and a pressure of 1 bar.

Why other options are wrong:
- Option A: Temperature $298 \mathrm{~K}$ and pressure $1 \mathrm{~atm}$ : This is incorrect because the standard pressure is 1 bar, not 1 atm, and the temperature is not fixed.
- Option B: Any temperature and pressure $1 \mathrm{~atm}$ : This is incorrect because the standard pressure is 1 bar, not 1 atm.
- Option D: Any temperature and pressure: This is incorrect because the pressure must be 1 bar.

Key Concepts:
- Standard conditions
- Molar enthalpy change
- Pressure units (bar, atm)

Memory Aids:
- Remember that standard pressure is 1 bar, not 1 atm.

Quick Tips:
- Pay attention to the units of pressure (bar vs. atm) when dealing with standard conditions.

96 Medium

96 of 244Question permalink

Application +4 | -1

The heat of combustion of ethanol determined by a bomb calorimeter is - 670.48 kcal

mo{l^{ - 1}}

at

{25^0}C

.What is ∆U at

{25^0}C

for the following reaction?

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}\left( l \right) + \frac{7}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)

View answer and explanation

Correct option: C

- 670.48\mathrm{kcal}

Option C (-670.48 kcal) is correct because at constant volume (bomb calorimeter), the heat of combustion is equal to the change in internal energy, ∆U.

Solution approach

Correct Answer + Core Reason: $- 670.48\mathrm{kcal}$ . In a bomb calorimeter, the volume is constant, so the heat of combustion is equal to the change in internal energy.

Step-by-Step Solution:

Step 1: Identify the given information: Heat of combustion = -670.48 kcal/mol at 25°C.

Step 2: Understand the relationship between heat of combustion at constant volume and change in internal energy (∆U): In a bomb calorimeter, the volume is constant. Therefore, the heat released during the combustion process is equal to the change in internal energy (∆U).

Step 3: Since the heat of combustion is given as -670.48 kcal/mol, the change in internal energy (∆U) is also -670.48 kcal/mol.

Step 4: Final Answer: ∆U = -670.48 kcal/mol

Why other options are wrong:
- Option A: This value is incorrect because it is half of the given heat of combustion and doesn't represent the total change in internal energy.
- Option B: This value is incorrect because it seems to be a slightly modified version of the correct answer, likely due to a calculation error or misunderstanding of the concept.
- Option D: This value is incorrect because it has the wrong sign. Combustion is an exothermic process, so the change in internal energy must be negative.

Key Concepts:
- Heat of combustion at constant volume
- Change in internal energy (∆U)
- Bomb calorimeter
- Exothermic reactions

Memory Aids:
- Bomb calorimeter = constant volume = ∆U = heat of combustion

Quick Tips:
- In a bomb calorimeter, the heat released is directly equal to the change in internal energy (∆U).

97 Medium

97 of 244Question permalink

Application +4 | -1

For the reaction

CC{l_{4\left( g \right)}} + 2{H_2}{O_{\left( g \right)}} \to C{O_{2\left( g \right)}} + 4HC{l_{\left( g \right)}}

at constant temperature ,

\Delta H - \Delta E

is

View answer and explanation

Correct option: D

2RT

Option D (2RT) is correct because it correctly calculates the difference between

\Delta H

and

\Delta E

using the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: 2RT. The difference between $\Delta H$ and $\Delta E$ is related to the change in the number of moles of gas.

Step-by-Step Solution:
Step 1: Recall the relationship between enthalpy change ( $\Delta H$ ) and internal energy change ( $\Delta E$ ):
$\Delta H = \Delta E + \Delta n_g RT$ \nwhere $\Delta n_g$ is the change in the number of moles of gas.
Step 2: Rearrange the equation to find $\Delta H - \Delta E$ :
$\Delta H - \Delta E = \Delta n_g RT$
Step 3: Determine the change in the number of moles of gas ( $\Delta n_g$ ) for the reaction:
$CC{l_{4(g)}} + 2{H_2}{O_{(g)}} \to C{O_{2(g)}} + 4HC{l_{(g)}}$
$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = (1 + 4) - (1 + 2) = 5 - 3 = 2$
Step 4: Substitute $\Delta n_g = 2$ into the equation:
$\Delta H - \Delta E = 2RT$

Why other options are wrong:
- Option A: This is incorrect because it has the wrong sign and magnitude. It might be obtained by calculating the change in moles of gas incorrectly.
- Option B: This is incorrect because it has the wrong magnitude. It might be obtained by calculating the change in moles of gas incorrectly.
- Option C: This is incorrect because it has the wrong sign. It might be obtained by subtracting the reactants from the products incorrectly.

Key Concepts:
- Enthalpy change
- Internal energy change
- Change in moles of gas

Memory Aids:
- Remember that $\Delta H = \Delta E + \Delta n_g RT$ .

Quick Tips:
- Only consider the change in the number of moles of gaseous species when calculating $\Delta n_g$ .

98 Medium

98 of 244Question permalink

Application +4 | -1

When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice in initial temperature, the change in entropy

(\Delta S)

is

View answer and explanation

Correct option: D

({C_V} - R)\ln 2

Option D

({C_V} - R)\ln 2

is correct because the change in entropy is calculated by considering both the volume and temperature changes using the appropriate formula.

Solution approach

Correct Answer + Core Reason: $({C_V} - R)\ln 2$ . The change in entropy is calculated using the formula $\Delta S = nC_V \ln{\frac{{T_2}}{{T_1}}} + nR \ln{\frac{{V_2}}{{V_1}}}$ .

Step-by-Step Solution:

Step 1: Identify the given information:
- Number of moles (n) = 1
- $T_2 = 2T_1$
- $V_2 = \frac{{V_1}}{{2}}$

Step 2: Use the formula for change in entropy:
- $\Delta S = nC_V \ln{\frac{{T_2}}{{T_1}}} + nR \ln{\frac{{V_2}}{{V_1}}}$

Step 3: Substitute the values:
- $\Delta S = (1)C_V \ln{\frac{{2T_1}}{{T_1}}} + (1)R \ln{\frac{{\frac{{V_1}}{{2}}}}{{V_1}}}$
- $\Delta S = C_V \ln{2} + R \ln{\frac{{1}}{{2}}}$
- $\Delta S = C_V \ln{2} - R \ln{2}$

Step 4: Simplify the expression:
- $\Delta S = (C_V - R)\ln{2}$

Step 5: Final Answer: $\Delta S = ({C_V} - R)\ln 2$

Why other options are wrong:
- Option A: This value is incorrect because it only considers the temperature change and not the volume change.
- Option B: This value is incorrect because it uses $C_P$ instead of $C_V$ and doesn't account for the volume change correctly.
- Option C: This value is incorrect because it only considers the volume change and not the temperature change.

Key Concepts:
- Change in entropy
- Heat capacity at constant volume ( $C_V$ )
- Ideal gas constant (R)
- Isothermal and Isobaric Processes

Memory Aids:
- $\Delta S = nC_V \ln{\frac{{T_2}}{{T_1}}} + nR \ln{\frac{{V_2}}{{V_1}}}$

Quick Tips:
- Remember to consider both temperature and volume changes when calculating the change in entropy.
- Use the appropriate formula for the given conditions.

99 Medium

99 of 244Question permalink

Understanding +4 | -1

Which of the following is correct for a reaction to be spontaneous?

View answer and explanation

Correct option: B

\Delta H = - ve,\,\,\Delta S = + ve

The correct answer is B (

\Delta H = - ve,\,\Delta S = + ve

) because a reaction is spontaneous when it results in a decrease in enthalpy (negative ∆H) and an increase in entropy (positive ∆S).

Solution approach

Correct Answer + Core Reason: $\Delta H = - ve,\,\Delta S = + ve$ . A spontaneous reaction is favored by a decrease in enthalpy and an increase in entropy.

Step-by-Step Solution:

Step 1: Define spontaneity in terms of Gibbs free energy: A reaction is spontaneous if the change in Gibbs free energy (∆G) is negative (∆G < 0).

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

Step 3: Analyze the conditions for spontaneity: For ∆G to be negative, ∆H should be negative (exothermic) and ∆S should be positive (increase in disorder).

Why other options are wrong:
- Option A: $\Delta G = + ve,\,\Delta H = + ve$ : A positive ∆G indicates a non-spontaneous reaction.
- Option C: $\Delta G = + ve,{\mkern 1mu} {\mkern 1mu} \Delta H = - ve$ : A positive ∆G indicates a non-spontaneous reaction.
- Option D: $\Delta H = + ve,\,\Delta S = - ve$ : A positive ∆H and negative ∆S will always result in a positive ∆G, indicating a non-spontaneous reaction.

Key Concepts:
- Spontaneity
- Gibbs free energy (G)
- Enthalpy (H)
- Entropy (S)

Memory Aids:
- Spontaneous = Negative ∆G
- $\Delta G = \Delta H - T\Delta S$

Quick Tips:
- Remember the Gibbs free energy equation and how ∆H and ∆S affect spontaneity.

100 Medium

100 of 244Question permalink

Remembering +4 | -1

Entropy change of vaporisation at constant pressure is given by:

View answer and explanation

Correct option: A

\Delta {S_{\left( v \right)}} = \frac{{\Delta {H_v}}}{T}

Option A is correct because the entropy change of vaporization at constant pressure is given by ΔS(v) = ΔHv/T.

Solution approach

Correct Answer + Core Reason: $\Delta {S_{\left( v \right)}} = \frac{{\Delta {H_v}}}{T}$ . This is the formula for entropy change during vaporization.

Step-by-Step Solution:

Step 1: Entropy Change: $\Delta S = \frac{Q_{rev}}{T}$ , where $Q_{rev}$ is the heat absorbed during a reversible process and T is the temperature.

Step 2: Vaporization: A phase transition from liquid to gas.

Step 3: Enthalpy of Vaporization: $\Delta H_v$ is the heat absorbed during vaporization at constant pressure.

Step 4: Entropy change of vaporization: $\Delta S_v = \frac{\Delta H_v}{T}$

Step 5: Conclusion: The entropy change of vaporization at constant pressure is given by $\Delta {S_{\left( v \right)}} = \frac{{\Delta {H_v}}}{T}$ .

Why other options are wrong:
- Option B: Incorrect; it uses internal energy instead of enthalpy.
- Option C: Incorrect; it divides by a change in temperature, which is not correct for a phase transition at constant temperature.
- Option D: Incorrect; the correct formula is given in option A.

Key Concepts:
- Entropy Change
- Enthalpy of Vaporization

Memory Aids:
- $\Delta S = \frac{Q_{rev}}{T}$ (fundamental equation for entropy change)

Quick Tips:
- Remember that phase transitions occur at constant temperature and pressure.

101 Medium

101 of 244Question permalink

Understanding +4 | -1

In which of the following condition a chemical reaction can not occur?

View answer and explanation

Correct option: C

∆H increase and ∆S decreases

Option C (∆H increase and ∆S decreases) is correct because it leads to a positive ∆G at all temperatures, making the reaction non-spontaneous.

Solution approach

Correct Answer + Core Reason: ∆H increase and ∆S decreases. A reaction is non-spontaneous if Gibbs free energy change (∆G) is positive, which occurs when ∆H is positive and ∆S is negative.

Step-by-Step Solution:

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

Step 2: Analyze each option:
- Option A: ∆H and ∆S increase and T∆S>∆H. In this case, ∆G is negative, so the reaction is spontaneous.
- Option B: ∆H and ∆S decrease and ∆H>T∆S. In this case, ∆G is negative, so the reaction is spontaneous.
- Option C: ∆H increase and ∆S decreases. In this case, ∆G is always positive, so the reaction is non-spontaneous.
- Option D: ∆H decreases and ∆S increases. In this case, ∆G is always negative, so the reaction is spontaneous.

Step 3: Conclude that only option C results in a non-spontaneous reaction.

Why other options are wrong:
- Option A: ∆H and ∆S increase and T∆S>∆H. This leads to a negative ∆G, indicating a spontaneous reaction, not a non-spontaneous one.
- Option B: ∆H and ∆S decrease and ∆H>T∆S. This also leads to a negative ∆G, indicating a spontaneous reaction.
- Option D: ∆H decreases and ∆S increases. This results in a negative ∆G, indicating a spontaneous reaction.

Key Concepts:
- Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$
- Spontaneity: ∆G < 0 (spontaneous), ∆G > 0 (non-spontaneous), ∆G = 0 (equilibrium)

Memory Aids:
- "Good Honey Tastes Sweet": Gibbs Free Energy (G), Enthalpy (H), Temperature (T), Entropy (S).

Quick Tips:
- Remember the sign conventions for ∆H and ∆S and their effect on ∆G.
- If ∆H is negative and ∆S is positive, the reaction is always spontaneous.
- If ∆H is positive and ∆S is negative, the reaction is never spontaneous.

102 Medium

102 of 244Question permalink

Understanding +4 | -1

A positive change in enthalpy occurs in :

View answer and explanation

Correct option: C

\mathrm{MgC}{\mathrm{O}_3}\left( s \right) \to \mathrm{MgO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)

Option C (

\mathrm{MgC}{\mathrm{O}_3}\left( s \right) \to \mathrm{MgO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)

) is correct because the decomposition of MgCO3 is an endothermic process, requiring energy input and resulting in a positive enthalpy change.

Solution approach

Correct Answer + Core Reason: $\mathrm{MgC}{\mathrm{O}_3}\left( s \right) \to \mathrm{MgO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)$ . Decomposition reactions are generally endothermic, requiring energy input.

Step-by-Step Solution:

Step 1: Define positive enthalpy change. A positive enthalpy change indicates an endothermic reaction, where heat is absorbed from the surroundings.

Step 2: Analyze the given reactions.
- Option A: Formation of water is an exothermic reaction.
- Option B: Formation of ammonia is an exothermic reaction.
- Option C: Decomposition of magnesium carbonate is an endothermic reaction.
- Option D: Formation of liquid water is an exothermic reaction.

Step 3: Conclusion: Therefore, a positive change in enthalpy occurs in the decomposition of magnesium carbonate.

Why other options are wrong:
- Option A: ${\mathrm{H}_2}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)$ : This is an exothermic reaction.
- Option B: ${\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)$ : This is an exothermic reaction.
- Option D: ${\mathrm{H}_2}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\mathrm{O}\left( l \right)$ : This is an exothermic reaction.

Key Concepts:
- Enthalpy change
- Endothermic reaction
- Exothermic reaction

Memory Aids:
- Endothermic = Energy in = Positive enthalpy change.

Quick Tips:
- Remember that decomposition reactions are generally endothermic.

103 Medium

103 of 244Question permalink

Remembering +4 | -1

In the exothermic reaction, the enthalpy of
reaction is always:

View answer and explanation

Correct option: C

negative

Option C (negative) is correct because exothermic reactions release heat, resulting in a negative enthalpy change.

Solution approach

Correct Answer + Core Reason: negative. Exothermic reactions release heat, leading to a negative enthalpy change.

Step-by-Step Solution:

Step 1: Define exothermic reaction. An exothermic reaction is a reaction that releases heat to the surroundings.

Step 2: Relate heat release to enthalpy change. When heat is released, the enthalpy of the products is lower than the enthalpy of the reactants, resulting in a negative enthalpy change.

Step 3: Conclusion: Therefore, in an exothermic reaction, the enthalpy change is always negative.

Why other options are wrong:
- Option A: zero: This is incorrect because exothermic reactions involve a change in enthalpy, not zero.
- Option B: positive: This is incorrect because exothermic reactions release heat, resulting in a negative enthalpy change.
- Option D: None of these: This is incorrect because the enthalpy of reaction is negative for exothermic reactions.

Key Concepts:
- Exothermic reaction
- Enthalpy change

Memory Aids:
- Exothermic = Exit heat = Negative enthalpy change.

Quick Tips:
- Remember the sign convention for enthalpy changes in exothermic and endothermic reactions.

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Remembering +4 | -1

Calorific value of carbohydrates is approximately:

View answer and explanation

Correct option: A

4.0\mathrm{ }kcal/g

Option A (4.0 kcal/g) is correct because it is the approximate calorific value of carbohydrates.

Solution approach

Correct Answer + Core Reason: $4.0\mathrm{ }kcal/g$ . This is the approximate calorific value of carbohydrates.

Step-by-Step Solution:

Step 1: Define calorific value.

Step 2: Calorific value is the amount of heat released when 1 gram of a substance is completely burned.

Step 3: The calorific value of carbohydrates is approximately 4.0 kcal/g.

Why other options are wrong:
- Option B: 16.0 kcal/g is incorrect because this is much higher than the calorific value of carbohydrates.
- Option C: 20 kcal/g is incorrect because this is much higher than the calorific value of carbohydrates.
- Option D: 9.0 kcal/g is incorrect because this is the approximate calorific value of fats.

Key Concepts:
- Calorific value
- Carbohydrates
- Fats

Memory Aids:
- Carbohydrates: ~4 kcal/g
- Fats: ~9 kcal/g
- Proteins: ~4 kcal/g

Quick Tips:
- Remember the approximate calorific values of the major food groups.

105 Medium

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Understanding +4 | -1

In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the corerct option for free expansion of an ideal gas under adiabatic condition from the following.

View answer and explanation

Correct option: C

q = 0,\,\,\Delta T = 0,\,\,w = 0

Option C (

q = 0,\,\Delta T = 0,\,w = 0

) is correct because in an adiabatic free expansion of an ideal gas, no heat is exchanged (q=0), no work is done (w=0), and the temperature remains constant (ΔT=0).

Solution approach

Correct Answer + Core Reason: $q = 0,\,\Delta T = 0,\,w = 0$ . In an adiabatic free expansion, there is no heat transfer, no work done, and no change in temperature.

Step-by-Step Solution:

Step 1: Define adiabatic process: A process in which no heat is exchanged with the surroundings, so $q = 0$ .

Step 2: Define free expansion: Expansion against zero external pressure, so no work is done, $w = 0$ .

Step 3: Apply the first law of thermodynamics: $\Delta U = q + w$ . Since $q = 0$ and $w = 0$ , $\Delta U = 0$ .

Step 4: For an ideal gas, internal energy depends only on temperature: $\Delta U = nC_v\Delta T$ . Since $\Delta U = 0$ , $\Delta T = 0$ .

Step 5: Conclusion: Therefore, $q = 0$ , $\Delta T = 0$ , and $w = 0$ .

Why other options are wrong:
- Option A: $q = 0,\,\Delta T \ne 0,\,w = 0$ : This is incorrect because $\Delta T$ must be 0 for a free expansion of an ideal gas.
- Option B: $q \ne 0,\,\Delta T = 0,\,w = 0$ : This is incorrect because $q$ must be 0 for an adiabatic process.
- Option D: $q = 0,\,\Delta T < 0,\,w \ne 0$ : This is incorrect because $\Delta T$ and $w$ must be 0 for a free expansion of an ideal gas.

Key Concepts:
- Adiabatic Process
- Free Expansion
- First Law of Thermodynamics
- Internal Energy of an Ideal Gas

Memory Aids:
- Adiabatic: No heat exchange (q=0)
- Free Expansion: No external pressure, no work (w=0)

Quick Tips:
- Remember the conditions for adiabatic and free expansion processes.

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Application +4 | -1

A gas is allowed to expand in an insulated container against a constant external pressure of 5 atm from an initial volume of 1.75 L to a final volume of 6.75 L. The change in internal energy

(\Delta U)

of the gas will be

View answer and explanation

Correct option: A

- 2.5\mathrm{ }kJ

Option A (-2.5 kJ) is correct because the change in internal energy is equal to the heat exchanged, which is the negative of the work done against the constant external pressure in an adiabatic process.

Solution approach

Correct Answer + Core Reason: $- 2.5\mathrm{ }kJ$ . The change in internal energy is equal to the heat exchanged, which is the negative of the work done.

Step-by-Step Solution:

Step 1: Identify the given information:
- External pressure, $P_{ext} = 5\, \mathrm{atm}$
- Initial volume, $V_1 = 1.75\, \mathrm{L}$
- Final volume, $V_2 = 6.75\, \mathrm{L}$
- Insulated container (adiabatic process), $q = 0$

Step 2: Calculate the change in volume:
$\Delta V = V_2 - V_1 = 6.75\, \mathrm{L} - 1.75\, \mathrm{L} = 5\, \mathrm{L}$

Step 3: Calculate the work done by the gas:
$w = -P_{ext} \Delta V = -5\, \mathrm{atm} \times 5\, \mathrm{L} = -25\, \mathrm{L \cdot atm}$

Step 4: Convert L atm to Joules:
$1\, \mathrm{L \cdot atm} = 101.325\, \mathrm{J}$
$w = -25\, \mathrm{L \cdot atm} \times 101.325\, \mathrm{J/(L \cdot atm)} = -2533.125\, \mathrm{J}$

Step 5: Convert Joules to kJ:
$w = -2533.125\, \mathrm{J} = -2.533\, \mathrm{kJ} \approx -2.5\, \mathrm{kJ}$

Step 6: Apply the first law of thermodynamics: $\Delta U = q + w$
Since the process is adiabatic, $q = 0$ , so $\Delta U = w = -2.5\, \mathrm{kJ}$

Why other options are wrong:
- Option B: $- 4.5\mathrm{ }kJ$ is incorrect. It likely involves an incorrect calculation of the work done.
- Option C: $- 1.25\mathrm{ }kJ$ is incorrect. It likely involves an incorrect calculation of the work done.
- Option D: $- 3.2\mathrm{ }kJ$ is incorrect. It likely involves an incorrect calculation of the work done.

Key Concepts:
- First law of thermodynamics
- Work done in expansion
- Adiabatic process

Memory Aids:
- Adiabatic: No heat exchange (q=0).

Quick Tips:
- Remember to convert units to be consistent (L atm to Joules, Joules to kJ).
- In an adiabatic process, $\Delta U = w$ .

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Understanding +4 | -1

Which correctly represents the physical significance of Gibbs energy change?

View answer and explanation

Correct option: D

- \Delta G = {W_{\mathrm{expansion}}}

Option D (

- \Delta G = {W_{\mathrm{expansion}}}

) is correct because the decrease in Gibbs free energy (-∆G) represents the maximum work that can be obtained from a system during expansion at constant temperature and pressure.

Solution approach

Correct Answer + Core Reason: $- \Delta G = {W_{\mathrm{expansion}}}$ . The decrease in Gibbs free energy is equal to the maximum work that can be obtained from a system during expansion at constant temperature and pressure.

Step-by-Step Solution:

Step 1: Recall the relationship between Gibbs free energy change and work.

Step 2: The change in Gibbs free energy ( $\Delta G$ ) represents the maximum amount of non-expansion work that can be extracted from a closed system at constant temperature and pressure.

Step 3: For expansion work, $- \Delta G = W_{\text{expansion}}$ .

Why other options are wrong:
- Option A: $- \Delta G = {W_{\mathrm{compression}}}$ is incorrect because -∆G represents expansion work, not compression work.
- Option B: $\Delta G = {W_{\mathrm{expansion}}}$ is incorrect because it should be -∆G, not ∆G.
- Option C: $\Delta G = - {W_{\mathrm{expansion}}} = {W_{\mathrm{non} - \mathrm{expansion}}}$ is incorrect because it mixes the sign and equates expansion work with non-expansion work incorrectly.

Key Concepts:
- Gibbs Free Energy
- Work Done by a System

Memory Aids:
- -∆G = Maximum expansion work.

Quick Tips:
- Remember that the decrease in Gibbs free energy (-∆G) represents the maximum useful work obtainable from a system at constant temperature and pressure.

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Remembering +4 | -1

The energy that opposes dissolution of a solvent
is

View answer and explanation

Correct option: B

lattice energy

Option B (lattice energy) is correct because lattice energy is the energy required to break apart an ionic lattice, opposing the dissolution process.

Solution approach

Correct Answer + Core Reason: lattice energy. Lattice energy opposes dissolution by holding the solute together.

Step-by-Step Solution:

Step 1: Define dissolution: The process of a solute dissolving in a solvent.

Step 2: Define lattice energy: The energy required to separate one mole of an ionic solid into its gaseous ions.

Step 3: In order for a solute to dissolve, the attractive forces within the solute (lattice energy) must be overcome.

Step 4: Hydration energy, on the other hand, favors dissolution by stabilizing the ions in solution.

Why other options are wrong:
- Option A: hydration energy. Hydration energy favors dissolution.
- Option C: internal energy. Internal energy is a general term and not specific to dissolution.
- Option D: bond energy. Bond energy is relevant for covalent compounds, not ionic solids.

Key Concepts:
- Dissolution
- Lattice energy
- Hydration energy

Memory Aids:
- Lattice energy holds the lattice together, opposing dissolution.

Quick Tips:
- Consider the forces that favor and oppose dissolution.

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Remembering +4 | -1

The heat of combustion of a substance is :

View answer and explanation

Correct option: B

Always negative

Option B (Always negative) is correct because combustion reactions release heat, making them exothermic with a negative enthalpy change.

Solution approach

Correct Answer + Core Reason: Always negative. Combustion is an exothermic process, releasing heat.

Step-by-Step Solution:

Step 1: Define heat of combustion. The heat of combustion is the energy released when a substance undergoes combustion with oxygen.

Step 2: Relate heat release to enthalpy change. Since heat is released, the enthalpy change is negative.

Step 3: Conclusion: Therefore, the heat of combustion is always negative.

Why other options are wrong:
- Option A: Always positive: This is incorrect because combustion releases heat, resulting in a negative enthalpy change.
- Option C: Numerically equal to the heat of formation: This is incorrect because the heat of combustion is related to the complete oxidation of a substance, while the heat of formation is related to the formation of a compound from its elements.
- Option D: Unpredictable: This is incorrect because the heat of combustion is a defined property for a given substance under specific conditions.

Key Concepts:
- Heat of combustion
- Enthalpy change
- Exothermic reactions

Memory Aids:
- Exothermic reactions release heat (negative enthalpy change).

Quick Tips:
- Remember that combustion is always exothermic.

110 Medium

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Application +4 | -1

A person requires 2870 kcal of energy to lead normal daily life. If heat of combustion of cane sugar is -1349 kcal, then his daily consumption of sugar is:

View answer and explanation

Correct option: A

728 g

Option A (728 g) is correct because the daily sugar consumption is calculated by dividing the required energy by the energy per gram of sugar.

Solution approach

Correct Answer + Core Reason: 728 g. The daily sugar consumption is calculated by dividing the required energy by the energy per gram of sugar.

Step-by-Step Solution:

Step 1: Identify the given information:
- Daily energy requirement = 2870 kcal
- Heat of combustion of cane sugar = -1349 kcal/mol

Step 2: Calculate the molar mass of cane sugar ( $C_{12}H_{22}O_{11}$ ):
$M = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342\, \mathrm{g/mol}$

Step 3: Calculate the energy released per gram of cane sugar:
$\frac{1349\, \mathrm{kcal/mol}}{342\, \mathrm{g/mol}} = 3.94\, \mathrm{kcal/g}$

Step 4: Calculate the daily consumption of sugar:
$\frac{2870\, \mathrm{kcal}}{3.94\, \mathrm{kcal/g}} = 728.4\, \mathrm{g} \approx 728\, \mathrm{g}$

Why other options are wrong:
- Option B: 0.728 g is incorrect. It's a result of miscalculation.
- Option C: 342 g is the molar mass of sugar, not the daily consumption.
- Option D: 0.342 g is incorrect. It's a result of miscalculation.

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

Memory Aids:
- Remember to convert moles to grams using molar mass.

Quick Tips:
- Always check the units to ensure they are consistent.

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Application +4 | -1

Heat of formation of

{H_2}O

is -188 kJ /mol and

{H_2}{O_2}

is -286 kJ/mol.The enthalpy change for the reaction

2{H_2}{O_2} \to 2{H_2}O + {O_2}

is

View answer and explanation

Correct option: A

196 kJ

Option A (196 kJ) is correct because it is calculated using Hess's Law, considering the enthalpy of formation of reactants and products.

Solution approach

Correct Answer + Core Reason: 196 kJ. The enthalpy change is calculated using Hess's Law: $\Delta H = \sum \Delta H_{f,products} - \sum \Delta H_{f,reactants}$

Step-by-Step Solution:

Step 1: Write the reaction: $2{H_2}{O_2} \to 2{H_2}O + {O_2}$

Step 2: Write the enthalpies of formation:
- $\Delta H_f({H_2}O) = -188 \text{{ kJ/mol}}$
- $\Delta H_f({H_2}{O_2}) = -286 \text{{ kJ/mol}}$
- $\Delta H_f({O_2}) = 0 \text{{ kJ/mol}}$ (by definition, since it's in its standard state)

Step 3: Apply Hess's Law:
$\Delta H = [2 \times \Delta H_f({H_2}O) + \Delta H_f({O_2})] - [2 \times \Delta H_f({H_2}{O_2})]$
$\Delta H = [2 \times (-188 \text{{ kJ/mol}}) + 0] - [2 \times (-286 \text{{ kJ/mol}})]$
$\Delta H = -376 \text{{ kJ/mol}} + 572 \text{{ kJ/mol}} = 196 \text{{ kJ/mol}}$

Step 4: Final Answer: 196 kJ

Why other options are wrong:
- Option B (- 196 kJ): This is incorrect because the sign is wrong. The calculation is done correctly, but the sign is flipped.
- Option C (984 KJ): This is incorrect and likely due to a miscalculation or misunderstanding of Hess's Law.
- Option D (-984 kJ): This is incorrect and likely due to a miscalculation or misunderstanding of Hess's Law.

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

Memory Aids:
- Remember the formula: $\Delta H = \sum \Delta H_{f,products} - \sum \Delta H_{f,reactants}$

Quick Tips:
- Always write out the balanced chemical equation.
- Remember that the enthalpy of formation of an element in its standard state is zero.

112 Medium

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Application +4 | -1

1 mole of an ideal gas at 25℃ is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is:

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the change in entropy for reversible expansion is calculated using

\Delta S = nR \ln(\frac{V_2}{V_1})

.

Solution approach

Correct Answer + Core Reason: $19.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$ . The change in entropy for reversible expansion is $\Delta S = nR \ln(\frac{V_2}{V_1})$ .

Step-by-Step Solution:

Step 1: Identify the given information: $n = 1 \text{ mole}$ , $V_2 = 10V_1$ , and $T = 25^{\circ}C = 298 \text{ K}$ .

Step 2: Apply the formula for entropy change during reversible expansion: $\Delta S = nR \ln(\frac{V_2}{V_1})$ .

Step 3: Substitute the values: $\Delta S = (1 \text{ mol}) \times (8.314 \text{ J/(mol K)}) \times \ln(\frac{10V_1}{V_1})$ .

Step 4: Simplify: $\Delta S = 8.314 \text{ J/K} \times \ln(10)$ .

Step 5: Calculate the entropy change: $\Delta S = 8.314 \text{ J/K} \times 2.303 \times \log_{10}(10) = 8.314 \text{ J/K} \times 2.303 \times 1 = 19.145 \text{ J/K}$ .

Step 6: Round to the correct number of significant figures: $\Delta S \approx 19.15 \text{ J/(K mol)}$ .

Final Answer: $19.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$

Why other options are wrong:
- Option B: $16.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$ is incorrect. This value might be obtained by using an incorrect value for $R$ or by miscalculating the logarithm.
- Option C: $22.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$ is incorrect. This value might be obtained by using an incorrect formula or by miscalculating the logarithm.
- Option D: None of these is incorrect because $19.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$ is the correct answer.

Key Concepts:
- Entropy change during reversible expansion: $\Delta S = nR \ln(\frac{V_2}{V_1})$

Memory Aids:
- Remember that entropy increases during expansion.

Quick Tips:
- Always use the natural logarithm ( $\ln$ ) in the formula for entropy change.
- Ensure the units are consistent.

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Remembering +4 | -1

In thermodynamics, a quantity whose value
simply depends upon the initial and final states
of the system is called

View answer and explanation

Correct option: B

a state function

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

Solution approach

Correct Answer + Core Reason: a state function. State functions depend only on the initial and final states of the system.

Step-by-Step Solution:

Step 1: Understand the definition of a state function.

Step 2: A state function is a property of a system that depends only on the current state of the system, not on how that state was achieved. Examples include pressure, volume, temperature, internal energy, enthalpy, and entropy.

Step 3: A path function, on the other hand, depends on the process or path taken to reach a particular state. Examples include heat and work.

Step 4: Therefore, a quantity whose value simply depends upon the initial and final states of the system is called a state function.

Why other options are wrong:
- Option A: A thermodynamic quantity is a general term and doesn't specifically define the path-independent nature.
- Option C: An adiabatic quantity refers to a process where no heat is exchanged, not a property dependent on initial and final states.
- Option D: A path function is the opposite of what the question describes; it depends on the path taken, not just the initial and final states.

Key Concepts:
- State function
- Path function
- Thermodynamics

Memory Aids:
- State functions: Think of \"state\" as a specific condition, independent of history.

Quick Tips:
- Remember the difference between state and path functions. Common examples of state functions are internal energy, enthalpy, entropy, pressure, volume, and temperature. Common examples of path functions are heat and work.

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Understanding +4 | -1

Heat energy change during the chemical reaction,

\mathrm{CO} + \frac{1}{2}{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2}

is known as :

View answer and explanation

Correct option: A

Heat of combustion of CO

Option A (Heat of combustion of CO) is correct because the reaction describes the complete burning of CO in oxygen to form CO2, which is the definition of heat of combustion.

Solution approach

Correct Answer + Core Reason: Heat of combustion of CO. The reaction represents the complete combustion of CO to CO2.

Step-by-Step Solution:

Step 1: Define heat of combustion.\nHeat of combustion is the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.

Step 2: Analyze the given reaction:
$\mathrm{CO}(g) + \frac{1}{2}{\mathrm{O}_2}(g) \to \mathrm{C}{\mathrm{O}_2}(g)$ \nThis reaction represents the complete combustion of carbon monoxide (CO) to form carbon dioxide (CO2).

Step 3: Conclusion: The heat energy change during this reaction is the heat of combustion of CO.

Why other options are wrong:
- Option B: Latent heat of $\mathrm{C}{\mathrm{O}_2}$ is incorrect because latent heat refers to the heat absorbed or released during a phase change, not a chemical reaction.
- Option C: Latent heat of vaporisation is incorrect because it refers to the heat required to change a liquid to a gas, not a chemical reaction.
- Option D: Heat of formation of $\mathrm{C}{\mathrm{O}_2}$ is incorrect because the reactants are CO and O2, not the elements in their standard states (C and O2).

Key Concepts:
- Heat of combustion
- Heat of formation
- Thermochemical reactions

Memory Aids:
- Combustion involves burning a substance in oxygen.

Quick Tips:
- Remember the definitions of heat of combustion and heat of formation.

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Understanding +4 | -1

A thermally isolated gaseous system can exchange energy with the surroundings. The mode of transference of energy can be:

View answer and explanation

Correct option: B

Work

Option B (Work) is correct because a thermally isolated system cannot exchange heat with the surroundings, but it can still exchange energy through work.

Solution approach

Correct Answer + Core Reason: Work. A thermally isolated system cannot exchange heat but can exchange energy through work.

Step-by-Step Solution:

Step 1: Define Thermally Isolated System: A thermally isolated system is one that does not exchange heat with its surroundings (q = 0).

Step 2: Energy Exchange: Energy can be exchanged through heat (q) and work (W), as described by the first law of thermodynamics: $\Delta U = q + W$ .

Step 3: Apply to the System: Since the system is thermally isolated, q = 0. Therefore, energy can only be exchanged through work.

Step 4: Conclusion: The mode of transference of energy is work.

Why other options are wrong:
- Option A: Heat cannot be exchanged because the system is thermally isolated.
- Option C: Heat and radiation are not possible modes of energy transfer for a thermally isolated system.
- Option D: None of these is incorrect because work is a possible mode of energy transfer.

Key Concepts:
- Thermally Isolated System
- First Law of Thermodynamics
- Heat
- Work

Memory Aids:
- Thermally isolated = no heat exchange

Quick Tips:
- Remember that a thermally isolated system can still exchange energy through work.

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Remembering +4 | -1

Work done in reversible adiabatic process is given by:

View answer and explanation

Correct option: B

\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)

Option B (

\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)

) is correct because this formula represents the work done in a reversible adiabatic process, relating it to the number of moles, gas constant, adiabatic index, and temperature change.

Solution approach

Correct Answer + Core Reason: $\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)$ . This formula gives the work done in a reversible adiabatic process.

Step-by-Step Solution:

Step 1: Adiabatic Process: An adiabatic process is one in which no heat is exchanged with the surroundings (Q = 0).

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

Step 3: Work Done in Reversible Adiabatic Process: The work done (W) in a reversible adiabatic process is given by the formula: $W = \frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)$ , where n is the number of moles, R is the ideal gas constant, $\gamma$ is the adiabatic index ( $\frac{{C_p}}{{C_v}}$ ), $T_1$ is the initial temperature, and $T_2$ is the final temperature.

Step 4: Conclusion: Therefore, the work done in a reversible adiabatic process is $\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)$ .

Why other options are wrong:
- Option A: $2.303RT\log \log \frac{{{V_2}}}{{{V_1}}}$ : This formula represents the work done in a reversible isothermal process, not an adiabatic process.
- Option C: $2.303RT\log \log \frac{{{V_1}}}{{{V_2}}}$ : This is the negative of the work done in a reversible isothermal process, not an adiabatic process.
- Option D: None of these: This is incorrect because option B provides the correct formula.

Key Concepts:
- Adiabatic Process
- Reversible Process
- Work Done
- Thermodynamics

Memory Aids:
- Adiabatic Work = nR/(gamma-1) * Delta T

Quick Tips:
- Remember the difference between isothermal and adiabatic processes. Isothermal processes occur at constant temperature, while adiabatic processes occur with no heat exchange.

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Remembering +4 | -1

The van’t Hoff reaction isotherm is:

View answer and explanation

Correct option: B

-∆ G = RT\,{\log _e}{K_p}

Option B (-∆G = RT logₑ Kₚ) is correct because it represents the relationship between Gibbs free energy change and the equilibrium constant.

Solution approach

Correct Answer + Core Reason: $-\Delta G = RT\\,{\log _e}{K_p}$ . This equation relates Gibbs free energy change to the equilibrium constant.

Step-by-Step Solution:

Step 1: Define the van't Hoff reaction isotherm.

Step 2: The van't Hoff reaction isotherm relates the change in Gibbs free energy ( $\Delta G$ ) to the equilibrium constant ( $K_p$ ) and temperature (T).

Step 3: The equation is given by $-\Delta G = RT\\,{\log _e}{K_p}$ or $\Delta G = -RT\\,\ln{K_p}$ .

Why other options are wrong:
- Option A: $\Delta G = RT{\log _e}{K_p}$ is incorrect because it has the wrong sign. It should be negative.
- Option C: $\Delta G = R{T^2}\ln \ {K_p}$ is incorrect because the temperature is not squared, and the sign is wrong.
- Option D: None of these is incorrect because option B is the correct equation.

Key Concepts:
- Gibbs free energy
- Equilibrium constant
- Van't Hoff reaction isotherm

Memory Aids:
- Remember that $\Delta G$ is negative for spontaneous reactions.

Quick Tips:
- Know the relationship between Gibbs free energy and the equilibrium constant.

118 Medium

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Analyzing +4 | -1

Given are two statements one is labelled as Assertion A and other is labelled as Reason R.
Assertion A: Magnesium can reduce

A{l_2}{O_3}

at a temperature below

1350^\circ \,C

, while above

1350^\circ \,C

aluminium can reduce

MgO

.
Reason R: The melting and boiling points of magnesium are lower than those of aluminium.
In light of the above statements, choose most appropriate answer from the options given below:

View answer and explanation

Correct option: B

Both A and R are correct, but R is NOT the correct explanation of A.

Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion. The Ellingham diagram explains the assertion.

Solution approach

Correct Answer + Core Reason: Both A and R are correct, but R is NOT the correct explanation of A. The relative reducing power of Mg and Al depends on temperature, which is related to the Gibbs free energy change of the oxide formation, not just the melting/boiling points.

Step-by-Step Solution:
Step 1: Analyze Assertion A: The assertion states that Mg can reduce $Al_2O_3$ at lower temperatures, while Al can reduce MgO at higher temperatures. This is true and can be explained by the Ellingham diagram.
Step 2: Analyze Reason R: The reason states that the melting and boiling points of magnesium are lower than those of aluminum. This is also true.
Step 3: Determine the relationship: While both statements are correct, the melting and boiling points do not directly explain why the reducing power of Mg and Al changes with temperature. The Ellingham diagram, which plots the Gibbs free energy of oxide formation as a function of temperature, provides the correct explanation.

Why other options are wrong:
- Option A: Both A and R are correct, and R is correct explanation of A. This is incorrect because the melting/boiling points are not the direct cause of the change in reducing power.
- Option C: A is correct R is not correct. This is incorrect because both A and R are correct.
- Option D: A is not correct, R is correct. This is incorrect because A is also correct.

Key Concepts:
- Ellingham Diagram
- Reducing Power
- Gibbs Free Energy of Formation

Memory Aids:
- Remember that the Ellingham diagram shows the temperature dependence of the stability of metal oxides.

Quick Tips:
- When comparing reducing power, consider the Gibbs free energy of formation of the oxides at different temperatures.

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Application +4 | -1

Given,

\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2} + 94.2\mathrm{kcal} \ldots \mathrm{.}\left( \mathrm{i} \right)

{\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2} \to {\mathrm{H}_2}\mathrm{O} + 68.3\mathrm{kcal} \ldots .\left( {\mathrm{ii}} \right)

\mathrm{C}{\mathrm{H}_4} + 2{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O} + 210.8\mathrm{kcal} \ldots \mathrm{.}\left( {\mathrm{iii}} \right)

The heat of formation in kcal will be :

View answer and explanation

Correct option: C

-20

Option C (-20) is correct because it correctly calculates the heat of formation of methane by using Hess's Law and the given combustion reactions.

Solution approach

Correct Answer + Core Reason: -20. This is the correct heat of formation of methane.

Step-by-Step Solution:

Step 1: Identify the target reaction: Formation of methane from its elements.
$C(s) + 2H_2(g) \rightarrow CH_4(g) \quad \Delta H_f = ?$

Step 2: Write down the given reactions and their enthalpies:
(i) $C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_1 = -94.2 \,kcal$
(ii) $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H_2 = -68.3 \,kcal$
(iii) $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_3 = -210.8 \,kcal$

Step 3: Apply Hess's Law to manipulate the given reactions to obtain the target reaction.
We want: $C(s) + 2H_2(g) \rightarrow CH_4(g)$
Add reaction (i) and 2 times reaction (ii):
$C(s) + O_2(g) + 2H_2(g) + O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H = \Delta H_1 + 2\Delta H_2 = -94.2 + 2(-68.3) = -230.8 \,kcal$
Reverse reaction (iii):
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad \Delta H = -\Delta H_3 = 210.8 \,kcal$
Add the two modified reactions:
$C(s) + 2H_2(g) \rightarrow CH_4(g) \quad \Delta H_f = -230.8 + 210.8 = -20 \,kcal$

Step 4: Final Answer: -20 kcal

Why other options are wrong:
- Option A: 45.9 is incorrect. It's likely a result of adding or subtracting the enthalpies in the wrong order or with incorrect signs.
- Option B: 47.8 is incorrect. It's likely a result of adding or subtracting the enthalpies in the wrong order or with incorrect signs.
- Option D: 47.3 is incorrect. It's likely a result of adding or subtracting the enthalpies in the wrong order or with incorrect signs.

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

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

Quick Tips:
- Write down the target reaction.
- Manipulate the given reactions to match the target reaction.
- Pay attention to the signs when reversing reactions.

120 Medium

120 of 244Question permalink

Analyzing +4 | -1

Statement - I: In an adiabatic expansion of an ideal gas, the entropy of the surrounding
remains unchanged independent whether the expansion is reversible or irreversible.

Statement - II: In an adiabatic process, the surrounding does not exchange heat with the
system and the process may be reversible or irreversible for system but it always remain
reversible for the sorrounding.

View answer and explanation

Correct option: A

If both Statement-I and Statement-II are true, and Statement-II is the correct
explanation of Statement-I

Option A is correct because both statements are true, and Statement-II correctly explains why the entropy of the surrounding remains unchanged in an adiabatic expansion.

Solution approach

Correct Answer + Core Reason: If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I. In an adiabatic process, no heat is exchanged with the surroundings, thus the entropy change of the surroundings is zero.

Step-by-Step Solution:

Step 1: Analyze Statement-I: In an adiabatic expansion, by definition, there is no heat exchange between the system and the surroundings (Q = 0). Therefore, the entropy change of the surroundings is zero, regardless of whether the expansion is reversible or irreversible. Thus, Statement-I is true.

Step 2: Analyze Statement-II: An adiabatic process is defined by no heat exchange (Q = 0) with the surroundings. The process within the system can be reversible or irreversible. However, since the surroundings do not exchange heat, the process is always considered reversible for the surroundings. Thus, Statement-II is true.

Step 3: Determine if Statement-II explains Statement-I: Since Statement-II explains why the entropy of the surroundings remains unchanged (because there is no heat exchange), Statement-II is the correct explanation of Statement-I.

Why other options are wrong:
- Option B: Statement-II is the correct explanation of Statement-I.
- Option C: Statement-II is not false.
- Option D: Statement-I is not false.

Key Concepts:
- Adiabatic process
- Entropy
- Reversible and Irreversible processes

Memory Aids:
- Adiabatic: No heat exchange (Q = 0).

Quick Tips:
- Remember that the entropy change of the surroundings is zero in an adiabatic process.

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