Q. No.For the reaction : \(A_2 + B_2 \xrightarrow {500~K} 2 AB\) : where, Log K = 2.2
Calculate the value of x:
1. 70
2. 60
3. 50
4. 80
| \(\mathrm{H}_{\mathrm{f}}{ }^0 \text { (KJ/mole) }\) | \(\mathrm{S}_{\mathrm{f}}^0(\mathrm{~J} / \mathrm{K} \text {-mole })\) | |
| AB | 32 | 240 |
| \(A_2\) | 6 | 224 |
| \(B_2\) | x | 238 |
Calculate the value of x:
1. 70
2. 60
3. 50
4. 80
Subtopic: Enthalpy & Internal energy | Spontaneity & Entropy |
Level 3: 35%-60%
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Which of the following is true for an endothermic reaction that is non-spontaneous at the freezing point of water and spontaneous at the boiling point of water?
| 1. | Both \(\Delta H\) and \(\Delta S\) are positive |
| 2. | \(\Delta H\) is negative but \(\Delta S\) is positive |
| 3. | \(\Delta H\) is positive but \(\Delta S\) is negative |
| 4. | Both \(\Delta H\) and \(\Delta S\) are negative |
Subtopic: Spontaneity & Entropy | Gibbs Energy Change |
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Level 3: 35%-60%
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Consider the table given below and choose the correct option:
The effect of temperature on spontaneity can be represented by which of the following?
1. (B) and (D) only
2. (A) and (D) only
3. (B) and (C) only
4. (A) and (C) only
| ΔH | ΔS | Temperature | Spontaneity | |
| (A) | + | - | any T | Spontaneous |
| (B) | + | + | low T | Non spontaneous |
| (C) | - | - | low T | Spontaneous |
| (D) | - | + | any T | Non spontaneous |
The effect of temperature on spontaneity can be represented by which of the following?
1. (B) and (D) only
2. (A) and (D) only
3. (B) and (C) only
4. (A) and (C) only
Subtopic: Spontaneity & Entropy | Gibbs Energy Change |
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Level 1: 80%+
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Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 JK–1mol–1 respectively. The temperature in Kelvin at which the given reaction will be at equilibrium is:
\(\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_5,~ \Delta \mathrm{H}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
1. 300
2. 400
3. 600
4. 700
\(\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_5,~ \Delta \mathrm{H}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
1. 300
2. 400
3. 600
4. 700
Subtopic: Thermodynamics' Properties and process | Spontaneity & Entropy | Gibbs Energy Change |
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Level 2: 60%+
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Ice at –5°C is heated to convert into vapour with temperature of 110°C at atmospheric pressure. The entropy change associated with this process can be obtained from which of the following?
[Where: \(T_f\) is melting point and \(T_b\) is boiling point]
1. \(\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\frac{\Delta \mathrm{H}_{\text {melting }}}{273}+\frac{\Delta \mathrm{H}_{\text {boiling }}}{373}\)
2. \(\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}}}{\mathrm{~T}} \mathrm{dT}+\frac{\Delta \mathrm{H}_{\mathrm{m}}, \text { fusion }}{\mathrm{T}_{\mathrm{f}}}+\int_{273 \mathrm{~K}}^{373 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}}+ \frac{\Delta \mathrm{H}_{\mathrm{m}, \text { vaporisation }}}{\mathrm{T}_{\mathrm{b}}}\)\(+\int_{373 \mathrm{~K}}^{383 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}}\)
3. \(\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\frac{\mathrm{q}_{\mathrm{rev}}}{\mathrm{~T}}\)
4. \(\begin{aligned} & \int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT} +\int_{273 \mathrm{~K}}^{373 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}+\int_{373 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT} \end{aligned}\)
[Where: \(T_f\) is melting point and \(T_b\) is boiling point]
1. \(\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\frac{\Delta \mathrm{H}_{\text {melting }}}{273}+\frac{\Delta \mathrm{H}_{\text {boiling }}}{373}\)
2. \(\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}}}{\mathrm{~T}} \mathrm{dT}+\frac{\Delta \mathrm{H}_{\mathrm{m}}, \text { fusion }}{\mathrm{T}_{\mathrm{f}}}+\int_{273 \mathrm{~K}}^{373 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}}+ \frac{\Delta \mathrm{H}_{\mathrm{m}, \text { vaporisation }}}{\mathrm{T}_{\mathrm{b}}}\)\(+\int_{373 \mathrm{~K}}^{383 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}}\)
3. \(\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\frac{\mathrm{q}_{\mathrm{rev}}}{\mathrm{~T}}\)
4. \(\begin{aligned} & \int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT} +\int_{273 \mathrm{~K}}^{373 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}+\int_{373 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT} \end{aligned}\)
Subtopic: Cp & Cv | Spontaneity & Entropy |
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The standard enthalpy and entropy changes of decomposition of N2O4 to NO2 are 55.0 kJmol–1 and 175.0 JK–1 mol–1 respectively. The standard free energy change for this reaction at 25°C in J mol–1 is:
| 1. | 2750 | 2. | 2850 |
| 3. | 2875 | 4. | 2900 |
Subtopic: Spontaneity & Entropy | Gibbs Energy Change |
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For a certain reaction, if \(\Delta_{\mathrm{r}} \mathrm{H}\) is \(400 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta \mathrm{S}=0.2 \mathrm{~kJ} / \mathrm{mol}\), calculate the minimum temperature (in Kelvin) at which the reaction becomes spontaneous.
1. 1000
2. 2000
3. 4000
4. 100
1. 1000
2. 2000
3. 4000
4. 100
Subtopic: Spontaneity & Entropy |
92%
Level 1: 80%+
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Identify correct match using Column I & Column II
| Column I | Column II | ||
| (i) | Spontaneous process | (a) | Isothermal and isobaric process |
| (ii) | \(\Delta H^\circ\) | (b) | \(\Delta H<0 \) |
| (iii) | \(\Delta T=0, \Delta P=0 \) | (c) | \(\Delta G<0 \) |
| (iv) | Exothermic process | (d) | (Bond energy of reactant) - (Bond energy of product) |
| I | II | III | IV | |
| 1. | c | d | a | b |
| 2. | b | a | c | d |
| 3. | d | b | c | d |
| 4. | a | d | b | c |
Subtopic: First Law of Thermodynamics | Spontaneity & Entropy | Gibbs Energy Change |
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Which of the following relations is not correct?
1. \(\Delta \mathrm{H}=\Delta \mathrm{U}-\mathrm{P} \Delta \mathrm{V}\)
2. \(\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}\)
3. \(\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \geq 0\)
4. \(\Delta G=\Delta H-T \Delta S\)
1. \(\Delta \mathrm{H}=\Delta \mathrm{U}-\mathrm{P} \Delta \mathrm{V}\)
2. \(\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}\)
3. \(\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \geq 0\)
4. \(\Delta G=\Delta H-T \Delta S\)
Subtopic: 2nd & 3rd Law of Thermodynamics | First Law of Thermodynamics | Spontaneity & Entropy |
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Calculate the minimum temperature at which the following reaction becomes spontaneous:
FeO(s) + C(graphite) → Fe(s) + CO(g)
Given:
ΔHf°(FeO) = –266.3 kJ mol⁻¹
ΔHf°(C) = 0
ΔHf°°(Fe) = 0
ΔHf°(CO) = –110.5 kJ mol⁻¹
ΔS°(FeO) = 57.49 J mol⁻¹ K⁻¹
ΔS°(C) = 5.74 J mol⁻¹ K⁻¹
ΔS°(Fe) = 27.28 J mol⁻¹ K⁻¹
ΔS°(CO) = 197.6 J mol⁻¹ K⁻¹
1. 365 K
2. 653 K
3. 1432 K
4. 964 K
Subtopic: Spontaneity & Entropy | Gibbs Energy Change |
67%
Level 2: 60%+
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