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Q. No.Consider the following reactions:
\(\mathrm{S}_{(\mathrm{~g})}+\frac{3}{2} \mathrm{{O}_2}_{(\mathrm{~g})} \rightarrow \mathrm{{SO}_3}_{(\mathrm{~g})}+2 \mathrm{x} \mathrm{~kcal}\)
\(\mathrm{{SO}_2}_{(\mathrm{~g})}+\frac{1}{2} \mathrm{{O}_2}_{(\mathrm{~g})} \rightarrow \mathrm{{SO}_3}_{(\mathrm{~g})}+\mathrm{y} \mathrm{~kcal}\)
The heat of formation of SO2(g) is given by:
1.
\(\dfrac{2 \mathrm{x}}{\mathrm{y}} \mathrm{kcal}\)
2.
y - 2x kcal
3.
2x + y kcal
4.
x + y kcal
Subtopic: Thermochemistry |
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Find out the magnitude of work done in the process \(\mathrm{ABCD}\) (in \(\text{kJ}\)):
\(\mathrm{(1~atm~Lit=101.3~J)}\)
\(\mathrm{(1~atm~Lit=101.3~J)}\)
| 1. | 304 | 2. | 324 |
| 3. | 360 | 4. | 388 |
Subtopic: First Law of Thermodynamics |
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Level 1: 80%+
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Calculate the enthalpy change for the combustion of octane if the heat capacity of a bomb calorimeter is 5 kJ K⁻¹ and the temperature of the calorimeter increases by 5°C during combustion in excess oxygen.
| 1. | 20 | 2. | 25 |
| 3. | 30 | 4. | 35 |
Subtopic: Enthalpy & Internal energy | Cp & Cv |
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One mole of an ideal gas expands from \(10 ~\mathrm{dm}^3\) to \(20 ~\mathrm{dm}^3\) through isothermal reversible process. Find \(\Delta U, q \) & \(w\):
| 1. | \(\Delta U=0, q=0, w=0\) | 2. | \(\Delta U=0, q \neq 0, w \neq 0\) |
| 3. | \(\Delta \mathrm{U} \neq 0, \mathrm{q}=0, \mathrm{w} \neq 0\) | 4. | \(\Delta U \neq 0, q \neq 0, w=0\) |
Subtopic: First Law of Thermodynamics |
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Which of the following statement(s) is/are correct for the adiabatic process?
1. (A) and (C)
2. (B) and (C)
3. (A) and (D)
4. (B) and (D)
| (A) | Molar heat capacity is zero. |
| (B) | Molar heat capacity is infinite. |
| (C) | Work done on gas is equal to the increase in internal energy. |
| (D) | The increase in temperature results in a decrease in internal energy. |
1. (A) and (C)
2. (B) and (C)
3. (A) and (D)
4. (B) and (D)
Subtopic: First Law of Thermodynamics |
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In an adiabatic process, the magnitude of work done in the case of one step & infinite steps follows the order:
1. \(\left|\mathrm{W}_{\text {rev }}\right|_{\text {expansion }}>\left|\mathrm{W}_{\text {Irrev }}\right|_{\text {expansion }}\)
2. \(\left|\mathrm{W}_{\text {rev }}\right|_{\text {expansion }}<\left|\mathrm{W}_{\text {Irrev }}\right|_{\text {expansion }}\)
3. \(\left|\mathrm{W}_{\text {rev }}\right|_{\text {expansion }}=\left|\mathrm{W}_{\text {Irrev }}\right|_{\text {expansion }}\)
4. Can't be predicted
1. \(\left|\mathrm{W}_{\text {rev }}\right|_{\text {expansion }}>\left|\mathrm{W}_{\text {Irrev }}\right|_{\text {expansion }}\)
2. \(\left|\mathrm{W}_{\text {rev }}\right|_{\text {expansion }}<\left|\mathrm{W}_{\text {Irrev }}\right|_{\text {expansion }}\)
3. \(\left|\mathrm{W}_{\text {rev }}\right|_{\text {expansion }}=\left|\mathrm{W}_{\text {Irrev }}\right|_{\text {expansion }}\)
4. Can't be predicted
Subtopic: Thermodynamics' Properties and process |
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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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An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path \(\mathrm{A} \rightarrow \mathrm{~B} \rightarrow \mathrm{C} \rightarrow \mathrm{D} \rightarrow \mathrm{~A}\) as shown in the three cases below:
Choose the correct option regarding change in internal energy, \(\Delta \mathrm{U} \):
1. \(\Delta \mathrm{U}\) (Case-III) > \(\Delta \mathrm{U}\) (Case-II) > \(\Delta \mathrm{U}\) (Case-I)
2. \(\Delta \mathrm{U}\) (Case-I) > \(\Delta \mathrm{U}\) (Case-II) > \(\Delta \mathrm{U}\) (Case-III)
3. \(\Delta \mathrm{U}\) (Case-I) > \(\Delta \mathrm{U}\) (Case-III) > \(\Delta \mathrm{U}\) (Case-II)
4. \(\Delta \mathrm{U}\) (Case-I) = \(\Delta \mathrm{U}\) (Case-II) = \(\Delta \mathrm{U}\) (Case-III)
Choose the correct option regarding change in internal energy, \(\Delta \mathrm{U} \):
1. \(\Delta \mathrm{U}\) (Case-III) > \(\Delta \mathrm{U}\) (Case-II) > \(\Delta \mathrm{U}\) (Case-I)
2. \(\Delta \mathrm{U}\) (Case-I) > \(\Delta \mathrm{U}\) (Case-II) > \(\Delta \mathrm{U}\) (Case-III)
3. \(\Delta \mathrm{U}\) (Case-I) > \(\Delta \mathrm{U}\) (Case-III) > \(\Delta \mathrm{U}\) (Case-II)
4. \(\Delta \mathrm{U}\) (Case-I) = \(\Delta \mathrm{U}\) (Case-II) = \(\Delta \mathrm{U}\) (Case-III)
Subtopic: Thermodynamics' Properties and process |
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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 bond dissociation enthalpy of X2 \(\Delta \mathrm{H}_{\mathrm{bond}}^{\mathrm{o}}\) in kJ mol–1 calculated from the given data is:
\(\begin{aligned} &\small \mathrm{MX}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) ,\Delta \mathrm{H}_{\text {lattice }}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}), \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})^{-}+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\small \mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) ,\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)
[Given: M+X – is a pure ionic compound and X forms a diatomic
molecule X2 in the gaseous state]
1. 100
2. 150
3. 200
4. 250
\(\begin{aligned} &\small \mathrm{MX}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) ,\Delta \mathrm{H}_{\text {lattice }}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}), \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})^{-}+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \small \mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\small \mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) ,\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)
[Given: M+X – is a pure ionic compound and X forms a diatomic
molecule X2 in the gaseous state]
1. 100
2. 150
3. 200
4. 250
Subtopic: Hess's Law | Thermochemistry |
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