Q. No.2A (g) + B (g) → 2D (g)
\(ΔU^⊖\)=−10 \(kJ mol^{−1}\) and \(ΔS^⊖\)=−44 \(J K^{−1}\) at 298 K.
Identify the correct option with \(ΔG^⊖\) for the reaction and spontaneity of the reaction at 298 K.
(Given : R=\(8.31 ~J mol^{−1} K^{−1}\))
| 1. | −1.635 kJ mol−1, spontaneous |
| 2. | +0.63568 kJ mol−1, non-spontaneous |
| 3. | −0.63568 kJ mol−1, spontaneous |
| 4. | +1.635 kJ mol−1, non-spontaneous |
In the reaction, both H and S are positive. The condition(s) under which the reaction would not be spontaneous are/is:
1. H>TS
2. S=H/T
3. H=TS
4. All of the above
The values of ΔH and ΔS for the given reaction are 170 kJ and 170 JK-1, respectively.
C(graphite) + CO2(g)→2CO(g)
This reaction will be spontaneous at:
1. 710 K
2. 910 K
3. 1110 K
4. 510 K
Calculate (in kJ/mol) for from the and the values provided at 27C
\(\begin{array}{ll}
4 C r(s)+3 O_2(g) \rightarrow 2 C r_2 O_3(s), \\ \Delta_r G^{\circ}=-2093.4 k J / m o l \\
S^{\circ}(\mathrm{J} / / \mathrm{K} \mathrm{~mol}): S^{\circ}(C r, s)=24, \\ S^{\circ}\left(O_2, g\right)=205, \quad S^{\circ}\left(C r_2 O_3, s\right)=81
\end{array}\)
1. -2258.1 kJ/mol
2. -1129.05 kJ/mol
3. -964.35 kJ/mol
4. None of the above
Find the condition under which the standard Gibbs free energy change, ΔG°, is negative.
1. The surroundings perform no electrical work on the system.2. The surroundings perform electrical work on the system.
3. The system performs electrical work on the surroundings.
4. The system performs no electrical work on the surroundings.
'The free energy change due to a reaction is zero when-
1. The reactants are initially mixed.
2. A catalyst is added
3. The system is at equilibrium
4. The reactants are completely consumed
For a given reaction, if ΔH = 35.5 kJ/mol and ΔS = 83.6 J/K·mol, at what temperature is the reaction spontaneous?
(Assume ΔH and ΔS remain constant with temperature.)
| 1. | T < 425 K | 2. | T > 425 K |
| 3. | All temperatures | 4. | T > 298 K |
The correct statement for a reversible process in a state of equilibrium is:
1. G = – 2.30RT log K
2. G = 2.30RT log K
3. Go = – 2.30RT log K
4. Go = 2.30RT log K
\(\mathrm{Ag_2 CO_3 (s) \rightarrow 2Ag^+ (aq) + CO^{2-}_3 (aq)}\)
[Given: \(\text R = 8.314 \text J\text K^{–1} \text {mol}^{–1}\) ; \(\Delta \text G^\circ=+63.3~\text{kJ}\) ]
1. \(3.2 \times 10^{26}\)
2. \(8.0 \times 10^{-12}\)
3. \(2.9 \times 10^{-3}\)
4. \(7.9 \times 10^{-2}\)
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