Q. No.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
1. G = – 2.30RT log K
2. G = 2.30RT log K
3. Go = – 2.30RT log K
4. Go = 2.30RT log K
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\(\mathrm{Ag_2 CO_3 (s) \rightarrow 2Ag^+ (aq) + CO^{2-}_3 (aq)}\)
[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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Given the following reaction:
\(4H(g)\)→ \(2 H_{2}\)\((g)\)
The enthalpy change for the reaction is -869.6 kJ. The dissociation energy of the H-H bond is:
1. -869.6 kJ
2. +434.8kJ
3. +217.4kJ
4. -434.8 kJ
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Give the following bond energies:
H—H bond energy: 431.37 kJ mol-1
C=C bond energy: 606.10 kJ mol-1
C—C bond energy: 336.49 kJ mol-1
C—H bond energy: 410.50 kJ mol-1
Enthalpy for the following reaction will be:
2. -243.6 kJ mol-1
3. -120.0 kJ mol-1
4. 553.0 kJ mol-1
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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
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1. 1.968 V
2. 2.0968 V
3. 1.0968 V
4. 0.0968 V
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The bond energy of H—H and Cl-Cl is 430 kJ
mol-1 and 240 kJ mol-1 respectively and ΔHf for HCl is -90 kJ mol-1. The bond enthalpy of HCl is:
1. 290
2. 380
3. 425
4. 245
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Two moles of an ideal gas is heated at a constant pressure of one atmosphere from 27C to 127C. If then q and for the process are respectively:
1. 6362.8 J, 4700 J
2. 3037.2 J, 4700 J
3. 7062.8 J, 5400 J
4. 3181.4 J, 2350 J
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
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The standard heat of combustion of propane is –2220.1 kJ mol–1. The standard heat of vaporisation of liquid water is 44.0 kJ mol–1. The enthalpy change for the reaction is–
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O(g)
| 1. | –2220.1 kJ | 2. | –2044.1 kJ |
| 3. | –2396.1 kJ | 4. | –2176.1 kJ |
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