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
Given the reaction:
X2O4(l) → 2XO2(g)
ΔU = 2.1 kcal, ΔS = 20 cal K–1 at 300 K
The value of ΔG is:
1. 2.7 kcal
2. –2.7 kcal
3. 9.3 kcal
4. –9.3 kcal
In which of the following reactions, the standard reaction entropy change
is positive, and standard Gibb's energy change
decreases sharply with increasing temperature?
1. | C(graphite) + \(\frac{1}{2}\)O2(g) → CO(g) |
2. | CO(g) + \(\frac{1}{2}\)O2(g) → CO2(g) |
3. | Mg(s) + \(\frac{1}{2}\)O2(g) → MgO(s) |
4. | \(\frac{1}{2}\)C(graphite) + \(\frac{1}{2}\)O2(g) → \(\frac{1}{2}\)CO2(g) |
Standard entropies of X2, Y2 and XY3 are 60, 40 and 50JK-1mol-1 respectively. For the reaction
to be at equilibrium, the temperature should be:
1. 750 K
2. 1000 K
3. 1250 K
4. 500 K
For vaporization of water at 1 atmospheric pressure, the values of ∆H and ∆S are 40.63 kJ mol–1 and 108.8 JK–1 mol–1, respectively. The temperature when Gibbs energy change (∆G) for this transformation will be zero, is:
1. 393.4 K
2. 373.4 K
3. 293.4 K
4. 273.4 K
Match List – I (Equations) with List – II (Type of processes) and select the correct option.
List – I | List – II | ||
Equation | Type of processes | ||
(a) | Kp > Q | (i) | Non-spontaneous |
(b) | ∆Gº < RT ln Q | (ii) | Equilibrium |
(c) | Kp = Q | (iii) | Spontaneous and endothermic |
(d) | T > \(\frac{\Delta H}{\Delta S}\) | (iv) | Spontaneous |
Options: | (a) | (b) | (c) | (d) |
1. | (iii) | (iv) | (ii) | (i) |
2. | (iv) | (i) | (ii) | (iii) |
3. | (ii) | (i) | (iv) | (iii) |
4. | (i) | (ii) | (iii) | (iv) |
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
1. ΔH = 0 and ΔS < 0
2. ΔH > 0 and ΔS > 0
3. ΔH < 0 and ΔS < 0
4. ΔH > 0 and ΔS < 0