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

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. $∆$G^o = – 2.30RT log K
4. $∆$G^o = 2.30RT log K

Hydrolysis of sucrose is given by the following reaction

Sucrose + H₂O $\rightleftharpoons$ Glucose + Fructose

If the equilibrium constant (K_c) is 2$\times$10¹³ at 300 K, the value of ${∆}_r{G}^{⊝}$ at the same temperature will be:

1. 8.314 J mol^–1 K^–1$\times$300 K$\times$ln (2$\times$10¹³)

2. 8.314 J mol^–1 K^–1$\times$300 K$\times$ln (3$\times$10¹³)

3. –8.314 J mol^–1 K^–1$\times$300 K$\times$ln (4$\times$10¹³)

4. –8.314 J mol^–1 K^–1$\times$300 K$\times$ln (2$\times$10¹³)

The equilibrium constant for a reaction is 10. The value of $∆\mathrm{G}°$ will be:

($\mathrm{R}$ $=$ $8.314$ $\mathrm{J}$ ${\mathrm{K}}^{-1}$ ${\mathrm{mol}}^{-1}$ $;$ $\mathrm{T}$ $=$ $300$ $\mathrm{K}$)

$1.$ $-5.74$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$2.$ $-$ $5.74$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$
$3.$ $+$ $4.57$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$4.$ $-57.4$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$