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 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

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)

For a given reaction, $∆$H =35.5 kJ mol^-1 and $∆$S = 83.6JK^-1 mol^-1. The reaction is spontaneous at:
(Assume that $∆$H and $∆$S do not vary with temperature)

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

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

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

The free energy change $(∆G°)$ is negative when -

1. The surroundings do no electrical work on the system.

2. The surroundings do electrical work on the system.

3. The system does electrical work on the surroundings.

4. The system does no electrical work on the surroundings.

In the reaction, $∆$H and $∆$S both are positive. The condition under which the reaction would not be spontaneous is -

1. $∆$H>T$∆$S                             

2. $∆$S=$∆$H/T

3. $∆$H=T$∆$S                               

4. All of the above