$2Zn+O_2\to 2ZnO$ $$ $$ $$ $∆G°=-616$ $J$
$2Zn+S_2\to 2ZnS$ $$ $$ $$ $$ $∆G°=-293$ $J$
$S_2+2O_2\to 2S{O}_2$ $$ $$ $$ $$ $∆G°=-408J$

$∆G°$ for the following reaction is:

$2ZnS+3O_2\to 2ZnO+2S{O}_2$

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^–3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol^–1, the pressure at which graphite will be transformed into diamond at 298 K is:

1. 11.08$\times {10}^8$ $\mathrm{Pa}$

2. $9.92\times {10}^7$ $\mathrm{Pa}$

3. $9.92\times {10}^6$ $\mathrm{Pa}$

4. 11.08$\times {10}^5$ $\mathrm{Pa}$

The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are –382.64 kJ mol^–1 and –145.6 JK^–1 mol^–1, respectively. Standard Gibbs energy change for the same reaction at 298 K is:

1. –339.3 kJ mol^–1

2. – 439.3 kJ mol^–1

3. –523.2 kJ mol^–1

4. –221.1 kJ mol^–1