On the basic of the following `Delta_(r)G^(Theta)` values at `1073K`:
`S_(1)(s) +2O_(2)(g) rarr 2SO_(2)(g) Delta_(r)G^(Theta) =- 544 kJ mol^(-1)`
`2Zn(s) +O_(2)(g) rarr 2ZnO(s) Delta_(r)G^(Theta) =- 480 kJ mol^(-1)`
`2Zn(s) +S_(2)(s) rarr 2ZnS(s) Delta_(r)G^(Theta) =- 293 KJ mol^(-1)`
Show that roasting of zinc sulphide to zince oxide is a spontaneous process.

Explanation

To show that the roasting of zinc sulfide (ZnS) to zinc oxide (ZnO) is a spontaneous process, we need to calculate the change in Gibbs free energy (ΔG) for the reaction and check if it is negative. The roasting reaction can be represented as:

\[ 2 \text{ZnS}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{ZnO}(s) + 2 \text{SO}_2(g) \]

### Step 1: Identify the Reactions and Their ΔG Values
We have the following reactions and their corresponding ΔG values at 1073 K:

1. \( \text{S}_1(s) + 2 \text{O}_2(g) \rightarrow 2 \text{SO}_2(g) \)
...