InterviewSolution
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InterviewSolution
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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. |
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Answer» Roasting of zinc sulphide to zinc oxide may be expressed as: `2ZnS(s) +3O_(2)(g) rarr 2ZnO(s) +2SO_(2)(g)` The given equations are: i. `S_(2)(s) +2O_(2)(g) rarr 2SO_(2)(g), Delta_(r)G^(Theta) =- 544kJ mol^(-1)` ii. `2Zn(s) +O_(2)(g) rarr 2ZnO(s), Delta_(r)G^(Theta) =- 480 kJ mol^(-1)` iii. `2Zn(s) +S_(2)(s) rarr 2ZnS(s), Delta_(r)G^(Theta) =- 293 kJ mol^(-1)` Subtract equation (iii) from equaiton (ii), we get iv. `2ZnS(s) +O_(2)(g) rarr 2ZnO(s) +S_(2)(s), Delta_(r)G^(Theta) =- 187 kJ mol^(-1)` Add equations (i) and (iv) `2ZnS(s) +3O_(2)(g) rarr 2ZnS(s) +2SO_(2)(g), Delta_(r)G^(Theta) =- 731 kJ mol^(-1)` Since `DeltaG^(Theta)` for the process is negative, the process of roasting of zinc sulphide is a spontaneous process. |
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