At 1400 K (above the boiling point of zinc), take Delta G deg = -930 + 0.405 T kJ for 2Zn(g) + O2 -> 2ZnO(s) and Delta G deg = -221 - 0.179 T kJ for 2C + O2 -> 2CO. Calculate Delta G deg in kJ per mole of ZnO for ZnO(s) + C(s) -> Zn(g) + CO(g) at 1400 K (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3mt207

At 1400 K (above the boiling point of zinc), take $Δ G^{\circ} = - 930 + 0.405 T$ kJ for $2 Zn (g) + O X_{2} 2 ZnO (s)$ and $Δ G^{\circ} = - 221 - 0.179 T$ kJ for $2 C + O X_{2} 2 CO$ . Calculate $Δ G^{\circ}$ in kJ per mole of ZnO for $ZnO (s) + C (s) Zn (g) + CO (g)$ at 1400 K (nearest integer).

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Required reaction = half of [(C to CO) minus (Zn to ZnO)]. At 1400 K: C-to-CO =

- 221 - 0.179 \times 1400 = - 471.6

kJ; Zn-to-ZnO =

- 930 + 0.405 \times 1400 = - 363.0

kJ.

Δ G^{\circ} = \frac{1}{2} (- 471.6 + 363.0) = - 54.3

kJ, i.e. -54 kJ/mol.