For the oxide formations per mole of O2 (in kJ, T in K): 2Zn + O2 -> 2ZnO, Delta G deg = -690 + 0.198 T; 2C + O2 -> 2CO, Delta G deg = -222 - 0.1752 T. Treating these as valid throughout, calculate Delta G deg at 1673 K for ZnO + C -> Zn + CO per mole of ZnO, in kJ/mol, to the nearest integer.

JEE Advanced levelNumerical answerDifficulty 3 of 3mt097

For the oxide formations per mole of $O X_{2}$ (in $kJ$ , $T$ in K): $2 Zn + O X_{2} 2 ZnO$ , $Δ G^{\circ} = - 690 + 0.198 T$ ; $2 C + O X_{2} 2 CO$ , $Δ G^{\circ} = - 222 - 0.1752 T$ . Treating these as valid throughout, calculate $Δ G^{\circ}$ at 1673 K for $ZnO + C Zn + CO$ per mole of $ZnO$ , in $kJ/mol$ , to the nearest integer.

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

Subtract the two formation equations and halve:

ZnO + C Zn + CO

has

Δ G^{\circ} = \frac{1}{2} (Δ G_{CO} - Δ G_{Z n O})

. At 1673 K,

Δ G_{Z n O} = - 358.75

and

Δ G_{CO} = - 515.11 kJ

. Then

Δ G^{\circ} = \frac{1}{2} (- 515.11 - (- 358.75)) = - 78.18 kJ/mol \approx - 78

, negative, so the reduction is feasible.