At 1500 K the standard Gibbs energies per mole of O2 are -887.2 kJ for 4/3 Al + O2 -> 2/3 Al2O3 and -514.6 kJ for 4/3 Cr + O2 -> 2/3 Cr2O3. Find Delta G deg in kJ at 1500 K for Cr2O3 + 2Al -> Al2O3 + 2Cr (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3mt043

At $1500 K$ the standard Gibbs energies per mole of $O X_{2}$ are $- 887.2 kJ$ for $\frac{4}{3} Al + O X_{2} \frac{2}{3} Al X_{2} O X_{3}$ and $- 514.6 kJ$ for $\frac{4}{3} Cr + O X_{2} \frac{2}{3} Cr X_{2} O X_{3}$ . Find $Δ G^{\circ}$ in $kJ$ at $1500 K$ for $Cr X_{2} O X_{3} + 2 Al Al X_{2} O X_{3} + 2 Cr$ (nearest integer).

Numerical answer type, answer in kJ, tolerance 0.5.

The reaction is the Al line minus the Cr line, scaled to

\frac{3}{2}

mol of

O X_{2}

(one

Cr X_{2} O X_{3}

contains 3 O atoms). Per mol

O X_{2}

- 887.2 - (- 514.6) = - 372.6 kJ

. Multiplying by 1.5 gives

- 558.90 kJ

, i.e. -559 kJ to the nearest integer.