Question

At what temperature will the equilibrium constant for the formation of $NOCl ( g )$ be $K _{ eq }= K _{ p }=1.0 \times 10^{3}$
$2 NO ( g )+ Cl _{2}( g ) \longrightarrow 2 NOCl ( g )$
$\Delta H ^{\circ}=-77.1\, kJ\, mol ^{-1}$
$\Delta S^{\circ}=-121.2 \,J \,mol ^{-1} K ^{-1}$

• A. 432 K
• B. 298 K
• C. 412 K
• D. 398 K

Answer 1

Answer: (A)

$\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}-2.303 \,R T\, \log K_{\text {eq }}=\Delta H^{\circ}-T \Delta S^{\circ}$

$T\left[\Delta S^{\circ}-2.303 \,R \,\log K_{\text {eq }}\right]=\Delta H^{\circ}$

$\therefore T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}-2.303\, R\, \log K_{e q}}$

$=\frac{-77.1 \,kJ}{-121.2\times10^{-3}kJ - 2.3030\times8.314\times10^{-3}\,kJ \,mol^{-1}\,K^{-1}\,log 1.0\times10^{3}} =432 K$