Take R = 8.314 J mol^-1 K^-1. For N2O4(g) 2NO2(g) take Delta H deg = +57.0 kJ mol^-1 and Delta S deg = +176 J mol^-1 K^-1, independent of temperature. At what temperature does K_p (with standard state 1 bar) equal 10?

JEE Advanced levelSingle correct optionDifficulty 3 of 3ce158

Take $R = 8.314 J mo l^{- 1} K^{- 1}$ . For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ take $Δ H^{\circ} = + 57.0 kJ mo l^{- 1}$ and $Δ S^{\circ} = + 176 J mo l^{- 1} K^{- 1}$ , independent of temperature. At what temperature does $K_{p}$ (with standard state 1 bar) equal $10$ ?

RT ln K = T Δ S^{\circ} - Δ H^{\circ}

, so

T = Δ H^{\circ} / (Δ S^{\circ} - R ln K) = 57000/ (176 - 8.314 \times 2.3026) = 363.4 K

. The value 324 K is where

K = 1

, 614 K uses

ln K = 10

and 292 K has the wrong sign for

R ln K