Use the data below (at 298 K) for CaCO3(s) -> CaO(s) + CO2(g) and assume Delta H and Delta S are temperature independent. Delta_fH deg/kJ mol^-1: CaCO3=-1206.9, CaO=-635.1, CO2=-393.5. S deg/J mol^-1K^-1: CaCO3=92.9, CaO=39.7, CO2=213.7. The minimum temperature at which decomposition is spontaneous is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3td160

Use the data below (at $298 K$ ) for $CaCO X_{3} (s) CaO (s) + CO X_{2} (g)$ and assume $Δ H$ and $Δ S$ are temperature independent. $Δ_{f} H^{\circ} / kJ mo l^{- 1}$ : $CaCO X_{3} = - 1206.9$ , $CaO = - 635.1$ , $CO X_{2} = - 393.5$ . $S^{\circ} / J mo l^{- 1} K^{- 1}$ : $CaCO X_{3} = 92.9$ , $CaO = 39.7$ , $CO X_{2} = 213.7$ . The minimum temperature at which decomposition is spontaneous is closest to

Δ H = - 635.1 - 393.5 + 1206.9 = + 178.3 kJ

and

Δ S = 39.7 + 213.7 - 92.9 = + 160.5 J K^{- 1}

Δ G < 0

requires

T > Δ H /Δ S = 178300/160.5 = 1111 K

. Using only

S^{\circ} (CO X_{2})

gives

834

, forgetting the kJ to J conversion gives

1.11