Question

Calculate ∆S⁰R for the reaction 3H₂(g) + N₂(g) = 2NH₃(g) at 700 K given the following:

S⁰_298K for H2g = 197.7 J K^-1 mol ^-1

S⁰_298K (N2g) = 191.6 J K^-1 mol ^-1

S⁰_298K (NH3g) = 130.7 J K^-1 mol ^-1

Cp (H₂g) in J K^-1 mol ^-1 = 22.7 + 4.4 x 10^-2 T – 1.1x10^-4 T²

Cp (N₂g) in J K^-1 mol ^-1 = 30.8 – 1.2 x 10^-2 T + 2.4 x10^-5 T²

Cp (NH₃g) in J K^-1 mol ^-1 = 29.3 + 1.1 x 10^-2 T + 4.2x10^-5 T²

Assume that T is in K for heat capacity. Previous solution attempt is incorrect

Answer 1

To perform the calculation of the standard entropy change at 700 K, the integral of the respective Cp / T for each substance must be added to the value of S (298K):

S 700K for H2g = 197.7 J / mol * K + ∫ 22.7 / T + 4.4 x 10-2 - 1.1x10-4 T (from 298 K to 700 K) = 197.7 + 15.01 = 212.71 J / mol * K

S 700K for N2g = 191.6 + ∫ 30.8 / T - 1.2 x 10-2 + ​​2.4 x10-5 T (from 298 K to 700 K) = 191.6 + 26.29 = 217.89 J / mol * K

S 700 K for NH3g = 130.7 + ∫ 29.3 / T + 1.1 x 10-2 + ​​4.2x10-5 T (from 298 K to 700 K) = 130.7 + 37.87 = 168.57 J / mol * K

Calculation of ΔSr:

ΔSr (700K) = n * S products - n * S reactive = 2 * 168.57 - 3 * 212.71 - 217.89 = -518.88 J / K