Question

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

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

S⁰298K for H2g = 197.7 J K-1 mol -1

S⁰298K (N2g) = 191.6 J K-1 mol -1

S⁰298K (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

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

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