Question

Calculate the entropy change for the reaction: N2(g) + 3H2(g) -> 2NH3(g)

Entropy data:
NH3 = 192.5 J/mol K
H2 = 130.6 J/mol K
N2 = 191.5 J/mol K

Answer 1

Answer:

ΔS⁰ = -198.3 J/K

Explanation:

N₂(g) + 3H₂(g)  → 2NH₃(g)

S⁰:   N₂(g) = 1mole(191.5J/mole·K) = 191.5J/K

S⁰: 3H₂(g) = 3moles(130.6J/mole·K) = 391.8J/K                                

S⁰: 2NH₃(g) = 2moles(192.5J/mole·K) = 385J/K

ΔS⁰ = ∑n·S⁰(Products) - ∑n·S⁰(Reactants

      =[385J/K] - [191.5J/K + 391.8J/K]

      = (385 - 191.5 - 391.8)J/K

      = -198.3J/K

Answer 2

Answer: The $\Delta S^o$ of the reaction is $-198.3Jmol^{-1}K^{-1}$

Explanation:

Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.

Mathematically,

$\Delta S_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}]$

For the given chemical equation:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

We are given:

$\Delta S^o_{NH_3}=192.5Jmol^{-1}K^{-1}\\\Delta S^o_{H_2}=130.6Jmol^{-1}K^{-1}\\\Delta S^o_{N_2}=191.5Jmol^{-1}K^{-1}$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{NH_3})]-[(1\times \Delta S^o_{N_2})+(3\times \Delta S^o_{H_2})]$

$\Delta S^o=[(2\times 192.5)]-[(1\times 191.5)+(3\times 130.6)]=-198.3Jmol^{-1}K^{-1}$

Hence, the $\Delta S^o$ of the reaction is $-198.3Jmol^{-1}K^{-1}$