Answer to Question #332265 in Chemistry for mcc

Question

What is the ∆S of N₂(g) + 3H₂(g) ---> 2NH₃(g)?

Given

N₂(g)

(J/mol x K) = 191.5

H₂(g)

(J/mol x K) = 130.6

NH₃(g)

(J/mol x K) = 192.3

Answer 1

At 298 K as a standard temperature:

S°(N₂, g) = 191.5 J mol⁻¹ K⁻¹

S°(H₂, g) = 130.6 J mol⁻¹ K⁻¹

S°(NH₃, g) = 192.3 J mol⁻¹ K⁻¹

Solution:

The entropy change (ΔS°) in a chemical reaction is given by the sum of the entropies of the products minus the sum of the entropies of the reactants.

ΔS° =∑nS°(products) − ∑mS°(reactants)

Balanced chemical equation:

N₂(g) + 3H₂(g) → 2NH₃(g)

From the balanced equation we can write the equation for ΔS° (the change in the standard molar entropy for the reaction):

ΔS° = 2×S°(NH₃, g) − [S°(N₂, g) + 3×S°(H₂, g)]

ΔS° = 2×192.3 − [191.5 + 3×130.6] = −198.7 (J mol⁻¹ K⁻¹)

ΔS° = −198.7 J mol⁻¹ K⁻¹

Answer: The entropy change is −198.7 J mol⁻¹ K⁻¹