Answer to Question #345725 in General Chemistry for aaa

Question

What is the entropy change of the reaction below at 298 K and 1 atm pressure? N2(g) + 3H2(g) → 2NH3(g) S 0 298 (J/mol•K) 191.5 130.6 192.3

(a) –198.7 J/K

(c) –129.7 J/K

(b) 76.32 J/K

(d) 303.2 J/K

Answer 1

Solution:

The entropy change 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) − 3×S°(H₂, g) − S°(N₂, g)

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

ΔS° = −198.7 J mol⁻¹ K⁻¹

Thus, the correct option is A

Answer: (a) –198.7 J/K