Question

What is the change in entropy for the given reaction?

2 H 2​(g) + O 2​(g) → 2 H2​O(g)

A) -89 J/K

B) -147 J/K

C) 321 J/K

D) 378 J/K

Answer 1

The equation for the change of entropy of the given reaction:

2 H₂​(g) + O₂(g) → 2 H₂O(g)

ΔS^o_r = Σ S^o[products] - Σ S^o[reactants]

ΔS^o_r = 2 x S^o(H₂O) - 2 x S^o(H₂) – ΔS^o(O₂)

The standard entropy of water gas is 188.8 J-K mol^-1 .

The standard entropy of oxygen gas is 205.2 J-K mol^-1.

The standard entropy of hydrogen is 130.7 J-K mol^-1.

Substitute the values, we get:

ΔS^or = 2 X (188.8 J-K mol^-1) - 2 X (130.7 J-K mol^-1) – (205.2 kJ mol^-1)

ΔS^or = -89 J-K mol ^–1

Option A is correct.