Answer to Question #332264 in Chemistry for mcc

Question

What is the total entropy change for the decomposition of hydrogen peroxide.

2H₂O₂(l) ---> 2H₂O(l) + O₂(g)

Given

H₂O₂(l)

(J/mol x K) = 109.6

(kJ/mol) = -187.8

H₂O(l)

(J/mol x K) = 69.9

(kJ/mol) = -285.8

O₂(g)

(J/mol x K) = 205.0

(kJ/mol) = 0

Answer 1

At 298K as a standard temperature:

S°(O₂, g) = 205.0 J mol⁻¹ K⁻¹

S°(H₂O, l) = 69.9 J mol⁻¹ K⁻¹

S°(H₂O₂, l) = 109.6 J mol⁻¹ K⁻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:

2H₂O₂(l) → 2H₂O(l) + O₂(g)

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

ΔS° = S°(O₂, g) + 2×S°(H₂O, l) − 2×S°(H₂O₂, l)

ΔS° = 205.0 + 2×69.9 − 2×109.6 = 125.6 (J mol⁻¹ K⁻¹)

ΔS° = 125.6 J mol⁻¹ K⁻¹

Answer: The total entropy change for the decomposition of hydrogen peroxide is 125.6 J mol⁻¹ K⁻¹