Question

Calculate the entropy change when 3.66 g of H2 reacts with O2 according to the reaction 2 H2(g) + O2(g) → 2 H2O(ℓ) at 298 K and 1 atm pressure. The standard molar enthalpy of formation of H2O(ℓ) at 298 K is −285.8 kJ/mol. The corresponding free energy of formation is −237.2 kJ/mol. Answer in units of J/K.

Answer 1

The entropy change will be ΔS = - 150.0 J/K.

Step-by-step explanation:

2 H₂(g) + O₂(g) → 2 H₂O(ℓ)

ΔH = -258.8 kJ/mol

ΔG = - 237.2 kJ/mol

1 mol H₂ ______ 2 g

x ______ 3.66 g

x = 1.83 mol

1 mol H₂ ________ 1 mol H₂O

1.83 mol H₂ ______ 1.83 mol H₂O

ΔH = 1.83 mol x -258.8 kJ/mol

ΔH = - 478.78 kJ

ΔG = 1.83 mol x - 237.2 kJ/mol

ΔG = 434.076 kJ

ΔG = ΔH - TΔS

ΔS = (ΔH - ΔG) / T

ΔS = [ - 478.78 kJ - (- 434.076 kJ)]/298 K

ΔS = - 150.0 J/K