Question

½O2(g) + H2(g) ⇌ H2O(g)

Calculate the ΔH, ΔS and ΔSuniverse for this reaction.

Determine the spontaneity of the reaction.

Answer 1

-241.826 kJ·mol⁻¹; -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

Step-by-step explanation:

½O₂(g) + H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹: 0 0 -241.826

S°/J·K⁻¹mol⁻¹: 205.0 130.6 188.7

1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

`\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}} +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}} -\frac{\Delta H_{\text{sys}}}{T}\\\\& = & -146.9 - (-241826)/(298)\\\\& = & -146.9 + 811.5\\& = & \mathbf{664.6 \,\, J\cdot K^(-1)mol^(-1)}\\\end{array}`

4. Spontaneity

`\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -241.826 - 298 * (-0.1469)\\& = & -241.826 + 43.776\\& = & \textbf{-198.050 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}`

ΔG is negative, so the reaction is spontaneous.