Question

½O2(g) + H2(g) ⇌ H2O(g)

Answer 1

Answer:

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

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)

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

4. Spontaneity

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

ΔG is negative, so the reaction is spontaneous.