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)
[tex]\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}[/tex]
4. Spontaneity
[tex]\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}[/tex]
ΔG is negative, so the reaction is spontaneous.
