Answer:
-121.301 kJ·mol⁻¹; 57.8 J·K⁻¹mol⁻¹; 464.9 J·K⁻¹mol⁻¹; spontaneous
Explanation:
2Cl(g) ⇌ Cl₂(g)
ΔHf°/kJ·mol⁻¹: 121.301 0
S°/J·K⁻¹mol⁻¹: 165.190 223.0
1. ΔᵣH
ΔᵣH = products -reactants = 0 - 121.301 = -121.301 kJ·mol⁻¹
2. ΔᵣS
ΔᵣS = products - reactants = 223.0 - 165.190 = 57.8 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}\\\\& = & 57.8 - \dfrac{-121301}{298}\\\\& = & 57.8 + 407.1\\& = & \mathbf{464.9 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}[/tex]
4. Spontaneity
[tex]\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -121.301 - 298 \times 0.0578\\& = & -121.301 - 17.23\\& = & \textbf{-138.5 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}[/tex]
ΔG is negative, so the reaction is spontaneous.
