Calculate ΔrG (in kJ mol–1) for the given reaction A2(g)+B2(g)⇌2AB(g) at 300 K and at partial pressure of 10–2 bar, 10–1 bar and 10–4 bar for A2 , B2 and AB respectively. Report answer as ΔrG×3.0100 in nearest integer.Given : ΔfH°(AB,g)=180 kJ mol−1;ΔfH°(A2,g)=60 kJ mol−1 ΔfH°(B2,g)=29.5 kJ mol−1;ΔfS°(AB,g)=210 J /K. mol ΔfS°(A2,g)=190 J / K. mol;ΔfS°(B2,g)=205 J /K. molGiven : 2.303R×300 =5750Jmol–1

A2(g)+B2(g)⇌2AB(g) ΔrS°=2×ΔfS°(AB,g)−ΔfS°(A2,g)−ΔfS°(B2,g) ΔrS°=2×210−190−205=25 J K−1mol−1 ΔrH°=2×ΔfH°(AB,g)−ΔfH°(A2,g)−ΔfH°(B2,g) ΔrH°=2×180−60−29.5=270.5 kJ mol−1 Qp=(PAB)2(PA2)(PB2)=(10−4)210−2×10−1=10−5 ΔrG=ΔrG°+RTlnKp=ΔrH°−TΔrS°+RTlnQp ΔrG=270.5−300×251000+2.303R×3001000log10−5 ΔrG=270.5−7.5−5750×51000=234.25 kJ mol−1≃234 kJ mol−1 ΔrG100×3=7 kJ mol−1

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Calculate $Δ_{r} G$ (in kJ mol–1) for the given reaction $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ at 300 K and at partial pressure of $10^{- 2}$ bar, $10^{- 1}$ bar and $10^{- 4}$ bar for $A_{2}$ , $B_{2}$ and AB respectively. Report answer as $\frac{Δ_{r} G \times 3.0}{100}$ in nearest integer.
Given : $Δ_{f} H ° (A B, g) = 180 k J m o l^{- 1}; Δ_{f} H ° (A_{2}, g) = 60 k J m o l^{- 1} Δ_{f} H ° (B_{2}, g) = 29.5 k J m o l^{- 1}; Δ_{f} S ° (A B, g) = 210 J / K . m o l Δ_{f} S ° (A_{2}, g) = 190 J / K . m o l; Δ_{f} S ° (B_{2}, g) = 205 J / K . m o l$
Given : $2.303 R \times 300 = 5750 J m o l^{- 1}$

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Detailed Solution

$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g) Δ_{r} S ° = 2 \times Δ_{f} S ° (A B, g) - Δ_{f} S ° (A_{2}, g) - Δ_{f} S ° (B_{2}, g) Δ_{r} S ° = 2 \times 210 - 190 - 205 = 25 J K^{- 1} m o l^{- 1} Δ_{r} H ° = 2 \times Δ_{f} H ° (A B, g) - Δ_{f} H ° (A_{2}, g) - Δ_{f} H ° (B_{2}, g) Δ_{r} H ° = 2 \times 180 - 60 - 29.5 = 270.5 k J m o l^{- 1} Q_{p} = \frac{{(P_{A B})}^{2}}{(P_{A_{2}}) (P_{B_{2}})} = \frac{{(10^{- 4})}^{2}}{10^{- 2} \times 10^{- 1}} = 10^{- 5} Δ_{r} G = Δ_{r} G ° + R T \ln K_{p} = Δ_{r} H ° - T Δ_{r} S ° + R T \ln Q_{p} Δ_{r} G = 270.5 - \frac{300 \times 25}{1000} + \frac{2.303 R \times 300}{1000} \log 10^{- 5} Δ_{r} G = 270.5 - 7.5 - \frac{5750 \times 5}{1000} = 234.25 k J m o l^{- 1} ≃ 234 k J m o l^{- 1} \frac{Δ_{r} G}{100} \times 3 = 7 k J m o l^{- 1}$