Calculate the equilibrium constant for the reaction given below at 400k-If -H0 - 77-2kJmol - 1 and -S0 - 122JK - 1mol - 1 PCl5g- PCl3g - Cl2g

We know the relation: [ Δ H TΔ S = #160; RTln K; 77.2 ×10^3 400 × 122 = #160; 8.314 × 400ln K; 2800 = ...

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Standard XII

Chemistry

Kohlrausch Law

Calculate the...

Question

Calculate the equilibrium constant for the reaction given below at 400k.
If $Δ H^{0} = 77.2 k J m o l^{- 1}$ and $Δ S^{0} = 122 J K^{- 1} m o l^{- 1}$
$P C l_{5 (g)} \to P C l_{3 (g)} + C l_{2 (g)}$

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Solution

We know the relation:
$\begin{matrix} Δ H - T Δ S = - R T ln K 77.2 \times 10^{3} - 400 \times 122 = - 8.314 \times 400 ln K 2800 = - 3325.6 ln K - 8.539 = ln K K = e^{- 8.539} K = 1.95 \times 10^{- 4} \end{matrix}$

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Calculate the equilibrium constant for the reaction given below at 400k.If ΔH0=77.2kJmol−1 and ΔS0=122JK−1mol−1PCl5(g)→PCl3(g)+Cl2(g)

We know the relation:ΔH−TΔS=−RTlnK77.2×103−400×122=−8.314×400lnK2800=−3325.6lnK−8.539=lnKK=e−8.539K=1.95×10−4

Calculate the equilibrium constant for the reaction given below at 400k.
If $Δ H^{0} = 77.2 k J m o l^{- 1}$ and $Δ S^{0} = 122 J K^{- 1} m o l^{- 1}$
$P C l_{5 (g)} \to P C l_{3 (g)} + C l_{2 (g)}$

We know the relation:
$\begin{matrix} Δ H - T Δ S = - R T ln K 77.2 \times 10^{3} - 400 \times 122 = - 8.314 \times 400 ln K 2800 = - 3325.6 ln K - 8.539 = ln K K = e^{- 8.539} K = 1.95 \times 10^{- 4} \end{matrix}$