Calculate the equilibrium constant for the following equilibrium … - Vidyadip

Question

Calculate the equilibrium constant for the following equilibrium system at 1120K. $\text{C}(\text{s})+\text{CO}_2(\text{g})+2\text{Cl}_2\stackrel{\text{K}_\text{p}}{\rightleftharpoons}2\text{COCl}_2(\text{g})$ Given the following equations and equilibrium constants: $\text{CO}(\text{s})+\text{Cl}_2(\text{g})\rightleftharpoons\text{COCl}_2(\text{g})$ $\text{Kp}_1 = 6.0 \times 10-3 \dots(1)$ $\text{CO}(\text{s})+\text{Cl}_2(\text{g})\rightleftharpoons\text{CO}_2(\text{g})$$\text{Kp}_2 = 1.3 \times 1014 \dots(2)$

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Answer

$\text{Multiply Eq. (1) by 2}$
$2\text{CO}(\text{g})+2\text{Cl}_2(\text{g})\rightleftharpoons2\text{COCl}_2(\text{g})$
$\text{K}'_{\text{p}1}=\left(6.0 \times 10^{-3}\right)^2=36 \times 10^{-6}$
$\text{K}'_{\text{p}1}=\frac{[\text{COCl}_2]^2}{[\text{CO}]^2[\text{Cl}_2]^2}=36\times10^{-6}\dots(3)$
$\text{K}'_{\text{p}2}=\frac{[\text{CO}]^2}{[\text{CO}_2]}=36\times10^{14}\dots(4)$
$\text{Multiply K'p}_1 \text{ and } \text{K'p}_2$
$\text{K}'_{\text{p}}=\frac{[\text{COCl}_2]^2}{[\text{CO}_2][\text{Cl}_2]^2}$
$=46.8\times10^8=4.68\times10^9$

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