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Standard XII

Chemistry

Pressure and Temperature

A container c...

Question

A container contains $0.2 m o l$ of $C O_{2}$ and $0.1 m o l$ $H_{2}$ and establish following equilibrium.
$C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$ . At equlibrium $H_{2} O$ is found to be 25% by mol. The equilibrium constant $K p$ for the reaction is

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Solution

Given,
$\begin{matrix} C O_{2} (g) + & H_{2} (g) ⇌ & C O (g) + & H_{2} O (g) 0.2 & 0.1 & 0 & 0 0.2 - x & 0.1 - x & x & x \end{matrix}$
According to the question, $\frac{x}{0.3} \times 100 = 25 \Rightarrow x = 0.075$ or $3 / 40$
So, $K_{p} = \frac{x^{2}}{(0.2 - x) (0.1 - x)} = \frac{(3 / 40)^{2}}{1 / 8 \times 1 / 40} = 1.8$

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A container contains 0.2 mol of CO2 and 0.1 mol H2 and establish following equilibrium. CO2(g)+H2(g)⇌CO(g)+H2O(g). At equlibrium H2O is found to be 25% by mol. The equilibrium constant Kp for the reaction is

Given, CO2(g)+H2(g)⇌CO(g)+H2O(g)0.20.1000.2−x0.1−xxx According to the question,x0.3×100=25⇒x=0.075 or 3/40 So,Kp=x2(0.2−x)(0.1−x)=(3/40)21/8×1/40=1.8

A container contains $0.2 m o l$ of $C O_{2}$ and $0.1 m o l$ $H_{2}$ and establish following equilibrium.
$C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$ . At equlibrium $H_{2} O$ is found to be 25% by mol. The equilibrium constant $K p$ for the reaction is