Question

The mixture of one mole of ${CO}_2$ and one mole of $H_2$ attains equilibrium at a temperature of ${250}^{0}C$ and total pressure of 0.1 atm for the change $C{O}_{2\left(e\right)}+H_{2\left(e\right)}\rightleftharpoons C{O}_{2\left(e\right)}+H_2{0}_e$, calculate $K_p$ is the analysis of final reaction mixture shows $0.16$ volume fraction of CO ?

• A. $0.46$
• B. $0.63$
• C. $0.22$
• D. $0.82$

Answer 1

The correct option is C $0.22$

$C{O}_{\left(g\right)}+{H_2}_{\left(g\right)}\rightleftharpoons {C{O}_2}_{\left(g\right)}+{H_{2}O}_{\left(g\right)}$

Initially $1mole$ $1mole$ $0$ $0$

At eqm $1-x$ $1-x$ $x$ $x$

Total moles at equilibrium= $1-x+1-x+x+x=2$

$P_{C{O}_2}=P_{H_2}=\frac{1-x}{2}\times 0.1$ and $P_{CO}=P_{H_{2}O}=\frac{x}{2}\times 0.1$

Since mole fraction of $CO=0.16$

$\therefore \frac{x}{2}=0.16\Rightarrow x=0.32$

$K_P=\frac{P_{CO}\times P_{H_{2}O}}{P_{C{O}_2}\times P_{H_2}}=\frac{{(\frac{x}{2}\times 0.1)}^2}{{(\frac{1-x}{2}\times 0.1)}^2}$

$={\left(\frac{x}{1-x}\right)}^2={\left(\frac{0.32}{1-0.32}\right)}^2$

$=0.22$