Question

A mixture of one mole of ${CO}_2$ and one mole of $H_2$ attains equilibrium at a temperature of ${250}^{o}C$ and a total pressure of 0.1 atm for the change ${CO}_2\left(g\right)+H_2\left(g\right)\rightleftharpoons CO\left(g\right)+H_{2}O\left(g\right)$. Calculate $K_P$ if the analysis of final reaction mixture shows 0.16 volume percent of $CO$.

• A. $K_P=1.03\times {10}^{-5}$
• B. $K_P=1.25\times {10}^{-4}$
• C. $K_P=5\times {10}^{-5}$
• D. $K_P=1\times {10}^{-3}$

Answer 1

Answer: (A)

$K_P=1.03\times {10}^{-5}$

${CO}_2\left(g\right)+H_2\left(g\right)\rightleftharpoons CO\left(g\right)+H_2\left(g\right)$
1 1 0 0 Moles at $t=0$

$(1-x)$ $(1-x)$ $x$ $x$ Moles at equilibrium
Given that volume % of $CO=0.16$

$\because$ Moles of $CO$= $x$

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

$\therefore \frac{x}{2}=\frac{0.16}{100}\Rightarrow \therefore x=0.0032$

$K_C=K_P=\frac{x^2}{{(1-x)}^2}$

$(\because △n=0,\therefore Volumetermsarenotneeded)$

$K_P=\frac{{\left(0.0032\right)}^2}{{(1-0.0032)}^2}=1.03\times {10}^{-5}$