Question

${CH}_{3}OH\left(g\right)$ can be prepared using equation,
$CO\left(g\right)+2H_2\left(g\right)\rightleftharpoons {CH}_{2}OH\left(g\right),$ $K_c={4\times 10}^2$
At equilibrium, a 5 L flask contains equal moles of $CO\left(g\right)$ and ${CH}_{3}OH\left(g\right)$. Hence, number of moles of $H_2$ at equilibrium is:

• A. 0.25 mol
• B. 0.10 mol
• C. 0.50 mol
• D. 0.125 mol

Answer 1

The correct option is A 0.25 mol

Solution ;-

$CO\left(g\right)+2H_2\left(g\right)\rightleftharpoons C{H}_{2}OH\left(g\right)$

$kc=4\times {10}^2$

$CO\left(g\right)+2H_2\left(g\right)\rightleftharpoons C{H}_{3}OH\left(g\right)$

initial

1moles 1moles

At equilibrium

$\frac{1-x}{5}\frac{1-2x}{5}\frac{x}{5}$

At equilibrium, moles of CO =

moles of $C{H}_{3}OH$

$1-x=x$

$1=2x$

$x=0.5$

$K_c=\frac{conc.ofproduct}{conc.ofreartant}$

$4\times {10}^2=\frac{x/5}{\left[\frac{1-2x}{5}{]}^2\right(\frac{1-x}{5})}$

$4\times {10}^2=\frac{0.5/5}{\left(\frac{1-2x}{5}{)}^2\right(\frac{0.5}{5})}$

$\frac{400}{25}(1-2x{)}^2=1$

$16(1-2x{)}^2=1$

$(1-2x{)}^2=\frac{1}{16}$

$(1-2x)=\frac{1}{4}$

$\downarrow$

moles of $H_2=\frac{1}{4}$

moles of $H_2=0.25$