Question

Methane gas and steam in equimolar amounts are taken in a flask and sealed where the following equilibrium was established:
$C{H}_4\left(g\right)+H_{2}O\left(v\right)\rightleftharpoons CO\left(g\right)+3H_2\left(g\right)$
$2H_2\left(g\right)+CO\left(g\right)\rightleftharpoons C{H}_{3}OH\left(g\right);K_p=\frac{2}{7}at{m}^{-2}$
At equilibrium if partial pressure of CO(g) as found to be 0.5 atm and totaI pressure inside the flask was 5 atm. then $K_p$ of the first reaction is: (approximately)

• A. $3.5at{m}^2$
• B. $7.5at{m}^2$
• C. $10.5at{m}^2$
• D. $12.5at{m}^2$

Answer 1

Answer: (B)

$7.5at{m}^2$

$\underset{p^o-p_1}{C{H}_2\left(g\right)}+\underset{p^o-p_1}{H_{2}O}\left(v\right)\rightleftharpoons \underset{p_1-p_2}{CO\left(g\right)}+\underset{3p_1-2p_2}{3H_2\left(g\right)};K_{p_1}$

$\underset{3p_1-2p_2}{2H_2\left(g\right)}+\underset{(p_1+p_2)}{CO\left(g\right)}\rightleftharpoons \underset{p_2}{C{H}_{2}OH\left(g\right)}{K}_{p_2}=\frac{2}{7}at{m}^2$

$p_1-p_2=0.5$ ........(1)

Also, $5=2p^o+2p_1-2p_2$ ........(2)

$\therefore p^o=2atm$

$K_{p_2}=\frac{2}{7}=\frac{p_2}{{\left\{\right(3p_1-3p_2)+p_2\}}^2(p_1-p_2)}=\frac{p_2}{(1.5+p_2{)}^2(0.5)}$

$\therefore p_2=0.677atm$

$p_1=1.177atm$

$\therefore K_{p_1}=\frac{\left(2.177{)}^3\right(0.5)}{(0.823{)}^2}=7.6at{m}^2$