${CH}_{3} OH,$ methanol can be prepared from CO and $H_{2}$
$CO (g) + 2 H_{2} (g) ⇌ {CH}_{3} OH (g)$ The value of $K_{p}$ at 500 K is $6.23 \times 10^{- 3}$
What total pressure (in atm) at equilibrium is required to convert $25 %$ of CO to ${CH}_{3} OH$ at 500 K If CO and $H_{2}$ comes from
${CH}_{4} (g) + H_{2} O (g) ⟶ CO (g) + 3 H_{2} (g)$
Give your answer excluding decimal places

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 10.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} CO + 2 H_{2} ⇌ {CH}_{3} OH \\ \begin{array}{ccc} a & 3 a & - \\ \frac{3}{4} a & \frac{5}{2} a & \frac{a}{4} \end{array} \\ {CH}_{4} + H_{2} O ⟶ CO + 3 H_{2} \end{array}$

$a 3 a$

$\begin{array}{l} K_{P} = \frac{\frac{1}{4} \times \frac{7}{2} \times \frac{7}{2}}{\frac{3}{4} \times {(\frac{5}{2})}^{2} \times P^{2}} = 6.23 \times 10^{- 3} \\ P = 10.24 atm \end{array}$