First slide

Chemical Equilibrium

Question

0.4 mole of CO taken in a 2 L flask is maintained along with a catalyst at 812 K so that the reaction : $CO (g) + 2 H_{2} (g) ⇌ {CH}_{3} OH (g)$ can takes place , $H_{2}$ is introduced until total pressure of system is 20 atm & 0.2 moles of $C H_{3} O H$ is formed at equilibrium then the correct statements among the following:

Moderate

A

$K_{c}$ of reaction is 100 $M^{- 2}$
B

$K_{p} of reaction is 0.022 {atm}^{- 2}$
C

In absence of catalyst final pressure is 33.3 atm
D

Total number of moles in absence of catalyst is 0.5

Solution

T = 812 K ,P = 20 atm , V = 2 Lit.
CO(g) +    $2 H_{2} (g) \to {CH}_{3} OH (g)$
0.4 – x                  a- 2x           x = 0.2 mol
PV = nRT
          a = 0.6
          Kc = 100
   $Kp = \underset{}{\underset{}{K_{c}}} (RT)^{Δ_{n}}$ , In absence of catalyst CH₃OH formation is negligible. Then total no.of mole = $n C O + n H_{2}$ = 1

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App