Question

Methane gas and steam in equimolar amounts are taken in a flask and sealed where the following equilibrium was established at equilibrium if partial pressure of CO as found to be 0.5atm and total pressure inside the flask was 5atm then Kp of the first reaction is

Answer 1

Answer : The equilibrium constant for the reaction is 0.75

Explanation :

The chemical equation for the reaction can be written as follows.

$CH_{4} (g) + H_{2}O (g) \leftrightarrow CO (g) + 3 H_{2} (g)$

We are taking equimolar amounts of methane and steam. So let's assume we have x moles of methane and steam initially.

Let us set up the ICE table for this reaction.

From the ICE table, we can say that

Equilibrium partial pressure of CO is y. But we have given that  equilibrium partial pressure of CO is 0.5 atm.

Therefore we have y = 0.5 atm

From ICE table we have,

total pressure = $P_{CH_{4}} + P_{H_{2}O} + P_{CO} + P_{H_{2}}$

$Pressure = (x-y)+(x-y)+(y)+(3y)$

But the total pressure is given as 5 atm.

Therefore , $5 = (x-y)+(x-y)+(y)+(3y)$

$5 = 2x + 2y$

$5 =2(x+y)$

$x+y = 2.5$

But y = 0.5, therefore we have

$2.5 = x + 0.5$

$x = 2.0 atm$

Partial pressures of the gases at equilibrium are as follows.

P[CH₄] = x-y = 2.0 - 0.5 = 1.5 atm

P[H₂O] = x-y = 2.0 - 0.5 = 1.5 atm

P[CO] = y = 0.5 atm

P[H₂} = 3y = 3(0.5) = 1.5 atm  

The equilibrium constant is calculated as,

$k_{eq} = \frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}$

Let us plug in the above equilibrium values.

$k_{eq} = \frac{(0.5)(1.5)^{3}}{(1.5)(1.5)}$

$k_{eq} = (0.5)(1.5)$

$k_{eq} = 0.75$

The equilibrium constant for the reaction is 0.75