Question

Consider the following reaction:
CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kp=0.0611 at 2000 K
A reaction mixture initially contains a CO partial pressure of 1346 torr and a H2O partial pressure of 1762 torr at 2000 K.

A.) Calculate the equilibrium partial pressure of CO2.

B.) Calculate the equilibrium partial pressure of H2

Answer 1

Solution:

Given-

Kp=0.0611 at 2000 K

P_CO = 1346 torr

P_H2O = 1762 torr

Kp = 0.0611 at 2000 K

Let’s write the equation-

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
we know the Kp expression

Kp = (P(products))/(P(reactants)) = ((P(CO2))*(P(H2)))/((P(CO))*(P(H2O)))

We use ICE chart

                   CO(g) + H2O(g) ⇌ CO2(g)   +   H2(g)

I           1346             1762              0                        0

C            -x                    -x              + x                     +x

E         ( 1346-x)       (1762-x)            x                           x

Substitute the above equilibrium value in Kp expression

Kp = ((x)*(x))/((1346-x)*(1762-x)) = 0.0611
0.0611 = x²/((1346-x)*(1762-x))
x² = 0.0611*((1346-x)*(1762-x)) = 0.0611* (2371652 -3108X + x²)

x² = 144907.9-189.8988X + 0.0611X²

x² - 0.0611X² + 189.8988X – 144907 = 0

0.9389 x² + 189.8988X – 144907 = 0

use the quadratic equation solver

a = 0.9389

b = 189.89

c = -144907

the lowest value from both x is cobsider

if there is one negative and one postive then consider positive one

x = -b^2+/- suqare root b^2-4ac divided by 2a

the plug the values of a, b and c

X = 304.53 torr

From the ICE equation

Partial pressure of CO2 & H2 = X

Answer

A) Partial pressure for CO2 at equilibrium P(CO2) = x = 304.53 Torr

B) Partial pressure for H2 at equilibrium P(H2) = x = 304.53 Torr