A reaction mixture initially contains 0.140 MCO and 0.140 MH2O. What will be the equilibrium concentration of CO?

0.013 M

Explanation:

From the question, we can make the following deductions; we are given mixture that contains two compounds, that is A and B, 0.140 M CO and 0.140 M H2O respectively. Then, we are asked to find the equilibrium concentration of Carbonmonoxide,CO.

So, the equation for the reaction is given below;

CO + H2O <> CO2 + H2.

Initially: we have 0.14M of CO, 0.14M of H2O and zero (0) concentration of CO2 and H2.

At time,t = CO =0.14 - x , H2O = 0.14 - x, CO2 and H2 = x.

The above reaction consist of the forward reaction and the backward reaction.

Therefore, the equilibrium Concentration of CO;

(Since we are giving that Kc = 102). Then, Kc=  [CO2][H2] ÷ [CO][H2O]. Where Kc is the equilibrium constant.

Therefore, 102 = [x^2] / [0.14 - x]^2.

==> 10.1= x/0.14 - x.

> 0.141 - 10.1 x = x.

x + 10.1 x = 0.141.

===> 11.1 x = 0.141.

===> x = 0.141 ÷ 11.1.

===> x = 0.127 M .

Then, at time,t CO = 0.14 - x.

= 0.14 - 0.127 = 0.013 M

Consider the following reaction:

CO(g) + H2O(g) <=> CO2(g) + H2(g)  

Kc = 102 @ 500K

A reaction mixture initially contains: 0.140M CO and 0.140M H2O

What will be the equilibrium concentration of CO and H2?

Explanation:

CO(g) + H2O(g) <=> CO2(g) + H2(g)  

0.140.1400

0.14-x0.14-xxx

K = 102 = x^2 / (0.14-x)^2

x = 0.108

It decreases by about 10%

Explanation: