Answer:
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
Answer:
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)
0.14.......0.14.............0.............0
0.14-x.....0.14-x.........x.............x
K = 102 = x^2 / (0.14-x)^2
x = 0.108
