<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.
<u>Explanation:</u>
For the given chemical reaction:
The expression of 
for above reaction follows:
![K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
........(1)
We are given:
![[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M](https://tex.z-dn.net/?f=%5BCO%5D_%7Beq%7D%3D%5BH_2O%5D_%7Beq%7D%3D%5BH_2%5D_%7Beq%7D%3D0.10M)
![[CO_2]_{eq}=0.40M](https://tex.z-dn.net/?f=%5BCO_2%5D_%7Beq%7D%3D0.40M)
Putting values in above equation, we get:
To calculate the molarity of solution, we use the equation:
Moles of hydrogen gas = 0.30 mol
Volume of solution = 2.0 L
Putting values in above equation, we get:
When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.
Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M
Now, the equilibrium is shifting to the reactant side. The equation follows:
Initial: 0.1 0.1 0.4 0.1
At eqllm: 0.1+x 0.1+x 0.4-x 0.25-x
Putting values in expression 1, we get:
Neglecting the negative value of 'x'
Calculating the concentrations of the species:
Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M
Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M
Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M
Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M
Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.
