<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:
![CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)](https://tex.z-dn.net/?f=CO%28g%29%2BH_2O%28g%29%5Crightleftharpoons%20CO_2%28g%29%2BH_2%28g%29)
The expression of
for above reaction follows:
........(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:
![K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B0.40%5Ctimes%200.10%7D%7B010%5Ctimes%200.10%7D%5C%5C%5C%5CK_c%3D4)
To calculate the molarity of solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Moles of hydrogen gas = 0.30 mol
Volume of solution = 2.0 L
Putting values in above equation, we get:
![\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20%7DH_2%3D%5Cfrac%7B0.30mol%7D%7B2L%7D%3D0.15M)
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:
![CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)](https://tex.z-dn.net/?f=CO%28g%29%2BH_2O%28g%29%5Crightleftharpoons%20CO_2%28g%29%2BH_2%28g%29)
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:
![4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522](https://tex.z-dn.net/?f=4%3D%5Cfrac%7B%280.25-x%29%280.4-x%29%7D%7B%280.1%2Bx%29%280.1%2Bx%29%7D%5C%5C%5C%5C3x%5E2%2B1.45x-0.06%3D0%5C%5C%5C%5Cx%3D0.038%2C-0.522)
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.