The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: co(g) + h2o(g) ⇌ co2(g) + h2(g) a study was made in which equilibrium was reached with [co] = [h2o] = [h2] = 0.10 m and [co2] = 0.40 m. after 0.30 mol of h2 is added to the 2.0−l container and equilibrium is reestablished, what are the new concentrations of all the components?

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.

Explanation:

For the given chemical reaction:

The expression of for above reaction follows:

        ........(1)

We are given:

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.

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