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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?

Respuesta :

Answer: 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:

[tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]

The expression of [tex]K_c[/tex] for above reaction follows:

[tex]K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]         ........(1)

We are given:

[tex][CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M[/tex]

[tex][CO_2]_{eq}=0.40M[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4[/tex]

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

[tex]\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M[/tex]

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:

                      [tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]

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:

[tex]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[/tex]

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.

batolisis

The new concentrations of all the components are as listed below

  • [ CO ] = 0.15 M
  • [ H₂O ] = 0.15 M
  • [ CO₂ ] = 0.35 M
  • [ H₂ ] = 0.25 M

Determine the new concentratios of the components

                   CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

Initial conc   0.1         0.1             0.4         0.1 + 0.2

equilibrium   0.1-x     0.1-x          0.4+x      0.3 + x

Calculate the value of X

K = [tex]\frac{[H_{2}][CO_{2} ] }{[H_{2}O ][CO]}[/tex]

4 = ( 0.3 + x ) ( 0.4 + x ) / ( 0.1 - x ) ( 0.1 - x )

4x² - 0.8 x + 0.04 = x² + 0.7x + 0.12

solve for x

x = -0.0486 or 0.5486

we will pick - 0.0486 as a reasonable value

Therefore the new equilibruim values of the components are

  • [ CO ] = 0.15 M
[ H₂O ] = 0.15 M[ CO₂ ] = 0.35 M [ H₂ ] = 0.25 M

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