Question

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous CH4 and H2O in a 0.32-L flask at 1200 K. At equilibrium the flask contains 0.26 mol of CO, 0.091 mol of H2, and 0.041 mol of CH4. What is the [H2O] at equilibrium

Answer 1

Answer: The $[H_{2}O]$ at equilibrium is 0.561 M.

Explanation:

The given data is as follows.

    Volume of flask = 0.32 L,

 No. of moles of $CH_{4}$ = 0.041 mol,

 No, of moles of CO = 0.26 mol,  No. of moles of $H_{2}$ = 0.091 mol

 Equilibrium constant, K = 0.26

Balanced chemical equation for this reaction is as follows.

     $CH_{4}(g) + H_{2}O(g) \rightleftharpoons CO(g) + 3H_{2}(g)$

Hence,

              K = $\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}$ ...... (1)

First, we will calculate the molarity or concentration of given species as follows.

$CH_{4} = \frac{0.041}{0.32}$ = 0.128 M,

$CO = \frac{0.26}{0.32}$ = 0.8125 M,

$H_{2} = \frac{0.091}{0.32}$ = 0.284 M

Therefore, using expression (1) we will calculate the $[H_{2}O]$ as follows.

     K = $\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}$

or,    $[H_{2}O] = \frac{[CO][H_{2}]^{3}}{[CH_{4}] \times K}$

             = $\frac{0.8125 \times (0.284)^{3}}{0.128 \times 0.26}$

             = $\frac{0.0187}{0.0333}$

             = 0.561 M

Thus, we can conclude that the $[H_{2}O]$ at equilibrium is 0.561 M.