Respuesta :
Answer: The [tex][H_{2}O][/tex] at equilibrium is 0.561 M.
Explanation:
The given data is as follows.
Volume of flask = 0.32 L,
No. of moles of [tex]CH_{4}[/tex] = 0.041 mol,
No, of moles of CO = 0.26 mol, No. of moles of [tex]H_{2}[/tex] = 0.091 mol
Equilibrium constant, K = 0.26
Balanced chemical equation for this reaction is as follows.
[tex]CH_{4}(g) + H_{2}O(g) \rightleftharpoons CO(g) + 3H_{2}(g)[/tex]
Hence,
K = [tex]\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}[/tex] ...... (1)
First, we will calculate the molarity or concentration of given species as follows.
[tex]CH_{4} = \frac{0.041}{0.32}[/tex] = 0.128 M,
[tex]CO = \frac{0.26}{0.32}[/tex] = 0.8125 M,
[tex]H_{2} = \frac{0.091}{0.32}[/tex] = 0.284 M
Therefore, using expression (1) we will calculate the [tex][H_{2}O][/tex] as follows.
K = [tex]\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}[/tex]
or, [tex][H_{2}O] = \frac{[CO][H_{2}]^{3}}{[CH_{4}] \times K}[/tex]
= [tex]\frac{0.8125 \times (0.284)^{3}}{0.128 \times 0.26}[/tex]
= [tex]\frac{0.0187}{0.0333}[/tex]
= 0.561 M
Thus, we can conclude that the [tex][H_{2}O][/tex] at equilibrium is 0.561 M.
