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Answer :
Answer:
Equilibrium concentration of [tex]H_{2}O[/tex] is 12.5 M
Explanation:
Given reaction: [tex]C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH[/tex]
Here, [tex]K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}[/tex]
where [tex]K_{c}[/tex] represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, [tex][C_{2}H_{4}]=0.015M[/tex] , [tex][C_{2}H_{5}OH]=1.69M[/tex] and [tex]K_{c}=9.0[/tex]
So, [tex][H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M[/tex]
Hence equilibrium concentration of [tex]H_{2}O[/tex] is 12.5 M
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