Question

Consider the following reaction, equilibrium concentrations, and equilibrium constant at

Answer 1

Answer:

Equilibrium concentration of $H_{2}O$ is 12.5 M

Explanation:

Given reaction: $C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH$

Here, $K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}$

where $K_{c}$ represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, $[C_{2}H_{4}]=0.015M$ , $[C_{2}H_{5}OH]=1.69M$ and $K_{c}=9.0$

So, $[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M$

Hence equilibrium concentration of $H_{2}O$ is 12.5 M