Question

Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of H2O(g).C2H4(g) + H2O(g) ↔ C2H5OH(g) Kc = 9.0 × 103[C2H4]eq = 0.015 M [C2H5OH]eq = 1.69 M

Answer 1

Answer:

The equilibrium concentration of water vapor is 0.0125 M.

Explanation:

$C_2H_4(g) + H_2O(g)\rightleftharpoons C_2H_5OH(g)$

The concentration of ethene at an equilibrium = $[C_2H_4]=0.015M$

The concentration of water vapor at an equilibrium = $[H_2O]=?$

The concentration of ethanol at an equilibrium = $[C_2H_5OH]=1.69 M$

the equilibrium constant of the reaction,  $K_c=9.0\times 10^{3}$

The expression of an equilibrium constant will be written as:

$K_c=\frac{[C_2H_5OH]}{[C_2H_4][H_2O]}$

$9.0\times 10^3=\frac{1.69 M}{0.015 M\times [H_2O]}$

$[H_2O]=\frac{1.69 M}{0.015 M\times 9\timers 10^3}=0.0125 M$

The equilibrium concentration of water vapor is 0.0125 M.