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

The equilibrium concentration of water vapor is 0.0125 M.

Step-by-step 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* 10^(3)`

The expression of an equilibrium constant will be written as:

`K_c=([C_2H_5OH])/([C_2H_4][H_2O])`

`9.0* 10^3=(1.69 M)/(0.015 M* [H_2O])`

`[H_2O]=(1.69 M)/(0.015 M* 9\timers 10^3)=0.0125 M`

The equilibrium concentration of water vapor is 0.0125 M.