Answer:
0.05110 mol/L is the concentration of water at equilibrium.
Explanation:
[tex] CH_4 (g)+H_2O(g)\rightleftharpoons CO (g)+3 H_2 (g)[/tex]
An equilibrium constant of the given reaction =[tex]K_c=0.28[/tex]
Volume of the container = V = 0.64 L
Concentration =[tex]\frac{Moles}{Volume}[/tex]
Concentration of CO=[tex][CO]=\frac{0.36 mol}{0.64 L}[/tex]
Concentration of[tex]H_2=[H_2]=\frac{0.081 mol}{0.64 L}[/tex]
Concentration of[tex]CH_4=[CH_4]=\frac{0.051 mol}{0.64 L}[/tex]
Concentration of[tex]H_2O=[H_2O]=?[/tex]
An expression of an equilibrium constant is given as:
[tex]K_c=\frac{[CO][H_2]^3}{[CH_4][H_2O]}[/tex]
[tex]0.28=\frac{\frac{0.36 mol}{0.64 L}\times (\frac{0.081 mol}{0.64 L})^3}{\frac{0.051 mol}{0.64 L}\times [H_2O]}[/tex]
[tex][H_2O]=0.05110 mol/L[/tex]
0.05110 mol/L is the concentration of water at equilibrium.
