Answer:
6.2846
Explanation:
Given that:-
Concentrations at equilibrium :-
[tex][CH_4]=0.126\ M[/tex]
[tex][H_2O]= 0.242\ M[/tex]
[tex][CO]= 0.126\ M[/tex]
[tex][H_2]= 1.15\ M[/tex]
The equilibrium reaction is:-
[tex]CH_4+H_2O\rightleftharpoons 3H_2+CO[/tex]
The expression for equilibrium constant is:
[tex]K_{c}=\frac {\left [ H_2 \right ]^3\left [ CO \right ]}{\left [ CH_4 \right ]\left [ H_2O \right ]}[/tex]
Applying the values as:-
[tex]K_c=\frac{1.15^3\times 0.126}{0.126\times 0.242}=6.2846[/tex]
The equilibrium constant for the reaction is:- 6.2846