Respuesta :
Answer: Equilibrium constant is 0.70.
Explanation:
Initial moles of [tex]CO[/tex] = 0.35 mole
Volume of container = 1 L
Initial concentration of [tex]CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M[/tex]
Initial moles of [tex]H_2O[/tex] = 0.40 mole
Volume of container = 1 L
Initial concentration of [tex]H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M[/tex]
equilibrium concentration of [tex]CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M[/tex] [/tex]
The given balanced equilibrium reaction is,
[tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]
Initial conc. 0.35 M 0.40M 0 0
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}[/tex]
[tex]K_c=\frac{x\times x}{(0.40-x)(0.35-x)}[/tex]
we are given : (0.35-x)= 0.18
x = 0.17
Now put all the given values in this expression, we get :
[tex]K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}[/tex]
[tex]K_c=0.70[/tex]
Thus the value of the equilibrium constant is 0.70.
