Respuesta :
Answer: Concentration of [tex]CO_2[/tex] at equilibrium= 1.386 M
Concentration of [tex]H_2[/tex] at equilibrium = 1.386 M
Concentration of [tex]CO[/tex] at equilibrium = 0.614 M
Concentration of [tex]H_2O[/tex] at equilibrium= 0.614 M
Explanation:
Moles of [tex]CO[/tex] = 1.00 mole
Moles of [tex]H_2O[/tex] = 1.00 mole
Moles of [tex]CO_2[/tex] = 1.00 mole
Moles of [tex]H_2O[/tex] = 1.00 mole
Volume of solution = 1.00 L
Initial concentration of [tex]CO[/tex] =[tex]\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M[/tex]
Initial concentration of [tex]H_2O[/tex] =[tex]\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M[/tex]
Initial concentration of [tex]CO_2[/tex] =[tex]\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M[/tex]
Initial concentration of [tex]H_2[/tex] =[tex]\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M[/tex]
The given balanced equilibrium reaction is,
[tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2[/tex]
Initial conc. 1.00M 1.00 M 1.00 M 1.00 M
At eqm. conc. (1.00-x) M (1.00-x) M (1.00+x) M (1.00+x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}[/tex]
Now put all the given values in this expression, we get :
[tex]5.10=\frac{(1.00+x)^2}{(1.00-x)^2}[/tex]
By solving the term 'x', we get :
x = 0.386
Concentration of [tex]CO_2[/tex] at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M
Concentration of [tex]H_2[/tex] = (1.00+x) M= (1.00+0.386)= 1.386 M
Concentration of [tex]CO[/tex] = (1.00-x) M = (1.00-0.386) M = 0.614 M
Concentration of [tex]H_2O[/tex] = (1.00-x) M = (1.00-0.386) M = 0.614 M
