Respuesta :
Answer: [tex]K_{eq}[/tex] at the temperature of the experiment is 0.56.
Explanation:
Moles of [tex]CO[/tex] = 0.35 mole
Moles of [tex]H_2O[/tex] = 0.40 mole
Volume of solution = 1.00 L
Initial concentration of [tex]CO[/tex] = [tex]\frac{0.35mol}{1.00L}=0.35M[/tex]
Initial concentration of [tex]H_2O[/tex] = [tex]\frac{0.40mol}{1.00L}=0.40M[/tex]
Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{0.19mol}{1.00L}=0.19M[/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 0.40 M 0 M 0M
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
Given: (0.35-x) = 0.19
x= 0.16 M
The expression for equilibrium constant for this reaction will be,
[tex]K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}[/tex]
Now put all the given values in this expression, we get :
[tex]K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}[/tex]
[tex]K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56[/tex]
Thus [tex]K_{eq}[/tex] at the temperature of the experiment is 0.56.
