Respuesta :
The balanced chemical reaction is written as:
H2 + CO2 <---> H2O + CO
To determine the moles of H2 at equilibrium, we use the ICE table as follows:
Initial concentration = 0.229/11.0 = 0.021 M
H2O CO H2 CO2
I 0.021 0.021 0 0
C -x -x x x
--------------------------------------------------------
E 0.021- x 0.021-x x x
Kc = 0.534 = (H2)(CO2) / (H2O)(CO)
0.534 = (x)(x) / (0.021-x) (0.021-x)
x = 8.87x10^-3 M
mole H2 = 8.87x10^-3 M (11.0 L) = 0.098 mol H2
H2 + CO2 <---> H2O + CO
To determine the moles of H2 at equilibrium, we use the ICE table as follows:
Initial concentration = 0.229/11.0 = 0.021 M
H2O CO H2 CO2
I 0.021 0.021 0 0
C -x -x x x
--------------------------------------------------------
E 0.021- x 0.021-x x x
Kc = 0.534 = (H2)(CO2) / (H2O)(CO)
0.534 = (x)(x) / (0.021-x) (0.021-x)
x = 8.87x10^-3 M
mole H2 = 8.87x10^-3 M (11.0 L) = 0.098 mol H2
Answer : The concentration of [tex]H_2[/tex] at equilibrium is 0.012 M
Solution :
First we have to calculate the concentration of [tex]CO[/tex] and [tex]H_2O[/tex].
Concentration of [tex]CO[/tex] = [tex]\frac{Moles}{Volume}=\frac{0.229}{11}=0.0208M[/tex]
Concentration of [tex]H_2O[/tex] = [tex]\frac{Moles}{Volume}=\frac{0.229}{11}=0.0208M[/tex]
The given equilibrium reaction is,
[tex]H_2O(g)+CO(g)\rightleftharpoons H_2(g)+CO_2(g)[/tex]
Initially conc. 0.0208 0.0208 0 0
At equilibrium. (0.0208-x) (0.0208-x) x x
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[H_2][CO_2]}{[H_2O][CO]}[/tex]
For given reaction the value of [tex]K_c[/tex] will be, [tex]\frac{1}{0.534}[/tex] (for reverse reaction).
[tex]\frac{1}{0.534}=\frac{(x)(x)}{(0.0208-x)\times (0.0208-x)}[/tex]
[tex]\frac{1}{0.534}=\frac{(x)^2}{(0.0208-x)^2}[/tex]
[tex]\frac{1}{0.534}=(\frac{(x)}{(0.0208-x)})^2[/tex]
[tex]\sqrt{\frac{1}{0.534}}=\frac{(x)}{(0.0208-x)}[/tex]
[tex]1.368=\frac{(x)}{(0.0208-x)}[/tex]
By solving the term x, we get
[tex]x=0.012[/tex]
Thus, the concentration of [tex]H_2[/tex] at equilibrium = x = 0.012 M
