Respuesta :
The question is incomplete, complete question is:
A 0.10 mol sample of each of the four species in the reaction represented above is injected into a rigid, previously evacuated 1.0 L container. Which of the following species will have the highest concentration when the system reaches equilibrium?
[tex]2H_2S(g)+CH_4(g)\rightleftharpoons CS_2(g)+4H_2(g)[/tex]
The equilibrium constant of the reaction = [tex]K_c=3.4\times 10^{-4}[/tex]
A) [tex]H_2S(g)[/tex] B) [tex]CH_4(g)[/tex] C) [tex]CS_2(g)[/tex] D) [tex]H_2(g)[/tex]
Answer:
The correct answer is option A.
Explanation:
[tex]2H_2S(g)+CH_4(g)\rightleftharpoons CS_2(g)+4H_2(g)[/tex]
The equilibrium constant of the reaction = [tex]K_c=3.4\times 10^{-4}[/tex]
0.10 moles of each gas was kept in 1.0 L of container. So. the con concentration of each gas becomes:
[tex][H_2S]=[CH_4]=[CS_2]=[H_2]=\frac{0.10 mol}{1 L}=0.10 M[/tex]
Reaction quotient of the reaction :
[tex]Q=\frac{[CS_2][H_2]^4}{[H_2S]^2[CH_4]}[/tex]
[tex]Q=\frac{0.10 M\times (0.10 M)^4}{(0.10 M)^2\times 0.10 M}=0.01[/tex]
[tex]Q>K_c[/tex]
The reaction will go backward:
[tex]2H_2S(g)+CH_4(g)\rightleftharpoons CS_2(g)+4H_2(g)[/tex]
Initially
0.10 M 0.10 M 0.10 M 0.10 M
At equilibrium :
(0.10+2x) M (0.10+x) M (0.10-x) M (0.10-4x) M
The expression of an equilibrium constant will be given as';
[tex]K-c=\frac{[CS_2][H_2]^4}{[H_2S]^2[CH_4]}[/tex]
[tex]3.4\times 10^{-4}=\frac{(0.10-x) M\times ((0.10-4x) M)^4}{((0.10+2x) M)^2\times (0.10+x) M}[/tex]
Solving for x:
x = 0.099
Concentration of species at equilibrium :
[tex][H_2S]=(0.10+2x) M=(0.10+2\times 0.099) M=0.298 M[/tex]
[tex][CH_4]=(0.10+x) M=(0.10+ 0.099) M=0.199 M[/tex]
[tex][CS_2]=(0.10-x) M=(0.10- 0.099) M=0.001 M[/tex]
[tex][CH_4]=(0.10-4x) M=(0.10-4\times 0.099) M=-0.296 M[/tex]
The highest concentration when the system reaches at equilibrium will be of hydrogen sulfide.
