The reaction 2H2S(g)⇌2H2(g)+S2(g) Kc=1.67×10−7 at 800∘C is carried out with the following initial concentrations: [H2S] = 0.300 M , [H2] =0.300 M , and [S2] = 0.00 M. Find the equilibrium concentration of [S2].
To find the equilibrium concentration of [S2], we can use the equilibrium constant expression and the given initial concentrations.
The equilibrium constant expression for the reaction is:
Kc = [H2]^2[S2] / [H2S]^2
Given:
[H2S] = 0.300 M
[H2] = 0.300 M
[S2] = 0.00 M (initial concentration)
Given that the reaction is at equilibrium, let the equilibrium concentration of [S2] be x M.
Substituting the values into the equilibrium constant expression:
1.67×10^-7 = (0.300)^2(x) / (0.300)^2
Simplifying and solving for x:
1.67×10^-7 = x / 0.300^2
x = 1.67×10^-7 * 0.300^2
x = 1.67×10^-7 * 0.090
x = 1.503×10^-8
Therefore, the equilibrium concentration of [S2] is approximately 1.503×10^-8 M.
To find the equilibrium concentration of S2 in the given reaction, we need to use the equilibrium constant (Kc) and the initial concentrations of the reactants.
The equilibrium constant equation for this reaction is:
Kc = [H2]^2[S2] / [H2S]^2
First, let's substitute the given values into the equilibrium constant equation:
Kc = (0.300 M)^2 * [S2] / (0.300 M)^2
Simplifying the equation:
Kc = [S2] / 0.09
Now, we can rearrange the equation to isolate [S2]:
[S2] = Kc * 0.09
Substituting the given value of Kc (1.67 × 10^(-7)):
[S2] = (1.67 × 10^(-7)) * 0.09
Calculating the equilibrium concentration of S2:
[S2] = 1.503 × 10^(-8) M
Therefore, the equilibrium concentration of [S2] is 1.503 × 10^(-8) M.
.........2H2S(g)⇌2H2(g)+S2(g)
I.......0.300...0.300....0
C........-2x......+2x...+x
E......0.3-2x......2x....x
Substitute the E line into Kc expression and solve for x = (S2)