Question

Part A

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] = 0.150 M , [H2] =0.200 M , and [S2] = 0.000 M. Find the equilibrium [S2].

Express your answer with the appropriate units.

Answer 1

The reaction of decomposition of H2S into H2 and S2 is given as follows:

2H₂S(g)   <===> 2H₂(g) + S₂(g)

The expression for equilibrium constant(Kc) is given as follows:

Kc = ([H₂]² x [S₂]) / [H₂S]²

Where, [H₂] = molar concentration of H₂ at equilibrium.

[S₂] = molar concentration of S₂ at equilibrium.

[H₂S] = molar concentration of H₂S at equilibrium.

Let 2a M be the amount of H₂S decomposed at equilibrium. Therefore, Substitute [H₂S] = (0.150 – 2a) ;

[H₂] = (0.200 + 2a); [S₂] = a and Kc = 1.67 x 10^-7

(1.67 x 10^-7) = (0.200 + 2a)² x (a)) / (0.150-2a)²

Since, Kc is very small, the value of a can be assumed to be very small compared with 0.200 and 0.150. Hence, value of a can be neglected in subtraction and addition.

(1.67 x 10^-7) = (0.200)² x (a)) / (0.150)²

(1.67 x 10^-7) = (0.04a) / (0.0225)

a = (1.67 x 10^-7) x (0.0225) / (0.04)

a = 0.939 x 10^-7 M

Therefore, equilibrium concentration of S₂ is 0.939 x 10^-7 M