Question

At a particular temperature, Kc = 1.6 x 10⁻² for 2 H₂S(g) + 2H₂(g) + S₂(g)

Calculate Kc for each of the following reactions:
(a) ½ S₂(g) + H₂(g) + H₂S (g)

Answer 1

The equilibrium constant K'c for the reaction ½ S₂(g) + H₂(g) ⇌ H₂S (g) is calculated to be 3906.25, using the given value of Kc as 1.6 x 10⁻² for the original reaction and reversing the reaction and dividing the coefficients by 2.

Step-by-step explanation:

The original reaction given is 2 H₂S(g) → 2 H₂(g) + S₂(g) with a Kc value of 1.6 x 10⁻². To find the Kc for the reaction ½ S₂(g) + H₂(g) ⇌ H₂S (g), we need to realize that this new reaction is the reverse of the original one, with the coefficients divided by 2. Therefore, the equilibrium constant for this new reaction, K'c, is the reciprocal of the original equilibrium constant, raised to the power corresponding to the division of coefficients. In this case:

K'c = (1/Kc)²

So K'c for the reaction ½ S₂(g) + H₂(g) ⇌ H₂S (g) is:

K'c = (1 / (1.6 x 10⁻²))² = (1 / (0.016))² = (62.5)² = 3906.25