Br₂(g) + I₂(g) ⇌ 2IBr(g) K_c = 1.2 × 10² at 150 °C Given the information above, what is K_c for the reaction 4IBr(g) ⇌ 2Br₂(g) + 2I₂(g) at 150 °C? - VCE - SSCE Chemistry - Question 27 - 2018 - Paper 1

To find K_c for the given reaction, we start with the known equilibrium constant for the reaction Br₂(g) + I₂(g) ⇌ 2IBr(g), which is K_c = 1.2 × 10².

1. Balanced Equations: The original reaction can be reversed to express IBr as the reactant:

   IBr(g) ⇌ 1/2 Br₂(g) + 1/2 I₂(g)

   This reverse reaction will thus have an equilibrium constant of:

   K_c' = rac{1}{K_c} = rac{1}{1.2 imes 10^2}

2. New Reaction: To get the reaction from 4IBr(g) to 2Br₂(g) + 2I₂(g), we multiply the reverse reaction by 2, leading to: $2IBr(g) ⇌ Br₂(g) + I₂(g)$

   For this multiplied reaction, the equilibrium constant becomes: K_c'' = (K_c')^2 = rac{1}{(1.2 imes 10^2)^2}

3. Calculating the New K_c:
   Now we can calculate:

   K_c'' = rac{1}{(1.2 imes 10^2)^2} = rac{1}{1.44 imes 10^4}

   Calculating that gives: $K_c'' ext{ (approx)} = 6.94 imes 10^{-5}$

Thus, rounding and considering significant figures, K_c for the reaction 4IBr(g) ⇌ 2Br₂(g) + 2I₂(g) at 150 °C is approximately $6.9 imes 10^{-5}$, which corresponds to option C.