The following reactions at 1000.0 K have the following K_p values: H₂ (g) + CL₂ (g) ⇄ 2 HCl(g) K_p = 3.8 * 10⁴
H₂ (g) + Br₂ (g) ⇄ 2 HBr(g) ) K_p =5.1 *10⁸
Br₂ (g) + CL₂ (g) ⇄ 2 Br Cl(g) K_p = 0.20
What is K_p for the reaction HCl(g) + HBr(g) ⇄ BrCl(g) + H₂(g) at 1000.0 K?

Question

Quizplus · Accepted Answer

The equilibrium constant for the target reaction can be found by dividing the equilibrium constant for the reaction forming HBr from H2 and Br2 by the equilibrium constant for the reaction forming HCl from H2 and Cl2, and then multiplying by the equilibrium constant for the reaction of Br2 with Cl2 to form 2 BrCl. Mathematically, this is represented as $K_{\text{HCl} \rightarrow \text{HBr}} = \frac{K_{\text{H2 + Br2} \rightarrow 2\text{HBr}}}{K_{\text{H2 + Cl2} \rightarrow 2\text{HCl}}} \times K_{\text{Br2 + Cl2} \rightarrow 2\text{BrCl}}$. Given $K_{\text{H2 + Br2} \rightarrow 2\text{HBr}} = 5.1 \times 10^{18}$, $K_{\text{H2 + Cl2} \rightarrow 2\text{HCl}} = 3.8 \times 10^{14}$, and $K_{\text{Br2 + Cl2} \rightarrow 2\text{BrCl}} = 0.20$, the calculation yields $K_{\text{HCl} \rightarrow \text{HBr}} = \frac{5.1 \times 10^{18}}{3.8 \times 10^{14}} \times 0.20 = 1.0 \times 10^{-7}$.

The Following Reactions at 1000