91Ó°ÊÓ

We are given the initial reaction and its equilibrium constant: \[ \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\] \[ K_p = 3.5 \times 10^4 \] Temperature: \(1495 K\).

There are a few basic rules for manipulating equilibrium constants when changing a reaction: 1. If the reaction is reversed, then the equilibrium constant is the reciprocal of the original constant. 2. If the reaction is multiplied by a factor, then the equilibrium constant is raised to the power of that factor. Now, let's apply these rules to find the equilibrium constants for the new reactions. a. For the first reaction: \[ \mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Br}_{2}(g) \] Compare it to the original reaction. This reaction is the reverse of the original and is multiplied by a factor of 1/2. Therefore, the equilibrium constant for this reaction is: \[ K_{p_{new}} = \frac{1}{K_p^{\frac{1}{2}}} \] \[ K_{p_{new}} = \frac{1}{(3.5 \times 10^4)^{\frac{1}{2}}} = 1.69 \times 10^{-3} \] b. For the second reaction: \[ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \] This reaction is the reverse of the original reaction. Therefore, the equilibrium constant for this reaction is: \[ K_{p_{new}} = \frac{1}{K_p} \] \[ K_{p_{new}} = \frac{1}{3.5 \times 10^4} = 2.86 \times 10^{-5} \] c. For the third reaction: \[ \frac{1}{2} \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g) \] This reaction is identical to the original reaction but multiplied by a factor of 1/2. Therefore, the equilibrium constant for this reaction is: \[ K_{p_{new}} = {K_p}^{\frac{1}{2}} \] \[ K_{p_{new}} = (3.5 \times 10^4)^{\frac{1}{2}} = 1.87 \times 10^2 \]

The equilibrium constants for the new reactions at 1495 K are: a. \( K_{p_{new}} = 1.69 \times 10^{-3} \) b. \( K_{p_{new}} = 2.86 \times 10^{-5} \) c. \( K_{p_{new}} = 1.87 \times 10^2 \)