Question

can be monitored visually by following changes in the color of the reaction mixture (Br2 is reddish brown, and H2 and HBr are colorless). A gas mixture is prepared at 700 K, in which 0.40 atm is the initial partial pressure of both H2 and Br2 and 0.90 atm is the initial partial pressure of HBr. The color of this mixture then fades as the reaction progresses toward equilibrium. Give a condition that must be satis- fied by the equilibrium constant K

Answer 1

Answer:

K > 5.1

Explanation:

Assuming we have the following equilibrium system:

$H_2 (g) + Br_2 (g)\rightleftharpoons 2 HBr (g)$

Since hydrogen is colorless and bromine is reddish brown and because the color of this mixture fades as the reaction progresses toward equilibrium, this means bromine is consumed in this reaction. Both hydrogen and bromine here are reactants, so their partial pressures will decrease (as the color of bromine disappears) and the partial pressure of HBr will increase.

According to the given equation, we may write the equilibrium constant for it as:

$K_{eq} = \frac{[HBr]^2}{[H_2][Br_2]}$

Let's assume that -x atm will be the change for each of the reactants, then we'd have a +2x atm change for the product, as it has a coefficient of '2'. At equilibrium we expect to have a total of:

$[H_2]_{eq} = [Br_2]_{eq} = 0.40 atm - x, [HBr]_{eq} = 0.90 atm + 2x$

We know that reaction proceeds to the right, meaning the reaction quotient, Q, is lower than the equilibrium constant. Find Q by substituting the initial conditions into the expression of K:

$Q=\frac{(0.90)^2}{0.40\cdot 0.40}=5.1$

Now, since K > Q, this means K > 5.1.