Answer to Question #114580 in General Chemistry for Kailey

Question

Consider the following reaction: Bra(g) + Cla(g) 2 BrCl(g) After 0.60 mol of Brz and 0.60 mol of Ch were placed in a 1.00 L flask, the reaction was allowed to reach equilibrium. After reaching equilibrium, the flask is found to contain 0.28 mol of BrCl. What is the value of K for this reaction?

Answer 1

Br₂(g) + Cl₂(g) = 2BrCl(g)

K = x_BrCl²/(x_Br2*x_Cl2), where x - molar fraction of component in the mixture.

Find the number of moles of components at equilibrium:

n_equilibrium(Br₂) = n_initial(Br₂) - n_react(Br₂) = n_initial(Br₂) - n_react(BrCl)/2 = 0.60 - 0.28/2 = 0.46 mol

n_equilibrium(Cl₂) = n_initial(Cl₂) - n_react(Cl₂) = n_initial(Cl₂) - n_react(BrCl)/2 = 0.60 - 0.28/2 = 0.46 mol

Calculate the molar fractions for each component (x = n_component/n_mixture):

x_BrCl = n(BrCl)/n_mixture = 0.28/1.2 = 0.233

x_Br2 = n_equilibrium(Br₂)/n_mixture = 0.46/1.2 = 0.383

x_Cl2 = n_equilibrium(Cl₂)/n_mixture = 0.46/1.2 = 0.383

Сalculate K for this reaction:

K = x_BrCl²/(x_Br2*x_Cl2) = 0.233²/(0.383*0.383) = 0.37

Answer: K for reaction Br₂(g) + Cl₂(g) = 2BrCl(g) is 0.37.