Question

1. 0.60 mol of Br2 and 0.60 mol of Cl2 are placed in a 1.00 L flask and allowed to reach equilibrium. (There is no BrCl at first.) 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

0.60 mol of Br2 and 0.60 mol of Cl2 are placed in a 1.00 L flask and allowed to reach equilibrium. (There is no BrCl at first.) After reaching equilibrium, the flask is found to contain 0.28 mol of BrCl. What is the value of K for this reaction?

Answer 2

Answer:

My answer is - 0.217

Explanation:

GIVEN = conc. of Br₂= 0.6 mol

          conc. of Cl₂ = 0.6 mol

conc. of BrCl after eqm. = 0.28 mol

TO FIND = K for the reaction

K = $\frac{xBrCl^{2} }{xbr2 * xcl2}$

K = [0.28]²/ [0.6] [0.6]

k= 0.217

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