A mixture of 0.600 mol bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br2 (g) + I2 (g) 2 IBr (g). When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M. What is the equilibrium constant for this reaction at 350°C?

Question

Chemistry

Winston Meadows

28 October, 20:58

A mixture of 0.600 mol bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br2 (g) + I2 (g) 2 IBr (g). When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M. What is the equilibrium constant for this reaction at 350°C?

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Answers (1)

Know the Answer?

Regan Ellis · Answer 1

The answer to your question is Ke = 1.4751

Explanation:

Data

bromine = 0.6 mol

iodine = 1.6 mol

volume = 1 L

Ke = ?

Iodine monobromide concentration = 1.190 M

Process

1. - Write the balanced chemical equation

Br₂ + I₂ ⇒ 2IBr

2. - Write the formula for Ke

Ke = [IBr]²/[Br₂][I₂]

3. - Substitution

Ke = [1.19]² / [0.6][1.6]

4. - Simplification

Ke = 1.4161 / 0.96

5. - Result

Ke = 1.4751