A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C. Br₂(g) + I₂(g) ⇄ 2IBr(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

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:
Kc = ([IBr]^2)/([Br2][I2])
At equilibrium, the concentration of iodine monobromide is given as 1.190 M. The initial concentration of bromine is 0.600 mol/L and that of iodine is 1.600 mol/L. Let x be the concentration of IBr at equilibrium, then the equilibrium concentrations can be expressed as follows:
[Br2] = (0.600 - x) mol/L
[I2] = (1.600 - x) mol/L
[IBr] = x mol/L
Substituting these values in the equilibrium constant expression, we get:
Kc = ([x]^2)/([(0.600 - x)][(1.600 - x)])
Given, at equilibrium [IBr] = 1.190 M, substituting this value and solving for x, we get:
x = 1.190 M
Therefore, [Br2] = (0.600 - 1.190) = -0.590 M (negative value indicates that the concentration of bromine is completely consumed to form IBr)
[I2] = (1.600 - 1.190) = 0.410 M
Substituting these values in the equilibrium constant expression, we get:
Kc = ([1.190]^2)/([(-0.590)][(0.410)]) = 282
Therefore, the equilibrium constant for this reaction at 350°C is 282. The answer is option D.

A Mixture of 0