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

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)

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. Br<sub>2</sub>(g) + I<sub>2</sub>(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? A) 3.55 × 10<sup>-3</sup> B) 1.24 C) 1.47 D) 282 E) 325

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?

Quizplus · Accepted Answer

We can set up the equilibrium expression as follows:

Kc = ([IBr]/[I2][Br2])

We are given that the concentration of IBr is 1.190 M at equilibrium. To find the concentrations of I2 and Br2 at equilibrium, we need to use the stoichiometry of the reaction. For every 2 moles of IBr that react, 1 mole of I2 and 1 mole of Br2 are produced. Therefore, the initial moles of I2 and Br2 are:

n(I2) = 1.600 mol
n(Br2) = 0.600 mol

At equilibrium, let x be the amount (in moles) of IBr that reacts. Then, the equilibrium concentrations can be expressed as:

[IBr] = 1.190 M
[I2] = (1.600 - x)/1.000 L
[Br2] = (0.600 - x)/1.000 L

Substituting these concentrations into the equilibrium expression, we get:

Kc = (1.190/[(1.600 - x)/1.000][(0.600 - x)/1.000])

We are given that the concentration of IBr at equilibrium is 1.190 M, so we can substitute this into the equilibrium expression to get:

1.190 = (1.190/[(1.600 - x)/1.000][(0.600 - x)/1.000])

Simplifying this expression, we get:

(1.600 - x)(0.600 - x) = 1.000

Expanding and rearranging, we get a quadratic equation:

0.36x^2 - 1.080x + 0.64 = 0

Solving for x using the quadratic formula, we get:

x = 0.928 mol

Substituting this value back into the equilibrium concentrations, we get:

[IBr] = 1.190 M
[I2] = 0.672 M
[Br2] = 0.172 M

Substituting these concentrations into the equilibrium expression, we get:

Kc = (1.190/[(0.672)(0.172)]) = 282

Therefore, the equilibrium constant for this reaction at 350°C is 282. The correct answer is D.

A Mixture of 0