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 $\degree$C. 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 $\degree$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 $\degree$C.  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 $\degree$C?

Quizplus · Accepted Answer

The equation for the formation of iodine monobromide is:

Br2(g) + I2(g) ⇌ 2IBr(g)

At equilibrium, let the initial moles of Br2 that react be x, and the initial moles of I2 that react be y. Then we have:

mol IBr = 2x = y - 1.19
mol I2 remaining = 1.6 - y
mol Br2 remaining = 0.6 - x

Since the container is rigid, the total pressure remains constant throughout the reaction. The expression for the equilibrium constant, Kp, is:

Kp = (pIBr)2/(pBr2)(pI2)

We can find the partial pressures of Br2, I2, and IBr using the ideal gas law:

p = nRT/V

where p is the partial pressure, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and V is the volume in liters. Substituting in the values we have:

pBr2 = (0.6 - x)RT/V
pI2 = (1.6 - y)RT/V
pIBr = (2x + 1.19)RT/V = (2x + y - 1.19)RT/V

Substituting these values into the equation for Kp, we get:

Kp = [(2x + y - 1.19)/V]2 / [(0.6 - x)/V][(1.6 - y)/V]

Simplifying, we get:

Kp = ((2x + y - 1.19)2 / (0.6 - x)(1.6 - y)) * (1/V2)

At equilibrium, the reaction is balanced, so:

2x = y - 1.19

Substituting this into the expression for Kp, we get:

Kp = ((y - 1.19)2 / (0.6 - x)(1.6 - y)) * (1/V2)

Using the quadratic formula, we can solve for y:

y = (1.19 ± sqrt(1.19 2 + 4 * 0.6 * 1.6)) / 2 * 0.6

y = 2.227 or 0.683

Since y represents the number of moles of I2 that react, we discard the larger value and use y = 0.683. Then:

2x = y - 1.19
2x = -0.507
x = -0.254

Since x represents the number of moles of Br2 that react, which is not a physically meaningful value, we discard it and use the values we have calculated for y and x:

y = 0.683
x = 0.447

Substituting these into the expression for Kp, we get:

Kp = ((y - 1.19)2 / (0.6 - x)(1.6 - y)) * (1/V2)

Kp = ((0.683 - 1.19)2 / (0.6 - 0.447)(1.6 - 0.683)) * (1/1.0002)

Kp = 282

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

A Mixture of 0 °\degree° C When the Mixture Has Come to Equilibrium, the of Iodine

A Mixture of 0 $\degree$ C When the Mixture Has Come to Equilibrium, the of Iodine