Question

The equilibrium constant in terms of pressures for the reaction COBr2(g) CO(g) + Br2(g) is Kp = 5.40 at 346 K. (a) A pure sample of gaseous carbonyl bromide, COBr2(g), is introduced into a rigid flask at a temperature of 346 K so that its original pressure is 0.123 atm. Calculate the fraction of this starting material that is converted to products at equilibrium.

Answer 1

97.8% of starting material

Step-by-step explanation:

In the reaction:

COBr₂(g) ⇄ CO(g) + Br₂(g)

Kp is defined as:

`Kp = \frac{P_(CO)P_{Br_(2)}}{P_(COBr_2)}` = 5.40

As initial pressure of COBr₂ is 0.123atm, the pressure at equilibrium of each is:

COBr₂(g) = 0.123atm - X

CO(g) = X

Br₂(g) = X

Where X is reaction coordinate, the amount of reactant that is converted to products

Replacing in Kp expression:

5.40 = X² / (0.123 - X)

0.6642 - 5.40X = X²

0 = X² + 5.40X - 0.6642

Solving for X:

X = -5.52 atm → False solution, there is no negative concentrations.

X = 0.1203 atm → Right solution

Thus, the fraction of starting material that is converted to products is:

0.1203atm / 0.123atm = 0.978 =

97.8% of starting material