Question

Consider the following reaction:
Br2(g) + Cl2(g) ⇌ 2BrCl(g), Kp=1.112 at 150 K.
A reaction mixture initially contains a Br2 partial pressure of 751 torr and a Cl2 partial pressure of 737 torr at 150 K.
Calculate the equilibrium partial pressure of BrCl.

Answer 1

Answer:

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g) , Kp=1.112 at 150 K

denoting BC as BrCl , B as Br₂ , C as Cl₂, p as partial pressure , then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √(Kp*pB*pC)

replacing values

pBC = √(Kp*pB*pC) = √(1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Answer 2

The partial pressure is the pressure exerted by each molecule in the reaction mixture. The partial pressure of BrCl is 784.52 torr.

What is equilibrium constant?

The equilibrium constant is given as the ratio of the pressure of product to reactant, raised to the stoichiometric coefficient.

The equilibrium pressure constant ($K_p$) is given as:

$K_p=\rm \dfrac{[BrCl]^2}{[Br_2][Cl_2]}$

Where, the given equilibrium constant, $K_p=1.112$

The partial pressure of bromine is, $\rm [Br_2]=751\;torr$

The partial pressure of chlorine is, $\rm [Cl_2]=737\;torr$

Substituting the values for the partial pressure of BrCl:

$\rm 1.112=\dfrac{[BrCl]^2}{751\;\times\;737}\\\\ BrCl=\sqrt{751\;\times\;737\;\times\;1.112}\\\\ BrCl=784.52\;torr$

The partial pressure of BrCl in the solution is 784.52 torr.

Learn more about vapor pressure, here:

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