# Consider the following reaction and calculate the equilibrium partial pressure of BrCl.: Br2(g)+Cl2(g)⇌2BrCl(g) Kp=1.11×10−4 at 150 K.A reaction mixture initially contains a Br2 partial...

Consider the following reaction and calculate the equilibrium partial pressure of BrCl.:

Br2(g)+Cl2(g)⇌2BrCl(g)

Kp=1.11×10−4 at 150 K.

A reaction mixture initially contains a Br2 partial pressure of 780 torr and a Cl2 partial pressure of 730 torr at 150 K.

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The equilibrium constant expression for this reaction is:

`K_p = (P_(BrCl))/((P_(Br_2))(P_(Cl_2))`

Initial partial pressures are:

BrCl = 0

Br2 = 780 torr

Cl2 = 730 torr

Equilibrium partial pressures, assuming the reactants each decreased by an amount x, are:

BrCl = 2x

Br2 = 780 torr -x

Cl2 = 730 torr -x

Kp = (2x)^2/(780-x)(730-x) = 1.110x10^(-4)

We will make the assumption that x is small enough to neglect subtracting it from the equilibrium pressures of the two reactants, leaving the equation:

(2x)^2 /(780)(730) = 1.110x10^-4

x = 3.97

The equilibrium partial pressure of BrCl is 2x, which is 2x3.97 = **7.94 torr**

This value meets the 5% rule for neglecting x in an earlier step because it's less than 5% of 730 and 780. If it was more than 5% then it would be necessary to include the term (-x) and solve using the quadratic formula.