Bromine and chlorine gases react reversibly to form bromine monochloride: Br2(g) + Cl2(g) 2BrCl(g) Since Br2, Cl2 and BrCl are all gases, this is a homogeneous equilibrium: every species present is in the same (gaseous) phase. A gaseous equilibrium mixture of Br2, Cl2 and BrCl i…

AS Chemistry 9701: Kp for the Homogeneous Br2 + Cl2 ⇌ 2BrCl Equilibrium

Bromine and chlorine gases react reversibly to form bromine monochloride:

$\text{Br}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\text{BrCl}(g)$

Since $\text{Br}_2$ , $\text{Cl}_2$ and $\text{BrCl}$ are all gases, this is a homogeneous equilibrium: every species present is in the same (gaseous) phase. A gaseous equilibrium mixture of $\text{Br}_2$ , $\text{Cl}_2$ and $\text{BrCl}$ is established in a sealed container at constant temperature. At equilibrium, the total pressure is $200\ \text{kPa}$ , and the mole fractions of $\text{Br}_2$ , $\text{Cl}_2$ and $\text{BrCl}$ are $0.20$ , $0.20$ and $0.60$ respectively.

(a) Calculate the partial pressure, in kPa, of each of the three gases at equilibrium. [3]

(b) Write the expression for $K_p$ , and use your answers to (a) to calculate its value. State whether $K_p$ has units, explaining your reasoning. [3]

(c) State and explain the effect, if any, on the value of $K_p$ of increasing the total pressure at constant temperature. [2]

Part (a): Calculating the partial pressures

All three species ( $\text{Br}_2(g)$ , $\text{Cl}_2(g)$ and $\text{BrCl}(g)$ ) are gases occupying the same container, so this is a homogeneous equilibrium (a single phase throughout), not a heterogeneous one.

The mole fractions sum to $0.20+0.20+0.60=1.00$ , as required.

Each partial pressure is the mole fraction of that gas multiplied by the total pressure: $p(\text{Br}_2) = 0.20\times200 = 40\ \text{kPa}$ $p(\text{Cl}_2) = 0.20\times200 = 40\ \text{kPa}$ $p(\text{BrCl}) = 0.60\times200 = 120\ \text{kPa}$

(Check: the three partial pressures should also sum to the total pressure: $40+40+120=200\ \text{kPa}$ . Consistent.)

Part (b): Writing the Kp expression and calculating its value

$K_p = \frac{p(\text{BrCl})^2}{p(\text{Br}_2)\,p(\text{Cl}_2)}$

Substituting the values from part (a): $K_p = \frac{(120)^2}{40\times40} = \frac{14400}{1600} = 9.0$

(Check: dividing independently, $14400\div1600 = 9.0$ . Consistent.)

Units: the exponent on $\text{BrCl}$ is 2, giving $\text{kPa}^2$ in the numerator, and the denominator is $\text{kPa}\times\text{kPa}=\text{kPa}^2$ as well. These cancel exactly, so $K_p=9.0$ has no units. This makes sense because the total number of moles of gas is the same on both sides of the equation ( $1+1=2$ reactant moles, $2$ product moles).

Part (c): Effect of increasing the total pressure

Increasing the total pressure has no effect on the value of $K_p$ . An equilibrium constant is only changed by a change in temperature (changing pressure, concentration, or adding a catalyst can shift the position of equilibrium, but the value of $K_p$ itself stays the same at constant temperature. (In fact, for this particular reaction the position of equilibrium would not even shift with pressure, since there are equal moles of gas) 2. On each side of the equation, so no direction is favoured by compressing the mixture.)

Final answers

(a) $p(\text{Br}_2)=40\ \text{kPa}$ , $p(\text{Cl}_2)=40\ \text{kPa}$ , $p(\text{BrCl})=120\ \text{kPa}$
(b) $K_p=9.0$ , no units
(c) Unchanged, only a temperature change alters $K_p$