The reaction of bromine gas with chlorine gas, shown here, has a K_p value of 7.20. If a closed vessel was charged with the two reactants, each at an initial partial pressure of 0.500 atm, and the product also at 0.500 atm, what would be the equilibrium partial pressure of BrCl(g) ?
Br₂(g) + Cl₂(g) 2BrCl(g)

Question

The reaction of bromine gas with chlorine gas, shown here, has a K_p value of 7.20. If a closed vessel was charged with the two reactants, each at an initial partial pressure of 0.500 atm, and the product also at 0.500 atm, what would be the equilibrium partial pressure of BrCl(g) ?
Br₂(g) + Cl₂(g)

The reaction of bromine gas with chlorine gas, shown here, has a K<sub>p</sub> value of 7.20. If a closed vessel was charged with the two reactants, each at an initial partial pressure of 0.500 atm, and the product also at 0.500 atm, what would be the equilibrium partial pressure of BrCl(g) ? Br<sub>2</sub>(g) + Cl<sub>2</sub>(g) 2BrCl(g) A) 0.500 atm B) 0.680 atm C) 0.859 atm D) 0.029 atm E) 0.987 atm

2BrCl(g)

Quizplus · Accepted Answer

The equilibrium constant, K, can be calculated from the K_p value using the equation: K = (PBrCl)2 / (PBr2 × PCl2) Substituting the given values, we get: 7.20 = (PBrCl)2 / (0.5 × 0.5) PBrCl = √(7.20 × 0.25) = 0.859 atm Therefore, the equilibrium partial pressure of BrCl(g) is 0.859 atm, which is choice C.

The Reaction of Bromine Gas with Chlorine Gas, Shown Here