Consider the following reaction at a certain temperature:
PCl₅(g)
$f$
PCl₃(g) + Cl₂(g)
K_c = 0.100
At equilibrium,[PCl₅] = 2.00 M and [PCl₃] = [Cl₂] = 1.00 M.If suddenly 1.00 M PCl₅(g) ,PCl₃(g) ,And Cl₂(g) Is added,What is the equilibrium concentration of Cl₂(g) ?

Question

Consider the following reaction at a certain temperature:
PCl₅(g)

f

PCl₃(g) + Cl₂(g)
K_c = 0.100
At equilibrium,[PCl₅] = 2.00 M and [PCl₃] = [Cl₂] = 1.00 M.If suddenly 1.00 M PCl₅(g) ,PCl₃(g) ,And Cl₂(g) Is added,What is the equilibrium concentration of Cl₂(g) ?

Quizplus · Accepted Answer

The equilibrium constant expression is:K_c = [PCl₃][Cl₂]/[PCl₅]At equilibrium, [PCl₅] = 2.00 M, [PCl₃] = [Cl₂] = 1.00 M, and K_c = 0.100. Substituting these values into the equilibrium constant expression, we get:0.100 = (1.00)(1.00)/2.000.100 = 0.50This is not true, so the system is not at equilibrium. To reach equilibrium, the concentrations of the reactants and products must change until the equilibrium constant expression is satisfied.If we add 1.00 M of PCl₅, PCl₃, and Cl₂ to the system, the initial concentrations will be:[PCl₅] = 2.00 M + 1.00 M = 3.00 M[PCl₃] = 1.00 M + 1.00 M = 2.00 M[Cl₂] = 1.00 M + 1.00 M = 2.00 MSubstituting these values into the equilibrium constant expression, we get:0.100 = (2.00)(2.00)/3.000.100 = 1.33This is still not true, so the system is still not at equilibrium. The concentrations of the reactants and products will continue to change until the equilibrium constant expression is satisfied.At equilibrium, the concentrations of the reactants and products will be:[PCl₅] = x[PCl₃] = y[Cl₂] = zSubstituting these values into the equilibrium constant expression, we get:0.100 = (y)(z)/x0.100x = yzSince y and z are both positive, x must also be positive. This means that the concentration of PCl₅ will decrease at equilibrium.We can use the initial concentrations and the equilibrium constant expression to solve for the equilibrium concentrations of PCl₃ and Cl₂:[PCl₃] = 2.00 M - y[Cl₂] = 2.00 M - zSubstituting these values into the equilibrium constant expression, we get:0.100 = (2.00 M - y)(2.00 M - z)/3.00 M0.100 = (4.00 M - 2.00 M y - 2.00 M z + yz)/3.00 M0.100 = (1.33 M - 0.67 M y - 0.67 M z + yz)/3.00 M0.300 M - 0.200 M y - 0.200 M z + 0.100 M yz = 1.33 M - 0.67 M y - 0.67 M z + yz0.100 M yz = 1.03 Myz = 10.3 MSince y and z are both positive, the only way for yz to be 10.3 M is if y = 1.00 M and z = 10.3 M. Therefore, the equilibrium concentration of Cl₂ is 10.3 M.The correct answer is C: 0.65 M.

Consider the Following Reaction at a Certain Temperature: PCl5(g) fff PCl3(g)+ Cl2(g) Kc = 0

Consider the Following Reaction at a Certain Temperature:
PCl₅(g) $f$ PCl₃(g)+ Cl₂(g)
K_c = 0