A reaction starts with 1.00 mol each of PCl3 and Cl2 in a 1.00-L flask. When equilibrium is established at 250 degree Celsius in the reaction PCl3(g)+ Cl2(g)--> PCl5(g), the amount of PCl5 present is 0.82 mol. What is Kc for this reaction?
PCl3 + Cl2 ==> PCl5.
initial:
PCl3 = 1 mole/1 L = 1 M
Cl2 = 1 M
PCl5 = 0
change:
PCl5 = +x
Cl2 = -x
PCl3 = -x
equilibrium:
PCl3 = 1-x
Cl2 = 1-x
PCl5 = x = 0.82M
Redo PCl3 = 1-x = 1-0.82 = ??
Redo Cl2 = 1-x = 1 - 0.82 = ??
Substitute the equilibrium values into Kc and solve for Keq.
keq is 1.53 and the Kc isproducts over reactants
To find the value of Kc for this reaction, we need to use the equilibrium concentrations of the reactants and products. In this case, the equilibrium concentration of PCl5 is given as 0.82 mol in a 1.00-L flask.
First, we need to calculate the equilibrium concentrations of PCl3 and Cl2. Since the initial moles of both PCl3 and Cl2 are 1.00 mol each, they both react to form the product PCl5.
Let's assume the equilibrium concentration of PCl3 and Cl2 is x mol/L.
Thus, the amount of PCl3 consumed is 1.00 mol - x mol, and the amount of Cl2 consumed is 1.00 mol - x mol.
According to the balanced chemical equation, the ratio of moles of PCl3 consumed to PCl5 formed is 1:1 and the ratio of moles of Cl2 consumed to PCl5 formed is also 1:1.
Therefore, at equilibrium, the concentration of PCl5 is 0.82 mol/L, the concentration of PCl3 is (1.00 mol - x) mol/L, and the concentration of Cl2 is (1.00 mol - x) mol/L.
The expression for the equilibrium constant, Kc, is given by:
Kc = [PCl5] / [PCl3] * [Cl2]
Substituting the equilibrium concentrations we found:
Kc = 0.82 mol/L / [(1.00 mol - x) mol/L] * [(1.00 mol - x) mol/L]
Since we know that the equilibrium concentration of PCl5 is 0.82 mol/L, we can substitute it into the expression:
Kc = (0.82 mol/L) / [(1.00 mol - x) mol/L] * [(1.00 mol - x) mol/L]
Now, we solve for x. To do this, we use the fact that at equilibrium, the initial moles minus the moles consumed should be equal to the moles of PCl5 formed:
1.00 mol - x mol = 0.82 mol
x mol = 1.00 mol - 0.82 mol
x mol = 0.18 mol
Substituting the value of x into the expression for Kc:
Kc = (0.82 mol/L) / [(1.00 mol - 0.18 mol) mol/L] * [(1.00 mol - 0.18 mol) mol/L]
Kc = (0.82 mol/L) / (0.82 mol/L) * (0.82 mol/L)
Kc = 1.0
Therefore, the value of Kc for this reaction is 1.0.