PCl3 + Cl2 = PCl5 Kc=26 @ 250 degree Celcius
1.0 mol of PCl3 , 1.0 mol of Cl2 , and 5.0 mol of PCl5 are placed in a 10.0L flask at 250 C and allowed to come to equilibrium. What is the concentration, in M,of cl2 of at equilibrium?
Choose one answer.
a. 3.4
b. 0.19
c. 0.16
d. 0.13
e. 0.20
PCl3 + Cl2 ==> PCl5
(PCl3) = 1.0 mol/10.0 L = xx M
(Cl2) = 1.0 mol/10.0 L = yy M
(PCl5) = 5.0 mol/10.0 L = zz M
First determine the reaction quotient, compare it with Keq, and determine which way the reaction will move to reach equilibrium. Then set up an ICE chart, substitute into the Keq expression, and solve for the unknown; use that to calculate each individual concn. I think the reaction will move to the left and I did a quickie calculation (but you should confirm it because I may have made a mistake in my quickie) and I'm looking at 0.13 M for the answer.
If you will go to the original post, I saw there how you had worked the problem and I commented on where your error(s) were.
To solve this question, we need to use the equilibrium constant expression (Kc) and the stoichiometry of the balanced chemical equation. Here's how you can get the answer:
1. Start with the balanced chemical equation:
PCl3 + Cl2 -> PCl5
2. Write the expression for the equilibrium constant (Kc):
Kc = [PCl5] / ([PCl3] * [Cl2])
3. Substitute the given value of Kc (Kc = 26) into the expression:
26 = [PCl5] / ([PCl3] * [Cl2])
4. Since the initial amounts of PCl3, Cl2, and PCl5 are given as 1.0 mol, 1.0 mol, and 5.0 mol respectively, and the volume of the flask is 10.0 L, we can calculate the initial concentrations of PCl3, Cl2, and PCl5:
[PCl3] = 1.0 mol / 10.0 L = 0.10 M
[Cl2] = 1.0 mol / 10.0 L = 0.10 M
[PCl5] = 5.0 mol / 10.0 L = 0.50 M
5. Substitute these values into the equilibrium constant expression:
26 = 0.50 / (0.10 * [Cl2])
6. Rearrange the equation to solve for [Cl2]:
[Cl2] = 0.50 / (0.10 * 26) = 0.19 M
Therefore, the concentration of Cl2 at equilibrium is 0.19 M. So, the correct answer is option b.