Asked by layla
A 0.239 mol sample of PCl5(g) is injected into an empty 2.85 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
Answers
Answered by
DrBob222
(PCl5) = 0.239 mol/2.85L = approx 0.08 but you need a better answer than that estimate.
.........PCl5 ==> PCl3 + Cl2
I.......0.08.......0......0
C........-x........x......x
E......0.08-x......x......x
Substitute the E line into the Kc expression for PCl5 and solve for x and evaluate 0.08-x. By the way you can't work this problem without the Kc NUMBER. I assume you just didn't type this into the post.
.........PCl5 ==> PCl3 + Cl2
I.......0.08.......0......0
C........-x........x......x
E......0.08-x......x......x
Substitute the E line into the Kc expression for PCl5 and solve for x and evaluate 0.08-x. By the way you can't work this problem without the Kc NUMBER. I assume you just didn't type this into the post.
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