The equilibrium constant, K, for the following reaction is 3.40×10-2 at 527 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 10.5 L container at 527 K contains 0.271 M PCl5, 9.60×10-2 M PCl3 and 9.60×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 19.5 L? [PCl5] = M [PCl3] = M [Cl2] = M
Answer
[PCl5] = 0.1309M
[PCl3] = 0.06671M
[Cl2] = 0.06671M
Explanation
PCl5(g) <-------> PCl3(g) + Cl2(g)
K = [PCl3][Cl2]/[PCl5] = 3.40 × 10-2
After increasing volume of the container
Initial concentration
[PCl5] = 0.271M/( 19.5L/10.5L) = 0.1459M
[PCl3] = 0.0960M/(19.5L /10.5L) = 0.05169M
[Cl2] = 0.0960M/(19.5L /10.5L) = 0.05169M
change in concentration
[PCl5] = -x
[PCl3] = +x
[Cl2] = +x
equilibrium concentration
[PCl5] = 0.1459 - x
[PCl3] = 0.05169 + x
[Cl2] = 0.05169 + x
so,
(0.05169 + x)2/(0.1459 - x) = 3.40 ×10-2
solving for x
x = 0.01502
Therefore, at equilibrium
[PCl5] = 0.1459 - 0.01502 = 0.1309M
[PCl3] = 0.05169 + 0.01502 = 0.06671M
[Cl2] = 0.05169 + 0.01502 = 0.06671M
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