The equilibrium constant, K, for the following reaction is 2.35×10-2 at 517 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 11.3 L container at 517 K contains 0.269 M PCl5, 7.94×10-2 M PCl3 and 7.94×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 5.27 L? [PCl5] = M [PCl3] = M [Cl2] = M
equilibrium constant K = 2.35 * 10^-2
Temperature T = 517 K
Initial volume V1 = 11.3 L
Initial Concentrations M1
[PCl5] = 0.269 M
[PCl3] = 7.94 * 10^-2 M
[Cl2] = 7.94 * 10^-2 M
Final volume V2 = 5.27 L
Final concentrations M2
M1*V1 = M2*V2
[PCl5] = 0.269*11.3/5.27 = 0.577 M
[PCl3] = 7.94 * 10^-2 * 11.3/5.27 = 0.170 M
[Cl2] = 7.94 * 10^-2 * 11.3/5.27 = 0.170 M
From the Le Chatelier's principle
equilibrium mixture is compressed (increase in pressure) and volume is decreased, equilibrium will shift towards with fewer moles (reactant side in left direction)
The balanced reaction with ICE TABLE
PCl5 = PCl3 + Cl2
I 0.577 0.170 0.170
C +x -x - x
E (0.577+x) (0.170-x) (0.170-x)
Equilibrium constant expression of the reaction
K = [PCl3] [Cl2] / [PCl5]
2.35 * 10^-2 = [0.170-x] [0.170-x] / [0.577+x]
(2.35 * 10^-2) (0.577+x) = [0.170-x]^2
0.0135595 + 0.0235x = 0.0289 + x^2 - 0.340x
x^2 - 0.3635x + 0.0153405 = 0
x = 0.0487
once equilibrium has been reestablished
[PCl5] = 0.577 + 0.0487 = 0.6254 M
[PCl3] = 0.170 - 0.0487 = 0.1213 M
[Cl2] = 0.170 - 0.0487 = 0.1213 M
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