The equilibrium constant, K, for the following reaction is
3.16×10-2 at 525
K.
An equilibrium mixture of the three gases in a
18.5 L container at 525 K
contains 0.232 M PCl5,
8.57×10-2 M
PCl3 and
8.57×10-2 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 8.42 L?
[PCl5] | = | M |
[PCl3] | = | M |
[Cl2] | = | M |
Answer
[PCl5] = 0.5645M
[PCl3]= 0.1336M
[Cl2] = 0.1336M
Explanation
PCl5(g) <------> PCl3(g) + Cl2(g)
After changing the volume
Initial concentration
[PCl5] = 0.232M/( 8.42L/18.5L) = 0.5097M
[PCl3] = 0.0857M/(8.42L/18.5L) = 0.1883M
[Cl2] = 0.0857M/( 8.42L/18.5L) = 0.1883M
Change in concentration
[PCl5] = -x
[PCl3] = + x
[Cl2] = +x
Equilibrium concentration
[PCl5] = 0.5097 -x
[PCl3] = 0.1883 + x
[Cl2] = 0.1883 + x
so,
(0.1883 + x)2/(0.5097 -x) = 0.0316
solving for x
x = - 0.05475
Therefore, at equilibrium
[PCl5] = 0.5097 -( - 0.05475) = 0.5645M
[PCl3] = 0.1883 + ( - 0.05475) = 0.1336M
[Cl2] = 0.1883 + ( - 0.05475) = 0.1336M
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