The equilibrium constant, K, for the following reaction is 5.37E2 at 540 K. PCl
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The equilibrium constant, K, for the following reaction is 5.37E2 at 540 K.
 PCl_{5}(g) PCl_{3}(g) + Cl_{2}(g)
An equilibrium mixture of the three gases in a 18.5 L container at 540 K contains0.167 M PCl_{5}, 9.48E2 M PCl_{3} and 9.48E2 M Cl_{2}. 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 9.64 L?
[PCl_{5}]  =  
[PCl_{3}]  =  
[Cl_{2}]  = 
It says you have to use the quadratic formula
Thank you for the opportunity to help you with your question!
PCl5= 5.203
PCl3=4.27
Cl2=0.16
wrong. look
Step 1. Determine the concentrations immediately after the volume has been reduced, before the system responds to the disturbance.
Rearranging the relationship M_{1}V_{1} = M_{2}V_{2} gives M_{2} = M_{1}  V_{1} 
V_{2} 
[PCl_{5}> = 0.167 M  18.5 L  = 0.320 M 
9.64 L 
[PCl_{3}> = 9.48E2 M  18.5 L  = 0.182 M 
9.64 L 
[Cl_{2}> = 9.48E2 M  18.5 L  = 0.182 M 
9.64 L 
Step 2. Predict the direction in which the reaction will proceed to reach equilibrium.
 PCl_{5}(g) [img src="http://cxp.cengage.com/contentservice/assets/T=1432835347244/owms01h/mediaarchives/GenChem/Image/Doublearrow.GIF">PCl_{3}(g) + Cl_{2}(g) K = 5.37E2
Le Chatelier's Principle predicts that reducing the volume of the system will cause the system to shift to the left, forming fewer moles of gas.
It is also possible to compare Q to K.
K =  [PCl_{3}>[Cl_{2}]  = 5.37E2 
[PCl_{5}] 
Q =  (0.182)(0.182)  = 0.103 > K 
(0.320) 
This also predicts a shift towards reactants as the system comes to equilibrium.
Step 3. Set up an ICE table to define equilibrium concentrations in terms of x.
PCl_{3} (g)
PCl_{5} (g)  +  Cl_{2} (g)  
Initial (M)  0.320  0.182  0.182 
Change (M)  + x   x   x 
Equilibrium (M)  (0.320 + x)  (0.182  x)  (0.182 x) 
Step 4. Substitute the equilibrium concentrations into the equilibrium constant expression and solve for x.
K =  [PCl_{3}>[Cl_{2}]  =  (0.182  x)(0.182 x )  = 5.37E2 
[PCl_{5}]  (0.320 + x) 
Rearrange to get an expression of the form ax^{2} + bx + c = 0 and use the quadratic formula (see the information page) to solve for x. This gives:
x = 4.23E2, 0.375
The second value leads to results that are not physically reasonable.
Step 5. Use x = 4.23E2 to solve for the final equilibrium concentrations.
[PCl_{5}] = 0.320 + 4.23E2 = 0.363 M
[PCl_{3}] = 0.182  4.23E2 = 0.140 M
[Cl_{2}] = 0.182  4.23E2 = 0.140 M
Step 6. Check to see that the system is at equilibrium.
Q =  [PCl_{3}][Cl_{2}]  =  (0.140)(0.140)  = 5.37E2 = K 
[PCl_{5}]  (0.363) 
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