1)Consider the following equilibrium: PCl3 (g) + Cl2 (g) PCl5 (g) Kc = 26 at 250
ID: 1004943 • Letter: 1
Question
1)Consider the following equilibrium:
PCl3 (g) + Cl2 (g) PCl5 (g) Kc = 26 at 250°C
1.0 mol of PCl3, 1.0 mol of Cl2, and 5.0 mol of PCl5 are placed in a 10.0 L flask at 250°C and allowed to come to equilibrium. What is the concentration, in M, of Cl2 at equilibrium?
2)Cyclohexane, C6H12, undergoes a molecular rearrangement in the presence of AlCl3 to form 20) methylcyclopentane, CH3C5H9, according to the equation:
C6H12 CH3C5H9
If Kc = 0.143 at 25°C for this reaction, find the equilibrium concentrations of C6H12 and CH3C5H9 if the initial concentrations are 0.200 M and 0.075 M, respectively.
Can someone please help me solve these as simple as possible thank you!
Explanation / Answer
PCl3 Cl2 PCl5
Initial 1 mol 1mol 5 mol
at equilibrium 1-X 1-X 5+X
[PCl3] = ( 1-X)/10 , [Cl2] = ( 1-X) /10 , [PCl5] = (5+X) /10
K = [PCl5]/[PCl3] [Cl2]
26 = ( 5+X)/10 / (( 1-X) /10 ) ( 1-X) /10)
2.6 = ( 5+X) / ( 1-X)^2
2.6X^2 -6.2X -2.4 = 0
X = -0.339
hence [Cl2] = ( 1-(-0.339) /10 = 0.1339 M
2) C6H12 CH3C5H9
Initial 0.2 0.075
at equilibrium 0.2-X 0.075+X
Kc = [CH3C5H9]/[C6H12]
0.143 = ( 0.075+X) / ( 0.2-X)
0.0286-0.143X = 0.075+X
X = - 0.0406
[C6H12] = 0.2-(-0.0406) = 0.2406 M
[CH3C5H9] = 0.075+(-0.0406) = 0.0344 M
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