Consider the following reaction:
A(g)?2B(g)
Find the equilibrium partial pressures of A and Bfor each of the
following different values of Kp. Assume that the initial
partial pressure of B in each case is 1.0 atm and that the initial
partial pressure of A is 0.0 atm. Make any appropriate simplifying
assumptions.
Kp= 1.8
Kp= 1.6×10?4
Kp= 1.8×105
A(g) 2B(g)
Initial 0 1
Equilibrium x/2 1-x
Kp = PB2/PA
Part 1) kp = 1.8
1.8 = (1-x)2/(x/2)
(1.8x) = 2 x (1-x)2
1.8x = 2 + 2x2 -4x
2x2-5.8x +2 = 0
x = 0.399
Therefore, at equilibrium,
PB = 1-x =1-0.399 = 0.601 atm
PA = x/2 =0.399/2 = 0.199 atm
Part 2) kp = 1.6 x 10-4
1.6 x 10-4 = (1-x)2/(x/2)
(1.6 x 10-4 x) = 2 x (1-x)2
1.6 x 10-4 x = 2 + 2x2 -4x
2x2 -4x +2 = 0
x = 1
Therefore, at equilibrium,
PB = 1-1 = 0 atm
PA = 1/2 = 0.5 atm
Part 3) kp = 1.8 x 105
1.8 x 105 = (1-x)2/(x/2)
(1.8 x 105 x) = 2 x (1-x)2
1.8 x 105 x = 2 + 2x2 -4x
2x2 -1.8 x 105 x +2 = 0
x = 0.000011
Therefore, at equilibrium,
PB = 1-0.000011 1 atm
PA = 0.000011 /2 = 5.5 x 10-6 atm
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