Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g) + D(g) equilibrium reaction arrow C(g) what is the value of K at the temperature for the following reaction? C(g) + D(g) equilibrium reaction arrow 2 B(g) What is the value of Kp at 45°C for the same reaction? Starting with 1.64 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?

We are allowed to reverse reactions, in that case we take the reciprocal of K as our new K' ; we can also add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3) and we want D as a reactant . That suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B as product in our equilibrium (1) . Finally we would add (2) and (3) to get (1) which is our final goal.

2C(g) ⇄ 2A(g) + 2B(g) K₂´= ( 1/ 3.61 )²

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2 A(g) + D(g) ⇄ C(g) K₃ = 7.19

C(g) + D(g) ⇄ 2B(g) K₁ = ( 1/ 3.61 )² x 7.19

K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves gases.

To compute the last part lets setup the following mnemonic ICE table to determine the quantities at equilibrium:

pressure (atm) C D B

initial 1.64 1.64 0

change -x -x +2x

equilibrium 1.64-x 1.64- 2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)² where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27