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)

Question

3 years ago

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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?

Chemistry

1 answer:

Firlakuza [10] · Answer 1 · 2022-03-20T07:57:19+03:00

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g) K₁= ? (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g) K₂= 3.61 (2)

2 A(g) + D(g) ⇄ C(g) K₃= 7.19 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3) algebraically to get our desired equilibrium (1).

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

₊

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