Question

Use the reactions below and their equilibrium constants to predict the equilibrium constant for the reaction 2A(s)⇌3D(g). A(s) ⇌ 12 B(g)+C(g), K1=0.0334 3D(g) ⇌ B(g)+2C(g), K2=2.35

Answer 1

Answer : The equilibrium constant for the reaction will be, $4.747\times 10^{-4}$

Explanation :

The following equilibrium reactions are :

(1) $A(s)\rightleftharpoons \frac{1}{2}B(g)+C(g)$           $K_1=0.0334$

(2) $3D(g)\rightleftharpoons B(g)+2C(g)$          $K_2=2.35$  

The final equilibrium reaction is :

$2A(s)\rightleftharpoons 3D(g)$         $K_{eqm}=?$

Now we have to calculate the value of $K_{eqm}$ for the final reaction.

First we have to multiply equation (1) by 2 that means we are taking square of equilibrium constant 1 and reverse the equation 2 that means we are dividing equilibrium constant 2, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{eqm}=\frac{(K_1)^2}{K_2}$

Now put all the given values in this expression, we get :

$K_{eqm}=\frac{(0.0334)^2}{2.35}$

$K_{eqm}=4.747\times 10^{-4}$

Therefore, the value of $K_{eqm}$ for the final reaction is, $4.747\times 10^{-4}$

Answer 2

The equilibrium constant for the reaction 2A(s)⇌3D(g) is 4.747 × 10⁻⁴

From the question,

We are to determine the equilibrium constant for the reaction

2A(s) ⇌ 3D(g)

This equilibrium constant, K, for this reaction is given by

$K = \frac{[D]^{3} }{[A]^{2} }$

From the given reactions and equilibrium constants, we have

A(s) ⇌ 1/2 B(g) + C(g), K1=0.0334

3D(g) ⇌ B(g) + 2C(g), K2=2.35

From equation (1)

A(s) ⇌ 1/2 B(g) + C(g)

The equilibrium constant, K₁ is given by

$K_{1}=\frac{[B]^{\frac{1}{2} }[C] }{[A]}$  --------- (X)

From equation (2)

3D(g) ⇌ B(g) + 2C(g)

The equilibrium constant, K₂ is given by

$K_{2}=\frac{[B][C]^{2} }{[D]^{3} }$ --------- (Y)

From (X)

$K_{1}=\frac{[B]^{\frac{1}{2} }[C] }{[A]}$

We can write that

$[C] =\frac{K_{1}[A] }{[B]^{\frac{1}{2} } }$

Put this into (Y)

$K_{2}=\frac{[B][C]^{2} }{[D]^{3} }$

We get

$K_{2}=\frac{[B]\frac{K_{1}^{2} [A]^{2}}{([B]^{\frac{1}{2} })^{2} } }{[D]^{3} }$

Then,

$K_{2}=\frac{[B]\frac{K_{1}^{2} [A]^{2}}{[B] } }{[D]^{3} }$

$K_{2}=\frac{{K_{1}^{2} [A]^{2}} }{[D]^{3} }$

This becomes,

$\frac{[D]^{3} }{[A]^{2} } =\frac{K_{1}^{2} }{K_{2} }$

Recall that the equilibrium constant for the reaction

2A(s) ⇌ 3D(g)

is

$K = \frac{[D]^{3} }{[A]^{2} }$

∴ $K = \frac{K_{1}^{2} }{K_{2} }$

From the question,

K₁ = 0.0334 and K₂ = 2.35

∴$K = \frac{0.0334^{2} }{2.35}$

$K = \frac{0.00111556}{2.35}$

K = 0.0004747

K = 4.747 × 10⁻⁴

Hence, the equilibrium constant for the reaction 2A(s)⇌3D(g) is 4.747 × 10⁻⁴

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