Question

Two reactions and their equilibrium constants are given. A + 2 B → 2CK1 = 2.17 2C → DK2 = 0.222 A+2B→2CK1=2.172C→DK2=0.222 Calculate the value of the equilibrium constant for the reactions.

Answer 1

This is an incomplete question, here is a complete question.

Two reactions and their equilibrium constants are given.

(1) $A+2B\rightleftharpoons 2C$;        $K_1=2.17$

(2) $2C\rightleftharpoons D$;        $K_2=0.222$

Calculate the value of the equilibrium constant for the reactions.

$D\rightarrow A+2B$ ;        $K_3=?$

Answer: The value of equilibrium constant for the given reaction is, 2.08

Explanation:

The given chemical equations are:

(1) $A+2B\rightleftharpoons 2C$;        $K_1=2.17$

(2) $2C\rightleftharpoons D$;        $K_2=0.222$

We need to calculate the equilibrium constant for the chemical equation, which is:

$D\rightarrow A+2B$ ;        $K_3=?$

• If the equation is revered then the equilibrium constant will be the reciprocal of the reaction.
• If the equation is added then the equilibrium constant will be the multiplied.

The value of equilibrium constant for the given reaction is:

$K_3=\frac{1}{K_1}\times \frac{1}{K_2}$

$K_3=\frac{1}{2.17}\times \frac{1}{0.222}$

$K_3=2.08$

Hence, the value of equilibrium constant for the given reaction is, 2.08

Answer 2

The equilibrium constant for the given reaction has been 2.07.

The given reaction and their equilibrium constant has been given as:

$\rm A\;+\;2B\;\leftrightharpoons 2C\;;K1=2.17 \\2C\leftrightharpoons D\;;K2=0.222$

The reaction for which equilibrium constant has to be calculated has been:

$\rm D\;\rightarrow\;A\;+\;2B$

Computation for Equilibrium Constant

The equation has been achieved from the given reactions by the reverse of reaction 1, leading to the production of A and 2B.

The reactant C has been eliminated in the reaction by the reverse of the reaction 2.

Thus, the equilibrium constant, K has been given as:

$K=\dfrac{1}{K1}\;+\;\dfrac{1}{K2}$

Substituting the values  in the equation for the calculation of K:

$K=\dfrac{1}{2.17}\;\times\;\dfrac{1}{0.222 }\\K=0.46\times 4.5\\K=2.07$

The equilibrium constant for the given reaction has been 2.07.

For more information about the equilibrium constant, refer to the link:

https://brainly.com/question/17960050