Chemistry: The equilibrium constant for the equilibrium, 3a+ 2b ↔ 2d + e, is 4.22 x 10-3 . what is the equilibrium constant for the equilibrium: d + (1/2)e ↔ (3/2)a + b? 2.1110-3237-2.1110-315.4

The equilibrium constant for the equilibrium, 3a+

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12.03.2022

Option D: 15.4

There are following rules for manipulating equilibrium constant:

1. On adding two equilibrium reactions, their equilibrium constant gets multiplied.

For example: $A\rightarrow B (K_{1})$

$C\rightarrow D(K_{2})$

On adding two reactions,

$A+C\rightarrow B+D(K=K_{1}\times K_{2})$

2. On subtracting two equilibrium reactions, their equilibrium constant gets divided.

For example: $A\rightarrow B (K_{1})$

$C\rightarrow D(K_{2})$

On subtracting two reactions,

$A-C\rightarrow B-D$

Or,

$A+D\rightarrow B+C(K=\frac{K_{1}}{K_{2}})$

3. If an equilibrium reaction is multiplied by any constant, it goes to the power of its equilibrium constant.

For example: $A\rightarrow B (K_{1})$

Thus,

$2A\rightarrow 2B (K=K_{1}^{2})$

4. On reversing an equilibrium reaction, the equilibrium constant of reversed reaction becomes inverse of the original value.

For example: $A\rightarrow B (K_{1})$

Thus,

$B\rightarrow A (K=\frac{1}{K_{1}})$

Now, the given equilibrium reaction is as follows:

$3A+2B\rightleftharpoons 2D+E (K=4.22\times 10^{-3})$

To get the desired reaction, first reverse the above reaction as follows:

$2D+E\rightleftharpoons 3A+2B\left ( K=\frac{1}{4.22\times 10^{-3}} \right )$

Now, multiply the above reaction with 1/2,

$D+1/2E\rightleftharpoons 3/2A+B\left ( K=\left (\frac{1}{4.22\times 10^{-3}} \right )^{1/2} \right )$

Thus,

$K=\left (\frac{1}{4.22\times 10^{-3}} \right )^{1/2}=15.4$

Therefore, equilibrium constant for the resultant reaction is 15.4 that is option D.

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