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? 2.1110-3 237 -2.11…

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asked Aug 25, 2019 176k views

1 Answer

Draw · Answer 1 · 2019-08-31T17:38:37+0000

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,

$A+C\rightarrow B+D(K=K_(1)* K_(2))$

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,

$A+D\rightarrow B+C(K=(K_(1))/(K_(2)))$

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,

$B\rightarrow A (K=(1)/(K_(1)))$

Now, the given equilibrium reaction is as follows:

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

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

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

Now, multiply the above reaction with 1/2,

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

Thus,

$K=\left ((1)/(4.22* 10^(-3)) \right )^(1/2)=15.4$

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