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: [tex]A\rightarrow B (K_{1})[/tex]
[tex]C\rightarrow D(K_{2})[/tex]
On adding two reactions,
[tex]A+C\rightarrow B+D(K=K_{1}\times K_{2})[/tex]
2. On subtracting two equilibrium reactions, their equilibrium constant gets divided.
For example: [tex]A\rightarrow B (K_{1})[/tex]
[tex]C\rightarrow D(K_{2})[/tex]
On subtracting two reactions,
[tex]A-C\rightarrow B-D[/tex]
Or,
[tex]A+D\rightarrow B+C(K=\frac{K_{1}}{K_{2}})[/tex]
3. If an equilibrium reaction is multiplied by any constant, it goes to the power of its equilibrium constant.
For example: [tex]A\rightarrow B (K_{1})[/tex]
Thus,
[tex]2A\rightarrow 2B (K=K_{1}^{2})[/tex]
4. On reversing an equilibrium reaction, the equilibrium constant of reversed reaction becomes inverse of the original value.
For example: [tex]A\rightarrow B (K_{1})[/tex]
Thus,
[tex]B\rightarrow A (K=\frac{1}{K_{1}})[/tex]
Now, the given equilibrium reaction is as follows:
[tex]3A+2B\rightleftharpoons 2D+E (K=4.22\times 10^{-3})[/tex]
To get the desired reaction, first reverse the above reaction as follows:
[tex]2D+E\rightleftharpoons 3A+2B\left ( K=\frac{1}{4.22\times 10^{-3}} \right )[/tex]
Now, multiply the above reaction with 1/2,
[tex]D+1/2E\rightleftharpoons 3/2A+B\left ( K=\left (\frac{1}{4.22\times 10^{-3}} \right )^{1/2} \right )[/tex]
Thus,
[tex]K=\left (\frac{1}{4.22\times 10^{-3}} \right )^{1/2}=15.4[/tex]
Therefore, equilibrium constant for the resultant reaction is 15.4 that is option D.
