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

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:

On adding two reactions,

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

For example:

On subtracting two reactions,

Or,

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

For example:

Thus,

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

For example:

Thus,

Now, the given equilibrium reaction is as follows:

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

Now, multiply the above reaction with 1/2,

Thus,

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

50.23 i believe is correct but don't quote me on that.