The equilibrium constant for the following reaction will be.... | CHEMISTRY - HETEROGENEOUS EQUILIBRIA | SolveeIt

The equilibrium constant for the following reaction will be.

$K_{sp}$ $10^{- 15},10^{- 23}10^{- 20}$

$10^{- 54}$

$A_{x}B_{y}$

$xA^{y +} + YB^{x -},K_{sp}$

$\left\lbrack A^{y +} \right\rbrack^{x}\left\lbrack B^{x -} \right\rbrack^{y}$

Answer

$A_{x}B_{y}$

Explanation

: $P_{4(s)} + 5O_{2(g)}$ $P_{4}O_{10(s)}$

Since concentration of solids is taken as 1, expression for equilibrium constant involves only oxygen.

$K_{c} = \frac{1}{\left\lbrack O_{2} \right\rbrack^{5}}$

Question: The equilibrium constant for the following reaction will be. \(K_{sp}\) \(10^{- 15},10^{- 23}10^{- ...