1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant for the following reactions are K1 and K 2, respectively. 2 P (g)+3 Cl 2(g) leftharpoons 2 PCl 3(g) PCl 3(g)+ Cl 2(g) leftharpoons PCl 5(g) Then, the equilibrium constant for the reaction 2 P (g)+5 Cl 2(g) leftharpoons 2 PCl 5(g) is

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Q. The equilibrium constant for the following reactions are $K_1$ and $K_ 2$, respectively.
$2 P (g)+3 Cl _{2}(g) \rightleftharpoons 2 PCl _{3}(g)$
$PCl _{3}(g)+ Cl _{2}(g) \rightleftharpoons PCl _{5}(g)$
Then, the equilibrium constant for the reaction
$2 P (g)+5 Cl _{2}(g) \rightleftharpoons 2 PCl _{5}(g)$ is

KVPYKVPY 2012Equilibrium

Solution:

If the equation is multiplied by $2$ the equilibrium constant for the new equation is the square of $K$, i.e. $K^{2}$.
If we add two equation, then the equilibrium constant of new equations is the product of the equilibrium constant of those equations.
For the reactions,
$2 P (g)+3 Cl _{2}(g) \rightleftharpoons 2 PCl _{3}(g)\,\,\,\,...(i)$
$PCl _{3}(g)+ Cl _{2}(g) \rightleftharpoons PCl _{5}(g)\,\,\,\,...(ii) $
$2 P (g)+5 Cl _{2}(g) \rightleftharpoons 2 PCl _{5}(g)\,\,\,\,...(iii)$
To obtain (iii) we multiply Eq. (ii) by $2$ and then add Eq. (i) to it.
Thus, equilibrium constant for Eq. (iii), $K=K_{1} K_{2}^{2}$