Question

Given the equilibrium constants for the following two reactions in aqueous solution at 25 ∘C HNO2(aq)H2SO3(aq)⇌⇌H+(aq) + NO2−(aq)2H+(aq) + SO32−(aq)Kc = 4.5 × 10−4Kc = 1.1 × 10−9 what is the value of Kc for the reaction 2HNO2(aq) + SO32−(aq)⇌H2SO3(aq) + 2NO2−(aq)

Answer 1

Answer : The value of $K_c$ for the final reaction is, 184.09

Explanation :

The equilibrium reactions in aqueous solution are :

(1) $HNO_2(aq)\rightleftharpoons H^+(aq)+NO_2^-(aq)$         $K_{c_1}=4.5\times 10^{-4}$

(2) $H_2SO_3(aq)\rightleftharpoons 2H^+(aq)+SO_3^{2-}(aq)$         $K_{c_2}=1.1\times 10^{-9}$

The final equilibrium reaction is :

$2HNO_2(aq)+SO_3^{2-}(aq)\rightleftharpoons H_2SO_3(aq)+2NO_2^-(aq)$         $K_{c}=?$

Now we have to calculate the value of $K_c$ for the final reaction.

Now equation 1 is multiply by 2 and reverse the equation 2, we get the value of final equilibrium reaction and the expression of final equilibrium constant is:

$K_c=\frac{(K_{c_1})^2}{K_{c_2}}$

Now put all the given values in this expression, we get :

$K_c=\frac{(4.5\times 10^{-4})^2}{1.1\times 10^{-9}}=184.09$

Therefore, the value of $K_c$ for the final reaction is, 184.09