Question

For the reactions, $A\rightleftharpoons B,K_c=1;B\rightleftharpoons C,K_c=3;C\rightleftharpoons D,K_c=5K_c.$ The $K_c$ value for the reaction $A\rightleftharpoons D$ is

• A. 15
• B. 5
• C. 3
• D. 1

Answer 1

The correct option is A 15

Equilibrium constant of concentration (K_c) of a chemical reaction at equilibrium can be defined as the ratio of the concentration of products to the concentration of the reactants, each raised to their respective stoichiometric coefficients.

Given $K_c$ values,

$\frac{\left[B\right]}{\left[A\right]}=1\Rightarrow \left(i\right)$
$\frac{\left[C\right]}{\left[B\right]}=3\Rightarrow \left(ii\right)$
$\frac{\left[D\right]}{\left[C\right]}=5\Rightarrow \left(iii\right)$

$\frac{\left[D\right]}{\left[A\right]}$ = $K_c$ for the reaction $A\rightleftharpoons D$

Multiplying (i), (ii) and (iii) we get,
$\frac{\left[D\right]}{\left[A\right]}=1\times 3\times 5=15$
so $K_c$ = $15$