1. Tardigrade
  2. Question
  3. Chemistry
  4. For the reaction, A leftharpoons nB the concentration of A decreases from 0.06 to 0.03 mol L-1 and that of B rises from 0 to 0.06 mol L-1 at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. For the reaction, $A \rightleftharpoons nB$ the concentration of $A$ decreases from $0.06$ to $0.03\, mol\, L^{-1}$ and that of $B$ rises from $0$ to $0.06\, mol\, L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction, respectively, are

KVPYKVPY 2013Equilibrium

Solution:

image
Thus, the reaction becomes $A \rightleftharpoons 2B$
$k_{c}=\frac{\left[B\right]^{2}}{\left[A\right]}=\frac{\left(0.06\right)^{2}}{0.03}=0.12$