In reversible reaction A B , the initial concentration of A and B are a and b in moles…

Chemistry · JEE Advanced · NTA ExamsEquilibrium

In reversible reaction \(A \rightleftharpoons B\), the initial concentration of A and B are a and b in moles per litre and the equilibrium concentration are (a – x) and (b – x) respectively ; express x in terms of k1, k2, a and b.
  1. \(\frac{\mathrm{k}_{1} \mathrm{a}-\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1}+\mathrm{k}_{2}}\)
  2. \(\frac{\mathrm{k}_{1} \mathrm{a}-\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1}-\mathrm{k}_{2}}\)
  3. \(\frac{\mathrm{k}_{1} \mathrm{a}-\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1} \mathrm{k}_{2}}\)
  4. \(\frac{\mathrm{k}_{1} \mathrm{a}+\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1}+\mathrm{k}_{2}}\)

Answer

(B) k _ 1 a - k _ 2 ~b k _ 1 - k _ 2

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Equilibrium questions · Browse all practice questions