In reversible reaction A B , the initial concentration of A and B are a and b in moles…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
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.
- \(\frac{\mathrm{k}_{1} \mathrm{a}-\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1}+\mathrm{k}_{2}}\)
- \(\frac{\mathrm{k}_{1} \mathrm{a}-\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1}-\mathrm{k}_{2}}\)
- \(\frac{\mathrm{k}_{1} \mathrm{a}-\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1} \mathrm{k}_{2}}\)
- \(\frac{\mathrm{k}_{1} \mathrm{a}+\mathrm{k}_{2} \mathrm{~b}}{\mathrm{k}_{1}+\mathrm{k}_{2}}\)
Answer
(B) k _ 1 a - k _ 2 ~b k _ 1 - k _ 2
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