JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For the reaction, \(3A+2B \to C + D\), the differential rate law can be written as
- A \(\frac{1}{3}\frac{{d\left[ A \right]}}{{dt}} = \frac{{d\left[ C \right]}}{{dt}} = k{\left[ A \right]^n}{\left[ B \right]^m}\)
- B \( - \frac{{d\left[ A \right]}}{{dt}} = \frac{{d\left[ C \right]}}{{dt}} = k{\left[ A \right]^n}{\left[ B \right]^m}\)
- C \( + \frac{1}{3}\frac{{d\left[ A \right]}}{{dt}} = - \frac{{d\left[ C \right]}}{{dt}} = k{\left[ A \right]^n}{\left[ B \right]^m}\)
- D \( - \frac{1}{3}\frac{{d\left[ A \right]}}{{dt}} = \frac{{d\left[ C \right]}}{{dt}} = k{\left[ A \right]^n}{\left[ B \right]^m}\)
Answer & Solution
Correct Answer
(D) \( - \frac{1}{3}\frac{{d\left[ A \right]}}{{dt}} = \frac{{d\left[ C \right]}}{{dt}} = k{\left[ A \right]^n}{\left[ B \right]^m}\)
Step-by-step Solution
Detailed explanation
For the reaction \(3A+2B\to C+D\) Rate of disappearance of \(A\,=\) Rate of appearance of \(C\) reation \( = \, - \,\frac{1}{3}\,\frac{{d[A]}}{{dt}}\, = \,\frac{{d[C]}}{{dt}}\, = \,k\,{[A]^n}\,{[B]^m}\)
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