JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The reaction \(2 NO + Br _2 \rightarrow 2 NOBr\) takes places through the mechanism given below : \(NO + Br _2 \Leftrightarrow NOBr _2 \text { (fast) }\) \(NOBr _2+ NO \rightarrow 2 NOBr \text { (slow) }\) The overall order of the reaction is \(.....\).
- A \(4\)
- B \(3\)
- C \(5\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(RDS : NOBr _2+ NO \rightarrow 2 NOBr\) \(\left. r = K \left[ NOBr _2\right] NO \right] \quad--- \text { (i) }\) \(Keq =\frac{\left[ NOBr _2\right]}{[ NO ]\left[ Br _2\right]}----- \text { (ii) }\) \(\text { From (i) and (ii) }\)…
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