For the reaction, A+B⟶ Products On doubling the initial concentration of A only, the rate of reaction is also doubled, therefore Rate ∝[A]1 Let initially rate law is Rate =k[A][B]y If concentration of and both are doubled, the rate gets changed by a factor of 8 . 8× Rate =k[2 A][2 B]y…( iii) [∵ Rate ∝[A]1] Dividing Eq. (iii) by Eq. (ii), 8=2×2y4=2y∴y=2 Hence, rate law is, rate =k[A][B]2
For the reaction, A+B → products, it is observed that (1) On doubling the initial concentration of A only, the rate of reaction is also doubled and…
NEET UG 2009 — Chemistry Physical Chemistry
For the reaction, A+B→ products, it is observed that (1) On doubling the initial concentration of A only, the rate of reaction is also doubled and (2) On doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is, given by
Held on 30 Apr 2009 · Verified 9 Jul 2026.
Options
- A
rate =k[A]2[ B]
- B
rate =k[A][B]2
- C
rate =k[A]2[ B]2
- D
rate =k[ A][B]
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