JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The rate law for a reaction between the substances A and B is given by Rate = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as
- A(m + n)
- B(n - m)
- C2( n - m)
- D
View written solutionFree
Correct answer: CORRECT ANSWER SHOULD BE $2^{N-M}$ (NOT LISTED AMONG THE OPTIONS)
-
Given rate law
-
Initial rate
Let the initial rate be
-
New concentrations
- Concentration of is doubled:
- Concentration of is halved:
So the new rate becomes
-
Simplify the new rate
-
Find the ratio of new rate to old rate
-
Compare with options
The required ratio is
But the given options are:
- A:
- B:
- C:
- D:
None of these matches .
-
Conclusion
The correct expression should be
Hence, the stored answer C is not correct.
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