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 con

Question

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 con

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)
C. `2^(n-m)`
D. `1/(2^(m+n))`

Monjur Alahi

8 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
c) `Rate_(1) = k[A]^(n)[B]^(m)`
`Rate_(2) = k[2A]^(n)[1//2B]^(m)`
`(Rate_(2))/(Rate_(1))= (k[2A]^(n)[1//2B]^(m))/(k[A]^n)[B]^(m)`
`=[2]^(n([1//2])^(m))= 2^(n).2^(-m)`
`=2(n-m)`

8 views Answered 2023-02-05 15:29