JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Chemical Kinetics

Rate law for a reaction between A and B is given by

R = k [A]^{n} [B]^{m}

If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction

(\frac{r _{2}}{r _{1}})

(A) $2^{(n - m)}$ ✓ Correct
(B)(n-m)
(C) $(m + n)$
(D) $\frac{1}{2 ^{m + n}}$

Answer: (A)

Solution

With the rate law

R = k [A]^{n} [B]^{m}

, doubling

[A]

multiplies the rate by

2^{n}

and halving

[B]

multiplies it by

(1/2)^{m} = 2^{- m}

. Hence

\frac{r _{2}}{r _{1}} = \frac{k ( 2 [ A ] ) ^{n} ([ B ] /2 ) ^{m}}{k [ A ] ^{n} [ B ] ^{m}} = 2^{n} \cdot 2^{- m} = 2^{(n - m)} .

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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