JEE Main 2025 Apr 4 Shift 1, Chemistry Q8: Rate Of A Chemical Reaction

JEE Main2025Apr 4, Shift 1Chemistry

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

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

Solution

Initial rate: $r_{1} = k [A]^{n} [B]^{m}$ .

After doubling [A] and halving [B]: $r_{2} = k (2 [A])^{n} (\frac{[ B ]}{2})^{m} = k \cdot 2^{n} \cdot 2^{- m} \cdot [A]^{n} [B]^{m}$ .

Dividing: $\frac{r _{2}}{r _{1}} = 2^{n} \cdot 2^{- m} = 2^{(n - m)}$ .

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JEE Main 2025 Apr 4 Shift 1, Chemistry Q8: Rate Of A Chemical Reaction

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