JEE Main 2025 · 4 April · morning · Single correct · +4 / −1
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 (r₂/r₁) is
- (a) (n-m)
- (b) 2⁽ⁿ⁻ᵐ⁾
- (c) 1/2ᵐ⁺ⁿ
- (d) (m+n)
Answer and solution
Answer (b) 2⁽ⁿ⁻ᵐ⁾
Solution
Chemistry ma’am · 3 steps
Write the rate twice, divide, and the constant k disappears; only the factors 2 and 1/2 remain.
- New concentrations. After the change, [A] becomes 2[A] and [B] becomes 1/2[B]. The rate constant k stays the same because the temperature has not changed.
r₂/r₁ = k(2[A])ⁿ(1/2[B])ᵐ/k[A]ⁿ[B]ᵐ = 2ⁿ×(1/2)ᵐ
- Combine the powers. (1/2)ᵐ = 2⁻ᵐ, so the ratio is 2ⁿ × 2⁻ᵐ = 2ⁿ⁻ᵐ.
- So. r₂/r₁ = 2⁽ⁿ⁻ᵐ⁾, which is option (b).
The idea to keep. When a concentration is multiplied by a factor, the rate is multiplied by that factor raised to its order. Halving a concentration gives a negative power of 2.
Answer from the NTA final answer key. Solution by PrepWiser.