From data: Doubling [A] increases rate by 3x; tripling [A] increases rate by 6x. Approximate the reaction order n?

From data: Doubling [A] increases rate by 3x; tripling [A] increases rate by 6x. Approximate the reaction order n?

• A. Approximately 1.6
• B. Approximately 2
• C. 0
• D. 3

Answer: (A)

The rate law for a reaction that depends on a single reactant A is rate = k [A]^n. When you double the concentration, the rate scales by 2^n. Since the rate increases by a factor of 3 when [A] is doubled, you have 2^n = 3, so n = log(3)/log(2) ≈ 1.585, about 1.6. You can check with the tripling data: 3^n should give a factor of 6, so n = log(6)/log(3) ≈ 1.631, also near 1.6. The two estimates agree, pointing to a fractional order. Therefore, the approximate reaction order with respect to A is about 1.6, meaning it lies between first and second order. Fractional orders often arise from complex reaction mechanisms where the dependence on concentration isn’t simply an integer power.

The rate law for a reaction that depends on a single reactant A is rate = k [A]^n. When you double the concentration, the rate scales by 2^n. Since the rate increases by a factor of 3 when [A] is doubled, you have 2^n = 3, so n = log(3)/log(2) ≈ 1.585, about 1.6.

You can check with the tripling data: 3^n should give a factor of 6, so n = log(6)/log(3) ≈ 1.631, also near 1.6. The two estimates agree, pointing to a fractional order.

Therefore, the approximate reaction order with respect to A is about 1.6, meaning it lies between first and second order. Fractional orders often arise from complex reaction mechanisms where the dependence on concentration isn’t simply an integer power.