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Q.For the reaction, 2A + B → products, when the concentrations of A and B both were doubled the rate of the reaction increased from 0.3 mol L^-1 s^-1 to 2.4 mol L^-1 s^-1. When the concentration of A alone is doubled, the rate increased from 0.3 mol L^-1 s^-1 to 0.6 mol L^-1 s^-1. What is the order of the reaction with respect to B?

Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2025Subjective· 3mImportance★★★★★
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First find the order w.r.t. A from the experiment where only [A] is doubled, then use the experiment where both are doubled to isolate the order w.r.t. B.

Let the rate law be Rate=k[A]m[B]n\text{Rate} = k[A]^m[B]^n.

Step 1 — order with respect to A. When [A] alone is doubled (B constant), the rate goes from 0.3 to 0.6 mol L^-1 s^-1, i.e. it exactly doubles:

0.60.3=2=2m  ⟹  m=1\dfrac{0.6}{0.3} = 2 = 2^m \implies m = 1

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