Question

A reaction is third order with respect to a reactant. If the concentration of the reactant is doubled, the rate of reaction increased by

• A. $2$ times
• B. $3$ times
• C. $4$ times
• D. $8$ times
• E. $5$ times

Hint:

If concentration becomes $m$ times, rate changes by $m^n$, where $n$ is order.

Answer 1

The Correct Option is D

Concept: Chemistry - Rate Law. [ \text{Rate} = k[A]^n ] where $n$ is order of reaction.
Step 1: Given order. Reaction is third order with respect to reactant. [ n=3 ]
Step 2: Double concentration. If concentration becomes $2[A]$: [ \text{New Rate}=k(2[A])^3 ]
Step 3: Simplify. [ \text{New Rate}=k \cdot 8[A]^3 ] [ =8 \times \text{Original Rate} ]
Step 4: Final answer. [ \boxed{8\text{ times}} ]
Hence, correct option is (D).