Question

The rate of a reaction is doubled when the concentration of the reactant is doubled. The order of the reaction is :

• A. 0
• B. 1
• C. 2
• D. 0.5

Hint:

Direct, linear proportionality between concentration and reaction rate signifies a first-order reaction ($n=1$).

Answer 1

The Correct Option is B

Step 1: Write the general rate law.
For a reaction with a single reactant A, the rate law is $\text{Rate} = k[A]^n$, where $n$ is the order of the reaction.
Step 2: Set up the ratio for the concentration change.
When concentration doubles from $[A]$ to $2[A]$, the new rate becomes $k(2[A])^n = 2^n \cdot k[A]^n$.
Step 3: Apply the given condition.
The problem states the rate doubles: $\dfrac{\text{Rate}_{\text{new}}}{\text{Rate}_{\text{old}}} = 2$. So $2^n = 2$.
Step 4: Solve for $n$.
$2^n = 2^1$ gives $n = 1$. The reaction is first order.
\[ \boxed{n = 1} \]