Question

For a chemical reaction A \(\rightarrow\) B, it was observed that the rate of reaction doubles when the concentration of A is increased four times. The order of the reaction is:

• A. 2
• B. 1
• C. \( \frac{1}{2} \)
• D. Zero

Hint:

Use the relation \( \text{Rate} \propto [A]^n \) and take ratio of rates to determine order. Apply logarithms if necessary.

Answer 1

The Correct Option is C

Rate law: \[ \text{Rate} \propto [A]^n \Rightarrow \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[A]_2}{[A]_1} \right)^n \] Given: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = 2,\quad \frac{[A]_2}{[A]_1} = 4 \Rightarrow 2 = 4^n \Rightarrow \log 2 = n \log 4 \Rightarrow n = \frac{\log 2}{\log 4} = \frac{1}{2} \] \[ \boxed{n = \frac{1}{2} \Rightarrow \text{Order of reaction is } \frac{1}{2}} \]