Question

For a chemical reaction $A \rightarrow B$ the rate of the reaction is $2 \times 10^{-3}$ mol dm$^{-3}$ s$^{-1}$ when the initial concentration is 0.05 mol dm$^{-3}$. The rate of the same reaction is $1.6 \times 10^{-2}$ mol dm$^{-3}$ s$^{-1}$ when the initial concentration is 0.1 mol dm$^{-3}$. The order of the reaction is

• A. 2
• B. 0
• C. 3
• D. 1

Hint:

If doubling concentration increases rate by 8 times ($2^3$), the order is 3.

Answer 1

The Correct Option is C

Step 1: Set up Rate Equations
$Rate_{1} = k(0.05)^{n} = 2 \times 10^{-3}$. $Rate_{2} = k(0.1)^{n} = 1.6 \times 10^{-2}$.
Step 2: Compare Rates
$\frac{Rate_{2}}{Rate_{1}} = \frac{k(0.1)^{n}}{k(0.05)^{n}} = \frac{1.6 \times 10^{-2}}{2 \times 10^{-3}}$.
Step 3: Solve for n
$(2)^{n} = 8 \Rightarrow (2)^{n} = 2^{3}$. Thus, $n = 3$.
Final Answer: (c)