Question

During a chemical reaction \( X \rightarrow Y \), the rates of reaction starting with initial concentrations of \( X \) as \( 4.0 \times 10^{-3} M \) and \( 2.0 \times 10^{-3} M \) are \( 4.8 \times 10^{-4} \, mol \, L^{-1}/s \) and \( 1.2 \times 10^{-4} \, mol \, L^{-1}/s \) respectively. What is the order of reaction with respect to \( X \)?

• A. 2
• B. 3
• C. 1.5
• D. 1

Hint:

To find order, compare rate ratios with concentration ratios using \( \frac{r_1}{r_2} = \left(\frac{[A]_1}{[A]_2}\right)^n \).

Answer 1

The Correct Option is A

Step 1: Write rate law.
Let the rate law be:
\[ \text{Rate} = k[X]^n \]
where \( n \) is the order of reaction with respect to \( X \).

Step 2: Use given data.
For first case:
\[ r_1 = 4.8 \times 10^{-4}, \quad [X]_1 = 4.0 \times 10^{-3} \]
For second case:
\[ r_2 = 1.2 \times 10^{-4}, \quad [X]_2 = 2.0 \times 10^{-3} \]

Step 3: Take ratio of rates.
\[ \frac{r_1}{r_2} = \left(\frac{[X]_1}{[X]_2}\right)^n \]
\[ \frac{4.8 \times 10^{-4}}{1.2 \times 10^{-4}} = \left(\frac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}}\right)^n \]
\[ 4 = (2)^n \]

Step 4: Solve for \( n \).
\[ 2^n = 2^2 \]
\[ n = 2 \]

Step 5: Conclusion.
Thus, the order of reaction with respect to \( X \) is:
\[ \boxed{2} \]