Question

For a first-order reaction, the graph between \( \log \frac{a}{(a - x)} \) (on y-axis) and time (in min, on x-axis) gave a straight line passing through the origin. The slope is \( 2 \times 10^{-3} \) min\(^{-1}\). What is the rate constant (in min\(^{-1}\))?

• A. \( 2 \times 10^{-3} \)
• B. \( \frac{2 \times 10^{-3}}{2.303} \)
• C. \( 4.606 \times 10^{-3} \)
• D. \( 0.5 \times 10^{-5} \)

Hint:

For first-order reactions, the rate constant \( k \) is obtained from the slope of the plot \( \log \frac{a}{(a - x)} \) vs. time using the formula: \( k = \text{slope} \times 2.303 \).

Answer 1

The Correct Option is C

Step 1: Understanding the First-Order Rate Equation 
For a first-order reaction, the integrated rate law is: \[ \log \frac{a}{(a - x)} = kt \] Comparing this with the equation of a straight line (\( y = mx \)), we see that: \[ \text{slope} = k \times \frac{2.303}{1} \] 

Step 2: Calculating the Rate Constant \( k \) 
Given: \[ \text{slope} = 2 \times 10^{-3} \text{ min}^{-1} \] Using the relation: \[ k = \text{slope} \times 2.303 \] \[ k = (2 \times 10^{-3}) \times 2.303 \] \[ k = 4.606 \times 10^{-3} \text{ min}^{-1} \]

 Step 3: Evaluating the Given Options 
- Option (1): Incorrect, as \( k \) is not directly equal to the slope.
- Option (2): Incorrect, as it divides by 2.303 instead of multiplying.
- Option (3): Correct, as \( k = 4.606 \times 10^{-3} \) min\(^{-1}\).
- Option (4): Incorrect, as the value is far too small.
Thus, the correct answer is 

Option (3).