Question

A \( \rightarrow \) B is a first-order reaction. The concentration of A is decreased from \( x \) mol \( L^{-1} \) to \( y \) mol \( L^{-1} \) in \( 100 \) min. What is the average velocity of the reaction in mol \( L^{-1} \) min\(^{-1}\)?

• A. \( \frac{|x - y|}{100} \)
• B. \( \frac{|y - x|^2}{100} \)
• C. \( \frac{100}{|x - y|} \)
• D. \( \frac{100}{|x + y|} \)

Hint:

For first-order reactions, average reaction rate is simply the change in concentration divided by time.

Answer 1

The Correct Option is A

Step 1: Definition of Average Reaction Rate The average rate of a reaction is given by: \[ \text{Rate} = \frac{\text{Change in concentration}}{\text{Time interval}} \] For a first-order reaction: \[ \text{Average Rate} = \frac{|C_{\text{initial}} - C_{\text{final}}|}{t} \] Step 2: Substituting Given Values Here: - Initial concentration = \( x \) mol \( L^{-1} \), - Final concentration = \( y \) mol \( L^{-1} \), - Time interval = \( 100 \) min. \[ \text{Rate} = \frac{|x - y|}{100} \] Conclusion Thus, the correct answer is: \[ \frac{|x - y|}{100} \]