Question

A → P is a first order reaction. The reaction was started at 10.00 AM. At 10.10 AM, the concentration of A was x mol L^-1. At 10.20 AM, the concentration of A was y mol L^-1. The half life (in min) of the reaction is equal to:

• A. $2.303 \log\left(\dfrac{x}{y}\right)$
• B. $\log\left(\dfrac{x}{y}\right)$
• C. $3.01$
• D. $3.01 \log\left(\dfrac{x}{y}\right)$

Hint:

In first order kinetics, logarithmic relations are crucial; memorize integrated rate law and half-life formula.

Answer 1

The Correct Option is D

For first-order reactions: $k = \dfrac{2.303}{t} \log\left(\dfrac{[A]_0}{[A]}\right)$ and $t_{1/2} = \dfrac{0.693}{k}$. 
Here, 20 minutes interval and k calculated from concentrations leads to $t_{1/2} = 3.01 \log(x/y)$.