Question

For a first order reaction,the slope of the graph between ${\ln }_e\left(\frac{[A{]}_0}{\left[A\right]}\right)(ony-axis)$ and $\text{time (min)}(onx-axis)$ is found out to be $0.5mi{n}^{-1}$.
The time at which $25\%$ of the intial reactant is remaining is:
Here,
$[A{]}_0$ is the initial concentration
$\left[A\right]$ is the concentration at time t

• A. $1.38min$
• B. $2.77min$
• C. $0.693min$
• D. $0.5min$

Answer 1

The correct option is B $2.77min$

For a first order:
$\left[A\right]=[A{]}_0{e}^{-kt}$

${\ln }_e\left(\frac{[A{]}_0}{\left[A\right]}\right)=kt$
Comparing with
$y=mx$
Here slope is the rate constant.
So,
$k=0.5mi{n}^{-1}$
At,
$\left[A\right]=0.25[A{]}_0$
${\ln }_e\left(\frac{[A{]}_0}{\frac{[A{]}_0}{4}}\right)=kt$

${\ln }_{e}4=0.5\times tt=\frac{2\times 0.693}{0.5}mint=2.77min$

Theory:
Concentration of Reactant vs Time (For First Order)
$\left[A\right]=[A{]}_0{e}^{-kt}$
${\log }_{10}\left[A\right]={\log }_{10}[A{]}_0-\frac{kt}{2.303}$
${\ln }_e\left(\frac{[A{]}_0}{\left[A\right]}\right)=kt\Rightarrow {\log }_{10}\left(\frac{[A{]}_0}{\left[A\right]}\right)=\frac{kt}{2.303}$
Concentration of Reactant vs Time (For Second Order)
$\frac{1}{\left[A\right]}=kt+\frac{1}{[A{]}_0}$