Question

Thermal decomposition of a compound follows first order kinetics. If $50$% of the compound decomposes in $120$ minutes. How much compound will decompose in $360min$?

• A. $12.5\%$
• B. $87.5\%$
• C. $25\%$
• D. $75\%$

Answer 1

The correct option is B $87.5\%$

For a first order reaction,
$k=\frac{0.693}{t_{\frac{1}{2}}}=\frac{\ln 2}{120}mi{n}^{-1}$
Since for first order,
$\left[A\right]=[A{]}_0{e}^{-kt}$

$\ln \left(\frac{[A{]}_0}{\left[A\right]}\right)=kt$

$\ln \left(\frac{[A{]}_0}{\left[A\right]}\right)=\frac{\ln 2}{120}\times 360$

$\ln \left(\frac{[A{]}_0}{\left[A\right]}\right)=3\ln 2$

​​​​​​​$\ln \left(\frac{[A{]}_0}{\left[A\right]}\right)=\ln 2^3$

$\left(\frac{[A{]}_0}{\left[A\right]}\right)=8$

$\left[A\right]=\frac{[A{]}_0}{8}$
$\%\text{Decomposed}=\frac{[A{]}_0-[A]}{[A{]}_0}\times 100\%$

$\%\text{Decomposed}=\frac{[A{]}_0-\frac{[A{]}_0}{8}}{[A{]}_0}\times 100\%$

$\%\text{Decomposed}=87.5\%$