Question

First order reaction, $(A\to B)$ requires activation of 70 kJ $mo{l}^{-1}$. When a 20% solution of A was kept at ${25}^{o}C$ for 20 minutes, 25% decomposition took place. Assume that activation energy remains constant in this range of temperature. The percent decomposition at the same time in a 30% solution maintained at ${40}^{o}C$ will be:

• A. 67.2%
• B. 71.4%
• C. 69.3%
• D. None of these

Answer 1

Answer: (A)

67.2%

For the change, $A\to B$
20% solution of A decomposes 25% in 20 minutes at ${25}^{o}C$.
Let amount of solute undergoing decay $=a=20$
$\therefore$ amount of solute left (a-X) after 20 minutes
$=20\times \left(75/100\right)=15$
$\therefore K_{25}=\frac{2.303}{t}log\frac{a}{a-X}=\frac{2.303}{20}log\frac{20}{15}$
$=1.44\times {10}^{-2}minut{e}^{-1}$
$\because 2.303log\frac{K_{40}}{K_{25}}=\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1\cdot T_2}\right]$
$\therefore 2.303log\frac{K_{40}}{0.0144}=\frac{70\times {10}^3}{8.314}\left[\frac{313-298}{313\times 298}\right]$
$\therefore K_{40}=5.58\times {10}^{-2}minut{e}^{-1}$
The reaction of carried out be 30% solution shows the left amount $(a-X)=m$ in 20 minute
$\because K_{40}=\frac{2.303}{t}log\frac{a}{(a-X)}$
$5.58\times {10}^{-2}=\frac{2.303}{20}log\frac{30}{m}$
$\therefore m=9.83$
$\therefore$ % decomposition $=\left[\right(a-m\left)/a\right]\times 100$
$=\left[\right(30-9.83\left)/30\right]\times 100=67.2$%.