Question

A first order reaction $A\to B$ requires activation energy of $80kJmo{l}^{-1}$. When a 20% solution of A at $27{}^{o}C$ was kept for 20 min, 50% decomposition took place. What will be the degree of decomposition in the same time for a 30% solution at $47{}^{\circ }C$?
Take $ln2=0.3,e^2=7.4,e=2.8$

• A. $1-\frac{1}{e^{0.84}}$
• B. $1-\frac{1}{e^{2.2}}$
• C. $1-\frac{1}{e^{3.3}}$
• D. $1-\frac{1}{e^{4.4}}$

Answer 1

The correct option is B $1-\frac{1}{e^{2.2}}$

$k_{1}t=ln\frac{1}{0.5}\dots \left(1\right)k_{2}t=ln\frac{1}{x}\dots \left(2\right)$
$\text{Where x is the fraction remaining = 1 - fraction decomposed}$
Dividing (2) by (1)
$\frac{k_2}{k_1}=\frac{ln\frac{1}{x}}{ln2}$
Since this is a first order reaction, the concentration does not play a role, just the ratio decomposed, so the 20% and 30% do not factor into our calculations.
$T_1=300K{T}_2=320Kln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\Rightarrow ln\frac{k_2}{k_1}=\frac{80\times {10}^3\times 3}{25}(\frac{1}{300}-\frac{1}{320})\Rightarrow ln\frac{k_2}{k_1}=2\Rightarrow \frac{k_2}{k_1}=e^2{e}^2=\frac{ln\frac{1}{x}}{ln\left(2\right)}\Rightarrow \frac{1}{x}=e^{2.2}\Rightarrow x=\frac{1}{e^{2.2}}$
$\therefore$ Degree of decomposition $=1-\frac{1}{e^{2.2}}$