Question

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

• A. % decomposition $=67.21$%
• B. % decomposition $=33.5$%
• C. % decomposition $=50$%
• D. None of these

Answer 1

The correct option is A % decomposition $=67.21$%

$A\to B$$E_a=70kJ/mole$$t_{1/2}=40min$
$k=\left(\frac{0.693}{40min}\right)$$k=2.8875\times {10}^{-4}$
$lo{g}_{10}\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$=\frac{70\times {10}^3}{2.303\times 8.314}\left[\frac{15}{298\times 313}\right]$
$\frac{k_2}{k_1}=5.88\times {10}^{-4}\times {10}^3={10}^{0.588}$
$\frac{k_2}{k_1}=3.87$
$\frac{k_2}{k_1}=\frac{log(a/a-x)}{lo{g}_{10}\left(a/3ka\right)}$
$3.87=\frac{log(a/a-x)}{log\left(4/3\right)}$
${10}^{.4835}=\left(\frac{a}{a-x}\right)$
$\left(\frac{a}{a-x}\right)={10}^{-.4835}$$\left(\frac{a}{a-x}\right)\times 100=32.84$
$\Rightarrow$% decomposition $=100-32.84=67.16$