Question

A first order reaction is 50% completed in 30 minutes at 27°Cand in 10 minutes at 47°C. Calculate the activation energy of the reaction.

• A. $46.8kJmol^{- 1}$
• B. $43.8kJmol^{- 1}$
• C. $50.8kJmol^{- 1}$
• D. $60.8KJmol^{- 1}$

Answer 1

Answer: (B)

$43.8kJmol^{- 1}$

Answer 2

Let us first calculate $k_{1}$ and $k_{2}$ at temperatures 27°C and 47°C. We know that $t_{1/2} = \frac{0.693}{k}$ or $k = \frac{0.693}{t_{1/2}}$

At 27°C, $t_{1/2} = 30\min$; $k_{1} = \frac{0.693}{30} = 0.0231$

At 47°C, $t_{1/2} = 10\min$; $k_{2} = \frac{0.693}{10} = 0.0693$

Now, $\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303R}\left\lbrack \frac{1}{T_{1}} - \frac{1}{T_{2}} \right\rbrack$; $\log\frac{0.0693}{0.0231} = \frac{E_{a}}{2.303 \times 8.314}\left\lbrack \frac{1}{300} - \frac{1}{320} \right\rbrack$

$\log 3 = \frac{E_{a}}{2.303 \times 8.314}\left\lbrack \frac{20}{300 \times 320} \right\rbrack$ $\Rightarrow 0.4771 = \frac{E_{a} \times 20}{2.303 \times 8.314 \times 300 \times 320}$

or$E_{a} = \frac{0.4771 \times 2.303 \times 8.314 \times 300 \times 320}{20}$ $= 43848Jmol^{- 1}\text{or}43.8kJmol^{- 1}$.