Question

The energy of activation and specific rate constant for a first order reaction at $25^{\circ} C$ are $100\, kJ / mol$ and $3.46 \times 10^{5} \sec ^{1}$ respectively. Determine the temperature at which half life of reaction is 2 hour.
$\underset{(\text{in CCl}_4)}{2 N _{2} O _{5}} \rightarrow \underset{(\text{in CCl}_4)}{2 N _{2} O _{4}}+ O _{2}$

• A. 300 K
• B. 302 K
• C. 304 K
• D. 306 K

Answer 1

Answer: (D)

$k _{1}=3.46 \times 10^{-5}$
$T =298\, K$
$k_2 = \frac{0.0693}{2 \times 60 \times 60 }\sec^{-1} = 9.6 \times 10^{-5} \sec ^{-1}$
$T =?$
Using Arrhenius equation :
$In \frac{ k _{2}}{ k _{1}}=-\frac{ E _{ a }}{ R }\left(\frac{1}{ T _{2}}-\frac{1}{ T _{1}}\right)$
$2.303 \log \left(\frac{9.6 \times 10^{-5}}{3.46 \times 10^{-5}}\right)--\frac{100 \times 10^{3}}{8.314}\left(\frac{1}{ T }-\frac{1}{298}\right)$
$\frac{1}{298}-\frac{8.314}{10^{5}} \times 2.303(\log 2.77)=\frac{1}{ T }$
$3.355 \times 10^{-3}-8.48 \times 10^{-5}=\frac{1}{ T }$
$T =306\, K$