Question

Rate constants of a reaction at 500 K and 700 K are \(0.04 \, \text{s}^{-1}\) and \(0.14 \, \text{s}^{-1}\), respectively. Then, the activation energy of the reaction is:
{Given:} \[ \log 3.5 = 0.5441, \, R = 8.31 \, \text{J K}^{-1} \, \text{mol}^{-1} \]

• A. 182310 J
• B. 18500 J
• C. 18219 J
• D. 18030 J

Answer 1

The Correct Option is C

We can use the Arrhenius equation to solve this problem. The two-point form of the Arrhenius equation is:
\[ \ln{\frac{k_2}{k_1}} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
where:
$k_1$ and $k_2$ are the rate constants at temperatures $T_1$ and $T_2$, respectively. 
$E_a$ is the activation energy. 
R is the ideal gas constant.
Given: $k_1 = 0.04$ s$^{-1}$ 
$k_2 = 0.14$ s$^{-1}$ 
$T_1 = 500$ K 
$T_2 = 700$ K 
R = 8.31 J K$^{-1}$mol$^{-1}$ 
$\log 3.5 = 0.5441$ which means $\ln 3.5 = 2.303 \times 0.5441 = 1.253$
Plugging the values into the equation:
\(\ln{\frac{0.14}{0.04}} = \frac{E_a}{8.31} \left( \frac{1}{500} - \frac{1}{700} \right)\)

\(\ln{3.5} = \frac{E_a}{8.31} \left( \frac{700 - 500}{500 \times 700} \right)\)

\(1.253 = \frac{E_a}{8.31} \left( \frac{200}{350000} \right)\)

 
\(1.253 \times 8.31 \times 350000 = E_a\)

\(E_a = \frac{1.253 \times 8.31 \times 350000}{200}\)

\(E_a = 18219  J\)