Question

The rate constant of a reaction is 1.5×10⁻³ at 25°C and 2.1×10⁻² at 60°C. The activation energy is:

• A. \( \dfrac{35}{333}R\log_e\!\left(\dfrac{2.1\times10^{-2}}{1.5\times10^{-2}}\right) \)
• B. \( \dfrac{298\times333}{35}R\log_e\!\left(\dfrac{21}{1.5}\right) \)
• C. \( \dfrac{298\times333}{35}R\log_e(2.1) \)
• D. (298)/(35)Rlogₑ((1.5)/(2.1))

Hint:

Use absolute temperatures while applying Arrhenius equation.

Answer 1

The Correct Option is B

Step 1: Arrhenius equation:
\[ \ln\frac{k_2}{k_1} = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Step 2: Substitute \(T_1 = 298\,\text{K},\; T_2 = 333\,\text{K}\).
Step 3: Rearranging gives option (B).