Problem 1: Calculate Ea from Two k Values

Data: k_{1} = $2.5 \times 10^{-3}$ $s^{-1}$ at 300 K; k_{2} = $5.0 \times 10^{-3}$ $s^{-1}$ at 310 K

Step 1: Write the two-temperature Arrhenius equation: $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Step 2: Calculate left side: $\log\frac{5.0 \times 10^{-3}}{2.5 \times 10^{-3}} = \log 2 = 0.301$

Step 3: Calculate 1/$T_{1}$ − 1/$T_{2}$: $\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = 1.075 \times 10^{-4} \text{ K}^{-1}$

Step 4: Solve for Ea: $0.301 = \frac{E_a}{19.147} \times 1.075 \times 10^{-4}$

$E_a = \frac{0.301 \times 19.147}{1.075 \times 10^{-4}} = \frac{5.763}{1.075 \times 10^{-4}} = 53{,}612 \text{ J/mol} \approx 53.6 \text{ kJ/mol}$

Problem 2: Find k at New Temperature

Given: k = $2 \times 10^{-3}$ $s^{-1}$ at 300 K; Ea = 50 kJ/mol. Find k at 350 K.

$\log\frac{k_{350}}{2 \times 10^{-3}} = \frac{50000}{19.147}\left(\frac{1}{300} - \frac{1}{350}\right)$

$= 2611.5 \times \frac{50}{105000} = 2611.5 \times 4.762 \times 10^{-4} = 1.244$

$\frac{k_{350}}{2 \times 10^{-3}} = 10^{1.244} = 17.53$

$k_{350} = 17.53 \times 2 \times 10^{-3} = 3.51 \times 10^{-2} \text{ s}^{-1}$

Problem 3: Calculate A from Ea and k

Given: k = $1.5 \times 10^{-3}$ $s^{-1}$ at 500 K; Ea = 80 kJ/mol.

$$A = k \cdot $e^{E_a/RT}$ = 1.5 \times 10^{-3} \times $e^{80000/(8.314 \times 500)}$$$

= 1.5 \times 10^{-3} \times $e^{19.25}$ = 1.5 \times 10^{-3} \times 2.28 \times 10^8 = 3.42 \times 10^5 \text{ s}^{-1}