Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 Â°C. Alternatively, you can simply extrapolate the straight-line plot of ln(k) vs. 1/T in your notebook to 1/298, read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717 (45 Â°C), 0.00284 (55 Â°C), 0.00492 (65 Â°C), 0.0165 (75 Â°C), 0.0396 (85 Â°C).

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = [tex]e^{-2.54738}[/tex]

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = [tex]5.61*10^-^5[/tex]