Question

Scoring Scheme: 3-3-2-1 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. k (25 °C) = my slope is: -16538 my k values are: 0.00057 0.0017 0.00525 0.0238 0.1386 my activation energy is: 137.5 and my temperatures are: 45, 55, 65, 75, and 80 in celsius. the order of temp and k values correspond to each other.

Answer 1

K = 2.7x10⁻⁵ at 25ºC

Step-by-step explanation:

A way to write Arrhenius equation is:

ln K = - Ea/R × (1/T) + lnA

If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:

Y = -13815X +35.817

R² = 0.9927

(Taking the last k point as 0.0386) (ln 0.0386), 0.1386 has no sense)

Your slope is -13815

-13815K = - Ea/R

-13815K×8.314J/molK = 114858J/mol = Ea

And your intercept =

lnA = 35.817

A = 3.59x10¹⁵

Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):

Y = -13815X +35.817

Y = -13815(1/298.15K) +35.817

Y = -10.5187

lnK = -10.5187

K = 2.7x10⁻⁵ at 25ºC