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.

K = 2.7x10⁻⁵ at 25ºC

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

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.