Question

The following data show the rate constant of a reaction measured at several different temperatures.
Temperature (K) Rate Constant (1/s)
300 4.04×10^-2
310 0.121
320 0.340
330 0.895
340 2.23
Use an Arrhenius plot to determine the activation barrier for the reaction.

Answer 1

K = A x e^(-Ea/RT)

where K = rate constant
A = freq. factor
Ea = activation energy
R = gas constant.

and you have values of K and T.....

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so what happens if you do this...

take logs of both sides...
ln(K) = ln ( A x e^(-Ea/RT) )

and since ln(axb) = lna + lnb...
ln(K) = ln(A) + ln(e^(-Ea/RT))

and since ln(e^(a)) = a...
ln(K) = ln(A) + -Ea/RT
ln(K) = -Ea/RT + ln(A)

and that looks similar to this...
y = mx + b...right?

where
y = ln(K)
-Ea/R = m
(1/T) = x
ln(A) = b..


so if we plot ln(K) on the y axis and (1/T) on the X axis we will get a line with slope = -Ea/R and intercept = ln(A).

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easiest way.. if you excel available.. make 4 columns

Temp.. K... 1/T... ln(K)
where the 1/T and ln(K) are calculated based on T and K...ok?

and then
1) highlight the 1/T and ln(K) columns
2) insert a scatter plot.. pick the one with just points.. no lines
3) right click on a point on the plot
4) add trendline.. make sure the box for set intercept to zero is UNchecked. and the display r^2 and equation boxes are checked.
5) click the close button and tada.. you have your equation..

Answer 2

Activation energy:According to the concept of activation energy, all the molecules cannot take part in the chemical reaction. it is only a certain number of moleculesthat may be called active molecules which could take part in a chemical reaction. thus, the reactants do not pass directly to the products but mustfirst acquire necessary energy to pass over an energy barrier known as the activated state.The amount of energy which the reactants must absorb to pass over this activated energy barrier is known as the actvation energy.

Answer 3

The given data is :

Rate constant at different temperatures has given:

The above Rate constant values and the temperature can be plotted in a graph which can't be represented here.. Theprocedure can be explained by the following way:

In the graph on the Y-axis we have to take log K values and on the X-axis we need to take 1/T in kelvin

the graph we obtain as straight line with negative slope. The slope for a straight line can be calculated as:

slope = Y2-Y1 / X2-X1

from the given data if we assume any two set of values and calculate their slope we obtain the slope of

a straight line is:

Slope = -4764.1

According to Arrhenius equation:

Slope of the graph = -Ea / 2.303 R

-4764.1Kx 2.303x 8.314x10^-3kJ/mol-K = -E_a

E_a= 91.218 kJ/mol