Calculating the activation energy (Arrhenius equation)

2

I know this question involves the Arrhenius equation. Basically i want to work out the activation energy. We were given the results and everything, and i keep seeming to get too low a number.

I have attached the results table. If anyone can give me a worked solution i would be extremely grateful.

Table of results.doc28.2KB

Reply 1

Library

OP

2

Will rep (reputation) anyone who can help me.

Reply 2

Library

OP

2

Ill start you off.

k = Ae^-Ea/RT

Ea = activation energy
R = 8.314 J/mol/K
T = absolute temperature in Kelvins
A =constant (this constant has little or no effect to temperature, therefore we can call it 1)
K = rate coefficient

Reply 3

charco

Study Forum Helper

18

deep breath...

Rate = k[A]^x^y

if [A]^x^y is kept constant for two reactions then

Rate1/Rate2 = k1/k2

as k = Ae^-Ea/RT

then Rate1/Rate2 = (Ae^-Ea/RT1)/(Ae^-Ea/RT2)

A cancels out

Rate1/Rate2 = (e^-Ea/RT1)/(e^-Ea/RT2)

from expts 1 and 2

Rate1/Rate2 = 0.49934

thus 0.49934 = (e^-Ea/RT1)/(e^-Ea/RT2)

take logs(e)

ln 0.49934 = Ea/RT2 - Ea/RT1 = Ea/R x (1/T2-1/T1)

ln 0.49934 x 8.314 = Ea [ (T1 - T2)/T1T2]

(-0.69447 x 8.314) x T1T2/(T1-T2) = Ea

(-5.7738 x 101099) /(-10) = Ea

Ea = 58372 J

Ea = 58.372 kJ

better check my sums!!

Reply 4

Library

OP

2

charco

deep breath...

Rate = k[A]^x^y

if [A]^x^y is kept constant for two reactions then

Rate1/Rate2 = k1/k2

as k = Ae^-Ea/RT

then Rate1/Rate2 = (Ae^-Ea/RT1)/(Ae^-Ea/RT2)

A cancels out

Rate1/Rate2 = (e^-Ea/RT1)/(e^-Ea/RT2)

from expts 1 and 2

Rate1/Rate2 = 0.49934

thus 0.49934 = (e^-Ea/RT1)/(e^-Ea/RT2)

take logs(e)

ln 0.49934 = Ea/RT2 - Ea/RT1 = Ea/R x (1/T2-1/T1)

ln 0.49934 x 8.314 = Ea [ (T1 - T2)/T1T2]

(-0.69447 x 8.314) x T1T2/(T1-T2) = Ea

(-5.7738 x 101099) /(-10) = Ea

Ea = 58372 J

Ea = 58.372 kJ

better check my sums!!

Wow, i really appreciate that. I am going to have to decipher it lol.

Reply 5

Nandoz

r u sure thts how u do it. coz we r goiing to do it sum other way
log(10) = constant + Ea/2.3RT

can sum 1 work out my Ea aswell PLEASE

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Please

Activation Energy Graph for the Arrhenius Equation.doc58.4KB

Reply 6

Jonquil

10

ok, for nandoz...

take the gradient of your graph.

gradient = y/x

gradient = approx. -1250

gradient = -Ea/R R = 8.31

-1250 = -Ea/8.31

-Ea = -1250 x 8.31 = -10400 in joules per mole, I think

Ea = +10.4 kJ/mol May be wrong

Reply 7

charco

Study Forum Helper

18

Nandoz

r u sure thts how u do it. coz we r goiing to do it sum other way
log(10) = constant + Ea/2.3RT

can sum 1 work out my Ea aswell PLEASE

--------------

Please

there's more than one way to skin a cat (with apologies to the animal rights people out there - no animals were actuallly harmed while making this statement)

Reply 8

Nandoz

props for that

...do i do the same with this 1?

Book1.zip2.9KB

Reply 9

Jonquil

10

`I guess so, but why isn't the graph the other way up!? :confused:

Reply 10

Nandoz

which 1?
the 1st 1 was worng so i did it again in exel

Reply 11

Nandoz

so wot do i do with this...?
log(10) = constant + Ea/2.3RT

Reply 12

charco

Study Forum Helper

18

this equation is derived from the Arrhenius equation:

k= Ae^-Ea/RT

take logs (either natural logs or base 10)

natural ogs give you:

lnk = lnA - Ea/RT

base 10 logs give:

log(10)k = log(10)A -Ea/2.303RT

in either case you can plot graphs as they both have the form y = mx + c ( a straight line graph)

OK s for your equation, log(10)k = log(10)A -Ea/2.303RT

so, rearranging gives:

log(10)k = -Ea/2.303RT + log(10)A

(similar to y = mx + c)

a plot of log(10) k (the y values) against -1/2.303RT (the x values) gives the Ea as the gradient of the line and log(10) A as the intercept with the y axis.

your graph is probably upside down because you have your axis the wrong way round