Calculating activation energy?

Question

the initial stage of gaseous ammonia and nitrogen dioxide follows second order kinetics. Given that the rate constant at 327degrees celcius is 3.85 x 10^2 & at 443 degrees celcius is 1.6 x 10^4.

i know that it should start off like this ....

log10(1.6 x 10^4/3.85 x 10^2) = (Ea / 2.303 x 8.314) x (716 - 600 / 600 x 716)

where Ea is activation energy in joules.

i get stuck from there ^^ but in the book the answer says 113.17kJ mol^-1

temp is in kelvin ( 273 to degrees celcius)...

any help would be GREATLY APPRECIATED!! :)i have several of these problems to do and im not sure where to start :/

Answer 1

The Arrhenius equation relates the rate constant, temperature and activation energy.

k=Ae^-(Ea/RT)

The problem is the collision factor (frequency factor) A.  This number is not easy to determine or even predict.  Therefore, by determining the rate constant at two different temperatures it is possible to combine the data and eliminate A.

It can be rewritten

ln k = -Ea/RT + lnA

By plotting the logs of a number of rate constants (k) at ehe reciprocals of their corresponding temperatures (T) it is possible to determine Ea from the slope.

Or, for two sets of data we can combine the equations by subtracting to eliminate A.

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

or

ln (k1/k2) = Ea/R x (1/T2 - 1/T1)

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

Now, plug in your numbers

Ea = ( 8.314 J/molK) ln(3.85 x 10^2 / 1.6 x 10^4) / (1/716K - 1/600K)

Ea = 114.8 kJ/mol

The difference between 114.8 and 113.2 is probably due to some round-off error somewhere, most likely in the conversion between natural and common logs.  With modern calculators, there is no reason to convert to common logs, just use natural logs.

(Before calculators, common logs were easier to use from tables and on slide rules.)

Answer 2

Your formula is correct, but why do you use common logarithm? It just makes the things more complicated.

Formulate rate constants using Arrhenius equation:

k₁ = A·exp{- Ea/(R·T₁)} at T₁=327°C = 600K

k₂ = A·exp{- Ea/(R·T₂)} at T₂=443°C = 760K

Then take their ratio::

k₂/k₁ = A·exp{- Ea/(R·T₂)} / A·exp{- Ea/(R·T₁}

<=>

k₂/k₁ = exp{ (Ea/R)·(1/T₁ - 1/T₂) }

<=>

ln(k₂/k₁) =  (Ea/R)·(1/T₁ - 1/T₂)  

Hence:

Ea = R · ln(k₂/k₁) / (1/T₁ - 1/T₂)  

= 8.314472J/molK · ln(1.6×10⁴ / 3.85×10²) / ( 1/600K - 1/760K)

= 114759 J/mol

= 114.76kJ/mol

Close to the result of the book, deviation can be explained by some round-off error.