Calculate the activation energy Ea in kilojoules per mole?
a). Calculate the activation energy Ea in kilojoules per mole.
b). Calculate the rate constant at 75°C.
PS: plz help me solve this problem step by step plz... thanks so much in advance
Answer:
Using the Arrhenius equation,
k=Ae^(-Ea/RT),
where A is collision frequency
Ea is activation energy
R is a constant equal to 8.314 J/K *mol
and T is temparature in Kelvin
which becomes a linear/slope equation when its natural logarithm is taken, yielding
ln k= ln A- Ea/RT, which when rearranged,
ln k= (-Ea/R)(1/T)+ln A (Equation1)
We plot this into a graph with the values:
2.76X10^ -5 then -10.497 then 298 then 0.00335
6.65 X10^-4 then -7.315 then 323 then 0.00309
NOTE: First row, k
Second row, ln k
3rd row: Temp
4th row: Reciprocal
You can get the slope, with 1/T at the x axis...
slope= -10.497-(-7.315)/(0.00335-0.00...
You get: -12238.461 K
multiply by R and you get: -101 750.569 J/mol
Divide this by 100 and you get the kJ/mol value...
Problem 2:
Get the ln of A by substituting the value of the slope you found earlier (-12238.461 K), and the values of 1/T and k of one of the conditions into the linear equation (Equation 1).
When you already know ln A, you can subsitute this value to the equation, together with 75 degree C(348K), with the slope, which will leave k as the unknown...
ln k= (slope)(1/348)+ln A
You have to try this one out yourself... Finding the answer yourself can be more satisfying than what you might think...
Hope that helped.
A 1st order reaction has a rate equation as
-d/dt of the conc. of the reactant=rate constant*the conc .of the reaction.Therefore on solving we get
rateconstant = 2.303/t*logCo/C ,here t=time taken for the conversion of Co to C.Again from Arrhenius equation we have
K=Ae^-Eactivation/RT ,wereT=temperature.therefore taking log we get logK=logA-Eactivation/2.303RT. we have
on subtracting the 2 equationsLog(k1/k2)=Eactivatio... we get the activation energy..For the 2nd part K2is unknown but the activation energy is known and the temperatures are known.so u get the answere.
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