[Rutherford]

How is the overall activation energy of a reaction calculated, like in the problem I attached?

[Rutherford]

The solution is E=E₂+0.5(E₁-E₄). How is it obtained?

[SinkingTako]

Ah this. Firstly, have you found the overall rate equation using steady state approximation? Then you can Find the Ea rather easily.

In this case, the rate constant should be K₂K₁^0.5/K₄^0.5

[Rutherford]

Actually it's k=k₂k₁^0.5/(2k₄)^0.5
While deriving the Ea, what to do with the preexponential factors? The same relation among the preexponential factors as the one among the rate constants gives the correct answer for Ea. Why is the relation the same?

[Rutherford]

k=k₂k₁^0.5/(2k₄)^0.5 can be expanded to:
lnA - Ea/RT = lnA₂ - Ea₂/RT + 0.5lnA₁ - Ea₁/2RT - 0.5ln2 - 0.5lnA₄ + Ea₄/2RT
From this expression the correct answer can be determined if:
lnA=lnA₂ + 0.5lnA₁ - 0.5ln2 - 0.5lnA₄, i.e.:
A=A₂A₁^0.5/(2A₄)^0.5
Why is this assumption valid?

[Rutherford]

Anyone knows a way how to solve this properly?

[Big-Daddy]

have you figured out the answer to this yet?

If not, I could try to write out a good explanation. basically, collect terms

[Rutherford]

I found that Arhenius equations shouldn't be used, instead the temperature dependence of the rate constant:
d(lnk)/dT=Ea/RT². No preexponential factor here, so if this gets rearranged for Ea, keeping in mind the relation between the rate constants, it's easy onwards.

Write what you had in mind.

[Big-Daddy]

Where did you find that idea? All it proves is that the change of rate constant with temperature depends only on EA. We're not considering a rate constant changing with temperature in this question, so that equation is not useful as far as I can see.

k(overall) = A(overall) * e^(-E_A(overall)/RT)
k(overall) = k₂*k₁^1/2*(2k₄)^-1/2
Then re-express k₂=A₂*e^(-E_A(2)/RT) and the same for the others, expand all brackets and collect the multiplication terms on one side and exponential terms on the other. The exponential terms will give E_A(overall)/RT and the plainly multiplied terms will give A(overall).

(BTW I assumed your SSA expression)

[Rutherford]

You used the assumption that I asked for its validness, right?

It was in the solution of one similar problem from Mendeleev Olympiad and if it's rearranged for Ea, I don't see why it couldn't be used.

[Big-Daddy]

You used the assumption that I asked for its validness, right?

Notsure what you mean by this ...

It was in the solution of one similar problem from Mendeleev Olympiad and if it's rearranged for Ea, I don't see why it couldn't be used.

It can be used generally, it is just the exact differential form of the Arrhenius equation. The integrated form is familiar ln(k₂/k₁)= -E_A/R * (1/T₂ - 1/T₁). But temperature dependence is not the question here. How are you going to get E₁, E₂, E₄ to come out of that equation? Remember they represent activation energy for different reactions, at (I'm certain it can be assumed) the same temperature.

[Rutherford]

Rearrange the equation for Ea, then logarithm the expression connecting the rate constants, expand it and then:
d(lnk)/dT=d(lnk2)/dT+0.5d(lnk1)/dT-0.5d(ln2k₄)/dT
If both sides are multiplied by RT² you get the activation energy. I don't know to mathematically prove that is derived.

My previous assumption was lnA=lnA2 + 0.5lnA1 - 0.5ln2 - 0.5lnA4, why is it valid?

[Big-Daddy]

Rearrange the equation for Ea, then logarithm the expression connecting the rate constants, expand it and then:
d(lnk)/dT=d(lnk2)/dT+0.5d(lnk1)/dT-0.5d(ln2k₄)/dT
If both sides are multiplied by RT² you get the activation energy. I don't know to mathematically prove that is derived.

My previous assumption was lnA=lnA2 + 0.5lnA1 - 0.5ln2 - 0.5lnA4, why is it valid?

I already explained it two posts above. Your taking the derivative with respect to T removes A, that's all - it doesn't change any relationships.

[Rutherford]

Good. Thanks for clearing this.