Question

In Arrhenius equation Arrhenuis constant 'A' and Ea are 4*10^-13 and 19.6*10^3 if the reaction is of first order at what temperature will the half life period will be 10 times of its initial value?

Answer 1

\[\large t _{1/2}^{*}=10 ~ t _{1/2}\]
\[\large \frac{ \ln(2) }{ k _{r}^{*} }=10 \frac{ \ln(2) }{ k _{r} }\]
\[\large \frac{ \ln(2) }{ A ~ \exp \left( \frac{ -E _{a} }{ R ~ T _{2} } \right) }=10 ~ \frac{ \ln(2) }{ A ~ \exp \left( \frac{ -E _{a} }{ R ~ T _{1} } \right) }\]
Just some notes for my self.

Answer 2

Except we need information...
We need to calculate the change then? solve for Delta T?

Answer 3

if we had a start temperature (T1) we could solve for T2 and get an answer... but if assume that is not the case I see 2 options:
1) Substitute an expression for the temperature.
2) Solve for Delta T.

Answer 4

We need atleast one temperature to solve it..

Answer 5

i think we need to find answer in terms of Temperature

Answer 6

All that go though my mind right now is that:
\[\large \frac{ d(\ln[k _{r}]) }{ d(T ^{-1}) }=\frac{ -E _{a} }{ R }\]

Answer 7

I don't got any idea how to get in our A.

Answer 8

i guess you can find a temperature thats relative to the one you're at now.


t(1/2)=ln2/k

t(1/2)*k/ln2=constant

\[\frac{ 10*t _{1/2}*k _{10} }{ \ln2 }=\frac{ t _{1/2}*k _{1} }{ \ln2 }\]

\[\frac{ [10*t _{1/2}] }{ t _{1/2}}=\frac{ k _{1} }{ k _{10} } \to \frac{ k _{1} }{ k _{10} }=10\]


\[ \ln( \frac{k _{1} }{ k _{10} })=\frac{ E _{a} }{ R }(\frac{ 1 }{ T _{1} }-\frac{ 1 }{ T _{10} })\]

\[\ln (\frac{ k _{1} }{ k _{10} })*\frac{ R }{ E _{a} }=\frac{ 1 }{ T _{1} }-\frac{ 1 }{T _{10} }\]

\[\frac{ E _{a} }{ R*\ln10 }=T _{1}-T _{10}\]

this way you can get a relative temperature difference

Answer 9

That or we can write out the half life as a function of temperature and determine how long it most take for the half-life to increase 10 fold:
\[\large t _{1/2}(T)=\frac{ \ln(2) }{ A } \exp \left( \frac{ -E _{a} }{ RT } \right)\]
A good old exponential function.

Answer 10

sounds good !

Answer 11

i'm just wondering, when you subbed in k into the first order equation, did you make sure e^(-Ea/RT) was at the bottom?

\[t _{1/2}=\frac{ \ln2 }{ A*e ^{\frac{ -Ea }{ RT }} }\]

Answer 12

Hmmm woops, then rewrite to perhaps?
\[\large t _{1/2}(T)=\frac{ \log(2) }{ A } ~ \exp \left( \frac{ E _{a} }{ RT } \right)\]

Answer 13

thats works haha