Solving Arrhenius Eqn: Double Reaction Rate at 10°C

[fk378]

Homework Statement

A common rule of thumb is that temperature 10 degrees C for many reactions will double the reaction rate. Derive under what conditions this is true.

Homework Equations

ln(k) - ln(2k) = (-Ea/R) [(1/T1)-(1/T2)]

The Attempt at a Solution

I know I have to start off by plugging in values for T1 and T2, but what exactly am I trying to solve for? My teacher said to solve for Ea, but that is the measure of kinetic energy needed for reaction...what do I solve for when looking for reaction rate?

Also, can I leave the rate constants in or should I plug in actual values for those as well?

 

[kuahji]

I had the exact same question on a quiz, but instead we were given the room temperature. Indeed you do want to solve for Ea. You'll want to plug in values for ln k & temp. Think arbitrary values for what you want to plug in for the numbers, as long as they all relate to each other.

 

[fk378]

But what value of Ea am I looking for? If I'm plugging in all my values into that equation that I typed up there, of course I'll just get a value for Ea.
And why am I solving for Ea anyway? How will I know what the reaction rate is if I know Ea?

 

[rocomath]

The equation you are using is incorrect.

http://alt1.mathlinks.ro/Forum/latexrender/pictures/1/b/a/1bacaba62a260d46b7600cc644b6bba500f54b3b.gif

 

[fk378]

I'm using the same equation as that...I'm just changing the locations of the corresponding k1/T1 and k2/T2 values.

 

[chemisttree]

Nice derivation, Roco. Rewrite the expression and solve for Ea. Use "T1+10" in place of "T2".