# chemistry help Rates aqa a level

Example
A reaction carried out at 25oC has a value of k = 3.3 × 10−3 mol-1 dm3 s
−1
ln A = 17.1
The gas constant R = 8.31 J K−1 mol−1
Calculate a value for the activation energy in kJ mol-1
Using Equation ln k = ln A EA/(RT)
Rearrange to EA = (ln A - ln k) x RT = (17.1 - - 5.71) x 8.31 x 298
= 56486 J mol-1
= 56.5 kJ mol-1

How is this correct ,where did the 5.71 come from
Original post by fa9089
Example
A reaction carried out at 25oC has a value of k = 3.3 × 10−3 mol-1 dm3 s
−1
ln A = 17.1
The gas constant R = 8.31 J K−1 mol−1
Calculate a value for the activation energy in kJ mol-1
Using Equation ln k = ln A EA/(RT)
Rearrange to EA = (ln A - ln k) x RT = (17.1 - - 5.71) x 8.31 x 298
= 56486 J mol-1
= 56.5 kJ mol-1

How is this correct ,where did the 5.71 come from

-5.71 is the natural log of k
Original post by charco
-5.71 is the natural log of k

what does that means?
Original post by fa9089
what does that means?

You only really need to know what a natural log is if you are doing A level maths.

Otherwise, know that it is a mathematical operation that you can find on your calculator and it is the inverse of the exponential function, e^x.

Your calculator should have a ln( ) button and if you type ln(3.3 x 10^-3), the calculator will give an answer of roughly -5.71. This number is what you use in the rearranged equation.
(edited 5 months ago)