What's your trouble with this?
b.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
ln(k2/0.0160) = (32900/0.08206)(1/298/15 - 1/403.15)
Solve for k2.
The activation energy of a certain reaction is 32.9 kJ/mol At 25 degrees C the rate constant is 0.0160 units s-1 units. At what temperature in degrees Celsius would this reaction go twice as fast?
Given that the initial rate constant is 0.0160 s-1 at an initial temperature of 25 degrees C , what would the rate constant be at a temperature of 130 degrees C for the same reaction described in Part A?
2 answers
Roses are red,
Violets are blue,
I'm terrible at Chemistry
And apparently, you are too <3
Violets are blue,
I'm terrible at Chemistry
And apparently, you are too <3