A reaction is followed and found to have a rate constant of 3.36 × 104 M-1s-1 at 344 K and a rate constant of 7.69 M-1s-1 at 219 K. Determine the activation energy for this reaction.
a. 12.5 kJ/mol
b. 11.5 kJ/mol
c. 42.0 kJ/mol
d. 58.2 kJ/mol
e. 23.8 kJ/mol
a. 12.5 kJ/mol
b. 11.5 kJ/mol
c. 42.0 kJ/mol
d. 58.2 kJ/mol
e. 23.8 kJ/mol
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ok so lets use this equation which relates 2 rates of reactions and their associated temperatures to the activation energy
ln (k1/k2) = (Ea/R)((1/T1-1/T2)
and subbing in what you gave me i got c. Hopefully thats the right answer
ln (k1/k2) = (Ea/R)((1/T1-1/T2)
and subbing in what you gave me i got c. Hopefully thats the right answer