The reaction,
C2H5Br --> C2H4 + HBr,
has a rate constant k = 0.0000100 s-1 at 363 °C and k = 0.000479 s-1 at 427 °C. What is the rate constant for the reaction at 725 K? Round your answer to 3 significant figures.
4 answers
I would use the Arrhenius equation along with k @ 363 C and k @ 427 C (remember to change those to Kelvin) to calculate activation energy and use that activation energy with the k @ 427C to calculate the new k @ 725 K.
so...
k=Ae^(Ea/RT)
Do I set it up as ratios? I.e.
0.0000100s-1=Ae^(-Ea/R * 636)
____________________________
0.000479s-1=Ae^(-Ea/R * 700)
A cancels out
...and then solve from there?
k=Ae^(Ea/RT)
Do I set it up as ratios? I.e.
0.0000100s-1=Ae^(-Ea/R * 636)
____________________________
0.000479s-1=Ae^(-Ea/R * 700)
A cancels out
...and then solve from there?
I would use the integrated form. I use
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
oooh, righteous! Thanks so much I appreciate the time you take out to help!
1 answer