Why does an increase in activation energy (Ea) cause a decrease in the rate constant (k)?

[Merisa15]

Advertisement - Members don't see this ad

Just watched Chad's video where he explains that high activation energy equates to a low rate constant. I don't understand why. Can someone explain?

Thank you.


Soooooo, the bigger the hill is (activation energy), the slower the reaction is to complete?

 

[deleted738762]

He told you the Arrhenius equation K= A ^Ea/RT .This relates Ea to K through an exponential function. They aren't inversely related because they aren't proportional but for simplicity's sake you can take of it as inversely proportional where if Ea gets bigger then K gets smaller. Just something you need to know.

However conceptually you can compare Ea better to the rate than the rate constant. Because if there is a bigger activation energy then the rate will decrease because it will take longer to overcome the activation energy.