Half Life for First Order vs Second Order

[Curtis Tam 1J]

On the homework problems, I noticed that for first order half life problems, the solution's manual can count the number of half-lives and multiply that number by the half-life for the reaction. However, for second order problems, it seems like that same operation cannot be carried out. Instead, we must use the full equation to solve for [A]t. Conceptually, why is there such a difference?

[Sarah_Stay_1D]

On the homework problems, I noticed that for first order half life problems, the solution's manual can count the number of half-lives and multiply that number by the half-life for the reaction. However, for second order problems, it seems like that same operation cannot be carried out. Instead, we must use the full equation to solve for [A]t. Conceptually, why is there such a difference?

Hi Curtis!

As far as I know there isn't a difference conceptually. You should be able to count the number of half-lives and multiply by the half life for any order reaction. You could always try multiplying for a second order reaction and see if you get the same answer as the book.

[Jeremiah Samaniego 2C]

I believe this difference arises because for first order reactions, concentrations and rates are directly proportional. On the other hand, for second order reactions, concentrations and rates have such a relationship that if you double the concentration, you would be raising the rate by a factor of 2=2⁴.

[Christine Wastila 1H]

1st order reaction: t_1/2=0.693/k
2nd order reaction: t_1/2=1/(k[A]₀)

both can be found on the constants and equations sheet :)

[Harjas Sabharwal 1G]

The half life for first order is constant whereas the half life for second order reactions are double the previous half life. Hence, you can not apply the method to second order reactions.

[Rohan Chaudhari- 1K]

Half life of first order reactions is linear while for second order it is not