half-lives of first order versus second order

[Shannon Wasley 2J]

When a question asks "how much time will elapse in a first order reaction for the concentration of A to decrease to 1/8[A] (not), I plug in ln(1/8) = -kt + (ln(1)) and I seem to get the right answer every time. However, if it were a second-order reaction and I tried using the same method, for example, (1/(1/8)) = kt + (1/1), I don't get the correct answer. This question might be stupid, but why is this? I might just completely not understand the concept of natural log but I am confused.

[Nicole Jacobs 1C]

for ln(1/2) and ln(2/1), they give the same value, just one is negative and one is positive, if that helps

[Juanyi Tan 2K]

I am not 100% sure, but maybe you forget to include the initial concentration for second order reaction.

[Seth_Evasco1L]

I believe you're able to use your method for the first order reaction because the half life equation doesn't involve the initial concentration of the reactant, so you could use any value for the initial concentration just as long as you have the concentration at time t=1/2 as 1/8 of that initial value.

However, for second order reactions, the half life equation does depend on the initial concentration, so the value you use in the initial concentration must be specific to the problem and cannot be generalized with [A]₀ and a concentration of 1/8 [A]₀.