Sapling #11 week 9/10

[luludaly2B]

Could someone help me with this problem? I'm not really sure where to start. Thanks!

Dinitrogen pentoxide, N2O5, decomposes by first‑order kinetics with a rate constant of 3.7×10−5 s−1 at 298 K.
What is the half‑life, in hours, of N2O5 at 298 K?

[Angelica Soriano 3L]

Hi, so we know that this reaction is a first-order reaction so we can use the equation:
t1/2 = 0.693/k

Now, we need to input the rate constant, k, to find the half life (t1/2). Once you find the half-life, the units will be in seconds so we have to convert it to hours.

Hope this helps!

[Arya Adibi 1K]

The problem tells you the reaction is first order. So you use the first order half life equation to find the time, and convert it to hours.

[LarisaAssadourian2K]

I agree with the previous posts! Since the question states that the reaction is first order, you would plug it into t_1/2= ln2/k, where k is 3.7×10−5 s−1. Be sure to convert to hours after you find your answer!

[MichaelRaad_1F]

We use the equations t1/2=0.0693/k and then use the equation ln[A]-ln[A]0/-k=t to solve this question.

[David Liu 1E]

i agree with above statements! It states in the question that it's a first order reaction, so we take the equation for a half life first order reaction.