Self-test 7B.5A

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

Self-test 7B.5A

Calculate (a) the number of half-lives and (b) the time required for the concentration of N2O to fall to one-eighth of its initial value in a first-order decomposition at 1000K.

According to Table 7A.1, k = 0.76 1/s for this reaction.
However, I keep on getting t = ln8 / k = ln8 / 0.912 s = 2.28 seconds total for this reaction, while the key says the answer should be 2.7 seconds.

t(1/2) = 0.693 / 0.76 s^-1 = 0.912... s
Since [A] = 1/8[A]initial, rearranging the integrated rate law for first order rxns gives t = ln8 / k

Can someone tell me what I'm doing wrong?

Chem_Mod
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Re: Self-test 7B.5A

You divided ln8 by the half-life (0.912 s) instead of k (0.76 s^-1). You had the right equation but put in the wrong number. Also, you have checked that your answer was wrong because your answer has units of 1/s, which is not a unit of time. Dividing by 1/s gives you s, a unit of time. Another way to approach this problem is to calculate the half-life and see how many half-lives pass before the amount of reactant is 1/8 its original value (3 half-lives since 1/8 = (1/2)^3). This gives you 3 half-lives * 0.912 s/half-life = 2.7 s. You can do this because for a first order process, the half-life is constant (ie it doesn't depend on the concentration).

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

Re: Self-test 7B.5A

Thank you, this helped a lot!