Question

The reaction 3NO ---> N2O + NO2 is found to obey the rate law Rate = k[NO]^2^. If the first half-life of the reaction is found to be 2.0 s, what is the length of the fourth half-life.

Answer 1

Answer:

8.0 s.

Explanation:

Hello,

In this, case, by knowing that the half-life time allows as to compute the rate constant in terms of the initial concentration as shown below:

$k=\frac{1}{[NO]_0*t_{/2}} = \frac{1}{[NO]_0*2.0}=\frac{1}{2[NO]_0}$

Therefore, the fourth half-life time implies that the concentration is 1/16 the initial one, therefore:

$t_{1/2}^{4th}=\frac{1}{k/16}= 8.0s$

That is analyzed considering that if the initial concentration is 1M the first half-time is at 0.5M, the second at 0.25M, the third at 0.125M and the fourth at 0.0625M which is 16 times smaller than 1M.

Best regards.