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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.

Respuesta :

sebassandin

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:

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

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

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

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.

onyebuchinnaji

The length of the fourth half-life of the given reaction is 8.0 s.

The reaction equation of the given elements;

  • [tex]3NO \ ---> \ N_2O \ + \ NO_2[/tex]
  • rate = [tex]K[NO]^2[/tex]
  • first half-life of the reaction, [tex]t_{1/2} = 2 .0 \ s[/tex]

The length of the fourth half-life is calculated as follows;

[tex]L = n \times t_{1/2}[/tex]

where;

  • L is the length of the time or age
  • n is the number of half-life

[tex]L = 4 \times 2\\\\L = 8.0 \ s[/tex]

Thus, the length of the fourth half-life of the given reaction is 8.0 s.

Learn more here:https://brainly.com/question/13685524

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