Respuesta :

Answer:

69.8%

Explanation:

k = 0.00813 sec-1

t = 44.1 s

What percentage of the compound has decomposed?

We can obtain this using the formular;

[A] = [A]o e^(-kt)

where;

Final Concentration = [A]

Initial concentration = [A]o

[A] / [A]o = e^(-kt)

[A] / [A]o  =  e^(-0.00813*44.1)

[A] / [A]o = e^(-0.359)

[A] / [A]o = 0.698

Percentage decomposed = Final Concentration / Initial concentration * 100

Percentage decomposed = [A] / [A]o   * 100%

Percentage decomposed = 0.698 * 100 = 69.8%

mickymike92

Based  on the data given,  the percentage decomposed in the first order reaction is 69.8%

What percentage of the compound has decomposed after 44.1  seconds?

From the data provided;

rate constant, k = 0.00813 sec-1

time, t = 44.1 s

Since the reaction is first order, we calculate using the formula below:

  • [A] = [A]₀ e^(-kt)

where;

  • [A] is final concentration
  • [A]₀ is initial concentration

Finding the ratio of final and initial concentration

[A]/[A]₀ = e^(-kt)

Substituting the values:

[A]/[A]₀ =  e^(-0.00813 * 44.1)

[A]/[A]₀ = e^(-0.359)

[A]/[A]₀ = 0.698

Percentage decomposed = [A]/[A]₀ * 100%

Percentage decomposed = 0.698 * 100%

Percentage decomposed = 69.8%

Therefore, the percentage decomposed in the first order reaction is 69.8%.

Learn more about percentage decomposition at: https://brainly.com/question/1563644

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