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In a first order decomposition, the constant is 0.00586 sec-1. what percentage of the compound is left after 3.32 minutes?

Sagot :

The correct answer for percentage of the compound is left after 3.32 minutes is 55 % .

We use the formula A(t) = A0e-kt for decomposition, where A0 is the initial amount of decomposed substance at t = 0, k is the decay constant, t is the time elapsed, and A(t) is the amount after time t.

The percentage of decomposed compound left after a time t is, [A(t)], divided the the initial amount, [A0], time 100, or A/A0*100:

A = A0e-kt

A/A0 = e^- 0.00586t, [k = 0.00299, divide both sides by A0]

A/Ao = e^(-0.00299*199), [t = 3.32 min*(60 sec/ 1 min) =199 sec]

A/A0 = . 0.551= 55%, (* 100, rounded)

The Arrhenius equation is k = Ae(-Ea/kT), where A is the frequency or pre-exponential factor and e(-Ea/RT) is the fraction of collisions at temperature T that have enough energy to overcome the activation barrier (i.e., have energy greater than or equal to the activation energy Ea).

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