In a first order decomposition, the constant is 0.00299 sec-1. What percentage of the compound is left after 6.2 minutes?

Question

In a first order decomposition, the constant is 0.00299 sec^-1. What percentage of the compound is left after 6.2 minutes?

The correct answer is 32.9%

Please show steps how to answer.

No way that I work this can I get 32.9%. I keep getting 39.2%..

Answer 1

Hi Janel S.,

For decomposition we use the formula A(t) = A₀e^-kt, where A₀ is the initial amount of decomposed substance at t = 0, k is the constant of decay, 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^-0.00299t, [k = 0.00299, divide both sides by A₀]

A/Ao = e^-0.00299*372, [t = 6.2 min*(60 sec/ 1 min) = 372 sec]

A/A₀ = .3288 = 32.9%, (* 100, rounded)

I hope this helps, Joe.