After 43.0 min, 15.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?

Answer 1

Whenever you have a first order decay or first order reaction, a very simple formula to use is...

Fraction remaining 0.5ⁿ where n = # of half lives elapsed. To me, at least, this is much easier than using the more commonly used integrated rate law. So, in this problem, the fraction remaining is 0.85 because it says 15% (0.15) has decomposed. So, we need to solve for n, the # of half lives elapsed.

0.85 = 0.5ⁿ

log 0.85 = n log 0.5

n = log 0.85/log. 0.5

n = 0.234 so this means that only 0.234 half lives have elapsed.

We know that this is equal to 43.0 minutes, so we can now easily find the half life.

43.0 min / 0.234 half lives = 184 min / half life

So, t_1/2 = 184 minutes

Answer 2

first-order rate equation = k * [A]^1

reads as rate constant (k) times the concentration raises to the first power

rate = k * [A]

the integrated rate law for a first-order reaction is

,

where [A] is the concentration at time t and [A]₀ is the initial concentration at zero time (the start)

The first-order rate law is confirmed if  is in fact a linear function of time

Notice that the slope of the linear line is negative and is the rate constant

the given information is t = 43.0 min

let [A]₀ = 100 following percent indicated

let [A] = 85 since 15% is gone at time T

use a calculator to take the natural log of both and write them down

ln [A]₀ = __________

ln [A] = __________

plug them and the time into the equation

do a bit of algebra to isolate and solve for k ... the rate constant which you'll need to solve for half-life

The first-order half-life is independent of the starting concentration and is given by 

use a calculator to take the natural log of 2 and write it down

ln 2 = __________

then divide by the rate constant you determined and you've solved for the half-life of the reaction