After 79.0 min, 39.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics? t1/2=minutes?

Answer 1

For a 1st order reaction

FR = 0.5ⁿ

FR = fraction remaining = 1 - 0.39 = 0.61

n = number of half lives that have elapsed = ?

0.61 = 0.5ⁿ and taking log of both sides gives us

-0.2147 = -0.301 n

n = number of half lives elapsed = 0.7133 half lives

This is equal to 79.0 minutes, therefore

79.0 minutes / 0.7133 half lives = 111 minutes (3 sig. figs.)

ln[A] = -kt + ln[A]_o

Let [A]_o = 100

Then [A] = 100 - 39 = 61

t = 79 min

ln 61 = -79k + ln 100

4.1109 = -79k + 4.6052

k = 0.006257 min^-1

t_1/2 = 0.693/k = 0.693 / 0.006257

t_1/2 = 111 min (3 sig. figs.)

t_1/2 = 0.693 / k

Answer 2

A=Pe^rt is the general formula

r=rate of decay or growth, t=time

1.00-.39 = e^79r

.61 =e^79r

take natural logs of both sides

ln.61=79r

r = ln.61/79 = about -.006257

1/2=.5 =e^-.006257t where t= half life time

t = ln.5/-.006257

t = about 110.78 minutes

= about 110 miinutes 47 seconds = half life when 50% has decomposed