Question

At a certain temperature, the half-life of the first order decomposition of iso-propanol (shown below) is 4.30 hr.



iso-propanol   → propanone + dihydrogen



Answer the following questions about the decomposition of iso-propanol and report all answers to three significant figures.



1. If the initial concentration of iso-propanol is 5.16×10^-3 M, calculate the time (in hr) required for the concentration of iso-propanol to decrease to 57.8 % of the initial concentration.




2. If the initial concentration of iso-propanol is 5.16×10^-3 M, calculate the concentration (in M) after 3.87 hr.

Answer 1

1. If the initial concentration of iso-propanol is 5.16×10^-3 M, calculate the time (in hr) required for the concentration of iso-propanol to decrease to 57.8 % of the initial concentration.

if first order, then

Half life = ln(2) / K

k = ln(2) / 4.3 = 0.16119 1/h

now

ln(A) = ln(A0) - kt

substitute data

ln(A/A0) = -kt

ln(0.578) = -0.16119*t

t = ln(0.578) /-0.16119

t = 3.4 hr

2. If the initial concentration of iso-propanol is 5.16×10^-3 M, calculate the concentration (in M) after 3.87 hr.

once again:

ln(A) = ln(A0) - kt

ln(A) = ln(5.16*10^-3) - 0.16119*3.87

A = exp(-5.89062) = 0.0027652M = 2.76*10^-3 M