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

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