Question 12.7: Determining the Half-Life for a First-Order Reaction (a) Est......
Determining the Half-Life for a First-Order Reaction
(a) Estimate the half-life for the decomposition of gaseous N_2O_5 at 55 °C from the concentration-versus-time plot in Figure 12.1.
(b) Calculate the half-life from the rate constant (1.7 × 10^{-3} s^{-1} ).
(c) If the initial concentration of N_2O_5 is 0.020 M, what is the concentration of N_2O_5after five half-lives?
(d) How long will it take for the N_2O_5 concentration to fall to 12.5% of its initial value?
STRATEGY
Because the decomposition of N_2O_5is a first-order reaction (Worked Example 12.6), we can determine its half-life either from the time required for [ N_2O_5 ] to drop to 1/2 of its initial value or from the equation t_{1/2} = 0.693/k. To find [ N_2O_5 ] after n half-lives, multiply its initial concentration by (1/2) n because [ N_2O_5 ] drops by a factor of 2 during each successive half-life.
(a) Figure 12.1 shows that the concentration of N_2O_5 falls from 0.020 M to 0.010 M during a time period of approximately 400 s. At 800 s, [ N_2O_5 ] has decreased by another factor of 2, to 0.0050 M. Therefore, t_{1/2} \ \approx 400 s
(b) Based on the value of the rate constant,
t_{1 / 2}=\frac{\ln 2}{k}=\frac{0.693}{1.7 \times 10^{-3} \mathrm{~s}^{-1}}=4.1 \times 10^2 \mathrm{~s}(6.8 \mathrm{~min})(c) At 5t_{1/2}, [ N_2O_5 ] will be (1/2)^5 = 1/32 of its initial value. Therefore,
\left[\mathrm{N}_2 \mathrm{O}_5\right]=\frac{0.020 \mathrm{M}}{32}=0.00062 \ \mathrm{M}(d) Since 12.5% of the initial concentration corresponds to 1/8 or (1/2)^3 of the initial concentration, the time required is three half-lives:
t=3 t_{1 / 2}=3\left(4.1 \times 10^2 \mathrm{~s}\right)=1.2 \times 10^3 \mathrm{~s} \ (20 \mathrm{~min})