Question 12.7: (a) Estimate the half-life for the decomposition of gaseous ......
(a) Estimate the half-life for the decomposition of gaseous N_{2}O_{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 \times 10^{-3} s^{-1})
(c) If the initial concentration of N_{2}O_{5} is 0.020 M, what is the concentration of N_{2}O_{5} after five half-lives?
(d) How long will it take for N_{2}O_{5} the concentration to fall to 12.5% of its initial value?
STRATEGY
Because the decomposition of N_{2}O_{5} is a first-order reaction (Worked Example 12.6), we can determine its half-life either from the time required for [latex]N_{2}O_{5}[/latex] to drop to one-half of its initial value or from the equation t_{1/2} = 0.693 / k To find [N_{2}O_{5}] after n half-lives, multiply its initial concentration by (1/2)^{n} \text{since} N_{2}O_{5} drops by a factor of 2 during each successive half-life.
(a) Figure 12.1 shows that the concentration of N_{2}O_{5} falls from 0.020 M to 0.010 M dur-ing a time period of approximately 400 s. At 800 s, N_{2}O_{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{0.693}{k} = \frac{0.693}{1.7 \times 10^{-3} s^{-1}} = 4.1 \times 10^{2} s (6.8 min)(c) At 5 t_{1/2} , [N_{2}O_{5}] \text{will be} (1/2)^{5} = 1/32 of its initial value. Therefore,
[N_{2}O_{5}] = \frac{0.020 M}{32} = 0.000 62 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 (4.1 \times 10^{2} s) = 1.2 \times 10^{3} s (20 min)