Question 14.4.6: Determine half-life, order, and k from concentration–time da......
Determine half-life, order, and k from concentration–time data.
The following data were collected for the decomposition of sulfuryl chloride at 383 ºC.
SO_{2}Cl_{2} → SO_{2} + Cl_{2}
a. What is the half-life of the reaction when [SO_{2}Cl_{2}] = 5.18 × 10^{-3} M?
b. What is the half-life of the reaction when [SO_{2}Cl_{2}] = 2.59 × 10^{-3} M?
c. Is the reaction first order? If so, what is the rate constant for this reaction?
| \pmb{[SO_{2}Cl_{2}]} ( M) | Time (min) |
| 5.18 × 10^{-3} | 0 |
| 2.59 × 10^{-3} | 166 |
| 1.30 × 10^{-3} | 332 |
You are asked to calculate the half-life of the reaction at different reactant concentrations, to determine whether the reaction is first order, and to calculate the rate constant for the reaction.
You are given the balanced equation for the reaction and the concentration of the reactant as a function of time.
a. The concentration of SO_{2}Cl_{2} is halved (5.18 × 10^{-3} M to 2.59 × 10^{-3} M) over the first 166 minutes of the reaction. Therefore, the half-life when [SO_{2}Cl_{2}] = 5.18 × 10^{-3} M is 166 minutes.
b. The concentration of SO_{2}Cl_{2} is halved again (2.59 × 10^{-3} M to 1.30 × 10^{-3} M) when another 166 minutes pass (166 min to 332 min). The half-life when [SO_{2}Cl_{2}] = 2.59 × 10^{-3} M is 166 minutes.
c. Because the half-life is independent of reactant concentration, the reaction is first order in SO_{2}Cl_{2}. The first-order rate constant is calculated using the equation in Table 14.4.3.
t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}
k = \frac{0.693}{t_{1/2}} = \frac{0.693}{166\text{ min}} = 4.17 × 10^{-3} min^{-1}
Table 14.4.3 Half-Life Equations for Reactions of Type A → S Products
| Reaction Order | Integrated Rate Law |
Substitute \pmb{t = t_{1/2}} and \pmb{[A]_{t} = ½[A]_{0}} |
Half-Life Equation |
| Zero order | [A]_{t }= [A]_{0} – kt | ½[A]_{0} = [A]_{0} – k(t_{1/2}) | t_{1/2} =\frac{[A]_{0}}{2k} |
| First order | \ln\frac{[A]_{t}}{[A]_{0}}= -kt | \ln(½[A]_{0}) = \ln[A]_{0} – kt_{1/2} | t_{1/2} =\frac{\ln 2}{k}=\frac{0.693}{k} |
| Second order | \frac{1}{[A]_{t}}=\frac{1}{[A]_{0}}+ kt | \frac{1}{(1/2[A]_{0})}= \frac{1}{[A]_{0}}+k(t_{1/2}) | t_{1/2} =\frac{1}{k[A]_{0}} |