Exercises
Exercise 1. The following reaction
C2H4 (g) + O3 (g) → C2H4O (g) + O2 (g)
has the time-concentration data in the table below. Use the information to answer the questions.
What is the value of the rate constant?
What is the half-life of O3?
Plot the data as [O3] vs. time, ln[O3] vs. time, and 1/[O3] vs. time. The data and plots are below.
From the plots, we see the order with respect to O3 is second order. The value of k is the slope of the regression line and is 1.5 x 103.
The half-life is found using the second order equation.
\(\displaystyle t_{1/2}\;=\;\frac{1}{k[A]_0}\;=\;\frac{1}{(1.5\times 10^3\;M^{-1}s^{-1})\times\;(2.13\times 10^{-5})}\;=\;\mathbf{31.3\;s}\)
Back to Determine Rate Law from Plot of Data
Exercise 2. Consider the reaction and the concentration/time data below to answer the questions.
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Plot the data as [N2O5] vs. time, ln[N2O5] vs. time, and 1/[N2O5] vs. time. The data and plots are below.
From the plots, we see the reaction is first order.
b) What is the value of the rate constant with units?
The value of the rate constant, from the regression equation, is 1.1 x 10-3 s-1.
c) Write the rate law for the reaction
Rate = 1.1 x 10-3 s-1[N2O5]
d) What is the concentration of N2O5 after 535 s?
Use the first order integrated rate law and solve for [A]t.
\(\displaystyle ln\frac{[A]_t}{[A]_0}\;=\;-kt\)
[A]0 = 0.0300 M, k = 1.1 x 10-3, t = 535 s
\(\displaystyle ln[A]_t\;=\;-1.1\times 10^{-3}s^{-1}\times\;535\;s\;+\;ln[0.0300\;M]\;=\;-4.095\)
Take the antilog.
eln[A]t = e-4.095
[A]t = 0.0167 M
Back to Determine Rate Law from Plot of Data
Exercise 3. Consider the following reaction and concentration/time data obtained from the lab to answer the questions.
2 NOBr (g) → 2 NO (g) + Br2 (g)
Plot the data as [NOBr] vs. time, ln[NOBr] vs. time, and 1/[NOBr] vs. time. The data and plots are below.
From the plots, the order with respect to [NOBr] is first order.
b) What is the value of the rate constant with units?
1.69 s-1
c) Write the rate law for the reaction
Rate = 1.69 s-1[NOBr]
d) What is the half-life of NOBr?
\(\displaystyle t_{1/2}\;=\;\frac{0.693}{1.69\;s^{-1}}\;=\;0.41\;s\)
e) How long, in seconds, will it take for the [NOBr] to decrease from 0.0400 M to 0.0165 M?
\(\displaystyle t\;=\;-\frac{ln\biggl(\frac{[NOBr]_t}{[NOBr]_0}\biggr)}{k}\;=\;-\frac{ln\biggl(\frac{[0.0165\;M}{[0.0400\;M]_0}\biggr)}{1.69\;s^{-1}}\;=0.52\;s\)