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Chapter 14: Problem 40

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

Short Answer

Expert verified
For a generic second-order reaction, plotting \(\frac{1}{[\mathrm{A}]}\) versus time will yield a straight line. The integrated rate equation for a second-order reaction is \[\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A}]_0}=kt,\] so the slope of the straight line is the rate constant, \(k\). The half-life (\(t_{1/2}\)) of first-order reactions is constant and can be expressed as \[t_{1/2}=\frac{\ln(2)}{k},\] whereas the half-life for second-order reactions depends on the initial reactant concentration, given by \[t_{1/2} = \frac{1}{k[\mathrm{A}]_0}.\]

Step by step solution

01

Understanding Second-Order Reactions

For a second-order reaction, the rate law can be represented as: \[rate = k[\mathrm{A}]^2\] where \(k\) is the rate constant and [\(\mathrm{A}\)] is the concentration of the reactant A.
02

Integrated Rate Equation

To obtain the integrated rate equation for the second-order reaction, we need to rewrite the rate law as a differential equation, and then integrate it. The result is: \[\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A}]_0}=kt\] where \([\mathrm{A}]_0\) is the initial concentration of A at time \(t = 0\).
03

(a) Quantity to plot

For a second-order reaction, plotting \(\frac{1}{[\mathrm{A}]}\) versus time will give us a straight line. This is because the integrated rate equation shows that the relationship between \(\frac{1}{[\mathrm{A}]}\) and \(t\) is linear: \[\frac{1}{[\mathrm{A}]} = kt + \frac{1}{[\mathrm{A}]_0}\]
04

(b) Slope of the straight line

In the integrated rate equation, the slope of the straight line is given by the rate constant, \(k\). Thus: \[slope=k\]
05

(c) Half-lives of first-order and second-order reactions

The half-life (\(t_{1/2}\)) of a reaction is the time it takes for the reactant concentration to reduce by half. For a first-order reaction, the half-life is constant and can be expressed as: \[t_{1/2}=\frac{\ln(2)}{k}\] For a second-order reaction, the half-life is not constant and depends on the initial concentration of the reactant as follows: \[t_{1/2} = \frac{1}{k[\mathrm{A}]_0}\] As we can see, the half-lives of first-order reactions are independent of the initial concentration, while the half-lives of second-order reactions depend on the initial concentration of the reactant.

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Most popular questions from this chapter

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right.\) ) and overall \(\Delta E\) are \(154 \mathrm{~kJ} / \mathrm{mol}\) and \(136 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

What are the differences between an intermediate and a transition state?

The following data were measured for the reaction $\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g):$ $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{BF}_{3}\right](M)} & {\left[\mathrm{NH}_{3}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.250 & 0.250 & 0.2130 \\ 2 & 0.250 & 0.125 & 0.1065 \\ 3 & 0.200 & 0.100 & 0.0682 \\ 4 & 0.350 & 0.100 & 0.1193 \\ 5 & 0.175 & 0.100 & 0.0596 \\ \hline \end{array} $$ (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g)\) \(\longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.036 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\) (b) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is 74 torr per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{hr}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)
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