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Chapter 13: Problem 21

Write an equation relating the concentration of a reactant \(\mathrm{A}\) at \(t=0\) to that at \(t=t\) for a first-order reaction. Define all the terms and give their units. Do the same for a second-order reaction.

Short Answer

Expert verified
For a first-order reaction, the concentration of reactant A at any time \(t\) is given by \([A]=[A]_0e^{-kt}\), and for a second-order reaction, it is given by \( [A]=\frac{[A]_0}{1+[A]_0kt}\). The terms \([A]\) and \([A]_0\) represent the concentrations of A at time \(t\) and \(t=0\) respectively, in units of molarity (\(\mathrm{M}\)). The term \(k\) represents the rate constant, in units of \(\mathrm{s}^{-1}\) for first-order and \(\mathrm{M}^{-1}\mathrm{s}^{-1}\) for second-order reactions. \(t\) represents the time, in seconds.

Step by step solution

01

First-order reaction

Let's denote the concentration of reactant A at time \(t=0\) as \([A]_0\), and the concentration at time \(t=t\) as \([A]\). For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant A. This can be expressed mathematically as: \( -\frac{d[A]}{dt}=k[A] \), where \(k\) is the rate constant (with units of \(\mathrm{s}^{-1}\)). Integrating this equation from \(t=0\) to \(t=t\), we get: \(ln[A]-ln[A]_0=-kt\). Rearranging this, we get the desired equation: \([A]=[A]_0e^{-kt}\), where \([A]\) and \([A]_0\) have units of molarity (\(\mathrm{M}\)), \(k\) has units of \(\mathrm{s}^{-1}\), \(t\) has units of time (\(s\)), and \(e\) is Euler's number.
02

Second-order reaction

Now let's look at a second-order reaction. For such a reaction, the rate of reaction is proportional to the square of the concentration of the reactant A, which can be expressed as: \( -\frac{d[A]}{dt}=k[A]^2 \). The rate constant \(k\) now has units of \(\mathrm{M}^{-1}\mathrm{s}^{-1}\). Again, integrating this from \(t=0\) to \(t=t\), we get: \(\frac{1}{[A]}-\frac{1}{[A]_0}=kt\). Hence, the equation relating the concentration of A at \(t=0\) to that at \(t=t\) is: \( [A]=\frac{[A]_0}{1+[A]_0kt}\).

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Most popular questions from this chapter

A certain reaction is known to proceed slowly at room temperature. Is it possible to make the reaction proceed at a faster rate without changing the temperature?

How does a catalyst increase the rate of a reaction?

What are the units of the rate constant for a thirdorder reaction?

Define half-life. Write the equation relating the halflife of a first-order reaction to the rate constant.

Chlorine oxide (ClO), which plays an important role in the depletion of ozone (see Problem 13.101 ), decays rapidly at room temperature according to the equation $$ 2 \mathrm{ClO}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{O}_{2}(g) $$ From the following data, determine the reaction order and calculate the rate constant of the reaction. $$ \begin{array}{ll} \hline \text { Time (s) } & {[\mathrm{ClO}](M)} \\ \hline 0.12 \times 10^{-3} & 8.49 \times 10^{-6} \\ 0.96 \times 10^{-3} & 7.10 \times 10^{-6} \\ 2.24 \times 10^{-3} & 5.79 \times 10^{-6} \\ 3.20 \times 10^{-3} & 5.20 \times 10^{-6} \\ 4.00 \times 10^{-3} & 4.77 \times 10^{-6} \\ \hline \end{array} $$
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