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Chapter 14: Problem 56

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The activation energy \(E_a\) is 98.6 kJ/mol and the rate constant at \(30^{\circ}C\), \(k_{30}\), is \(2.89\times 10^2 s^{-1}\).

Step by step solution

01

Setting up the first equation

For the first step, let’s use the Arrhenius equation in its integrated form since we know two sets of \(k\) and \(T\) values. The Arrhenius equation is: \[ k=A \cdot e^{-E_a /(RT)}\] Where \(A\) is the Arrhenius constant, \(E_a\) is the activation energy, \(R\) is the gas constant and \(T\) is the temperature. First, rewrite this as: \(ln(k_1) = ln(A) - \frac{{E_a}}{{R \cdot T_1}}\). With \(k_1 = \frac{0.693}{{t_{1/2_{1}}}}\) and \(T_1 =20^\circ C = 293.15K\)
02

Setting up the second equation

Now we need to set up another equation for \(T_2\) and \(k_2\). This equation will look just like the first equation, but \(T_2 = 40^\circ C = 313.15K\) and \(k_2= \frac{0.693}{{t_{1/2_{2}}}}\). Thus, we have: \(ln(k_2) = ln(A) - \frac{{E_a}}{{R \cdot T_2}}\).
03

Eliminate \(A\) from the two equations

Now, in order to get \(E_a\), we need to eliminate \(A\) from the two equations. So, subtract Eq. 2 from Eq. 1: \(ln(k_1)-ln(k_2) = \frac{{E_a}}{{R}} \left(\frac{1}{{T_2}} - \frac{1}{{T_1}}\right)\). This simplifies to: \(ln\left( \frac{{k_1}}{{k_2}} \right) = \frac{{E_a}}{{R}} \left(\frac{1}{{T_2}} - \frac{1}{{T_1}}\right)\). Plug in the known values and solve for \(E_a\).
04

Calculate \(k_3\) at \(30^{\circ}C\)

To find the rate constant \(k\) at different temperatures we use the Arrhenius equation with the calculated \(E_a\). This will give: \[ k = A \cdot e^{-E_a /(RT)}\] where \(T = 30^{\circ}C = 303.15K\]. Substitute the known values of \(A\) and \(E_a\) to get \(k\).

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Most popular questions from this chapter

The reaction \(A \longrightarrow\) products is first order in A. Initially, \([\mathrm{A}]=0.800 \mathrm{M}\) and after 54min, \([\mathrm{A}]=0.100 \mathrm{M}.\) (a) At what time is \([\mathrm{A}]=0.025 \mathrm{M} ?\) (b) What is the rate of reaction when \([\mathrm{A}]=0.025 \mathrm{M} ?\)

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ Which of these sets of data corresponds to a (a) zero-order, (b) first-order, (c) second-order reaction?

If the plot of the reactant concentration versus time is nonlinear, but the concentration drops by \(50 \%\) every 10 seconds, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.

For the reaction \(A \longrightarrow\) products the following data are obtained. $$\begin{array}{cll} \hline {\text { Experiment 1 }} & &{\text { Experiment 2 }} \\ \hline [\mathrm{A}]=1.204 \mathrm{M} & t=0 \mathrm{min} & {[\mathrm{A}]=2.408 \mathrm{M}} & t=0 \mathrm{min}\\\ {[\mathrm{A}]=1.180 \mathrm{M}} & t=1.0 \mathrm{min} & {[\mathrm{A}]=?} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=0.602 \mathrm{M}} & t=35 \mathrm{min} & {[\mathrm{A}]=?} & t=30 \mathrm{min} \\ \hline \end{array}$$ (a) Determine the initial rate of reaction in Experiment 1. (b) If the reaction is second order, what will be \([\mathrm{A}]\) at \(t=1.0\) min in Experiment 2? (c) If the reaction is first order, what will be \([\mathrm{A}]\) at 30 min in Experiment 2?

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.
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