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Chapter 14: Problem 79

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) $$ \begin{array}{cc} {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{M})} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \end{array} $$ Determine graphically the rate law for the reaction and calculate the rate constant.

Short Answer

Expert verified
The rate of reaction is first order with respect to concentration of N2O5 as a linear relationship is observed between the initial rate and the concentration of N2O5. The rate law equation for the reaction is \(rate = k[\mathrm{N}_{2} \mathrm{O}_{5}]\). The rate constant (k) can be calculated to be approximately \(k = \frac{0.95 \times 10^{-5}}{0.92} M^{-1}s^{-1}\).

Step by step solution

01

Find the relationship between concentration and reaction rate.

First, plot the given data using the concentration of N2O5 on the x-axis and the initial rate on the y-axis. If the points roughly form a straight line passing through the origin, then the reaction is first order in N2O5. By looking at the provided data, we may infer that the reaction is first order as the initial rate increases linearly with an increase in concentration.
02

Determine the rate law.

As the relation is linear, the form of the rate law equation we have is \[rate = k[\mathrm{N}_{2} \mathrm{O}_{5}]^n\]. From the graph, we found that reaction is first order with respect to N2O5 (n = 1). So, the rate law for the reaction becomes \[rate = k[\mathrm{N}_{2} \mathrm{O}_{5}]\].
03

Calculate the rate constant (k).

Now, using one of the given data points from the table, for example the first one where \[[\mathrm{N}_{2} \mathrm{O}_{5}] = 0.92 M\] and the initial rate is \(0.95 \times 10^{-5} M/s\), insert these values into the rate law equation. Solving for k gives \(k = \frac{rate}{[\mathrm{N}_{2} \mathrm{O}_{5}]} = \frac{0.95 \times 10^{-5}}{0.92} M^{-1}s^{-1} \). The value of k (rate constant) obtained should align closely for the other data points, as long as they line is relatively straight.

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Most popular questions from this chapter

What are the units of the rate of a reaction?

Sketch a potential-energy-versus-reaction-progress plot for the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-296.06 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{Cl}(g) \\\ \Delta H^{\circ}=242.7 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

The equation for the combustion of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \longrightarrow 4 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ Explain why it is unlikely that this equation also represents the elementary step for the reaction.

Consider this elementary step: $$ X+2 Y \longrightarrow X Y_{2} $$ (a) Write a rate law for this reaction. (b) If the initial rate of formation of \(\mathrm{XY}_{2}\) is \(3.8 \times 10^{-3} \mathrm{M} / \mathrm{s}\) and the initial concentrations of \(X\) and \(Y\) are \(0.26 M\) and \(0.88 M\), what is the rate constant of the reaction?

Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However, for a given enzyme, the rate drops off abruptly at a certain temperature. Account for this behavior.
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