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Chapter 13: Problem 40

Variation of the rate constant with temperature for the first-order reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(g)+\mathrm{O}_{2}(g) $$ is given in the following table. Determine graphically the activation energy for the reaction. $$ \begin{array}{cc} \hline T(K) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 298 & 1.74 \times 10^{-5} \\ 308 & 6.61 \times 10^{-5} \\ 318 & 2.51 \times 10^{-4} \\ 328 & 7.59 \times 10^{-4} \\ 338 & 2.40 \times 10^{-3} \\ \hline \end{array} $$

Short Answer

Expert verified
The graph of ln(k) versus 1/T will yield a straight line with a slope of -Ea/R. Calculating the slope and then multiplying by -R will result in the activation energy for the reaction in J/mol. Remember to convert to kJ/mol by dividing by 1000.

Step by step solution

01

Data converstion

Firstly, translate rate constant (k) and temperature (T) data into logarithmic form to fit the equation in order to plot it. The Arrhenius equation can be rewritten in logarithmic form as: ln(k) = -Ea/R * 1/T + ln(A). Here 'ln' refers to natural logarithm. The transformed data will look like: \[ T(K) \quad ln(k) \] \[ 298 \quad ln(1.74 \times 10^{-5}) \] \[ 308 \quad ln(6.61 \times 10^{-5}) \] \[ 318 \quad ln(2.51 \times 10^{-4}) \] \[ 328 \quad ln(7.59 \times 10^{-4}) \] \[ 338 \quad ln(2.40 \times 10^{-3}) \]
02

Plot the graph

Plot the graph of ln(k) versus 1/T. In this graph, the y-axis is ln(k) and x-axis is 1/T (which will be in units of 1/K). Each pair of corresponding values in the table forms a point on the graph.
03

Derive the slope

Draw the best possible straight line that fits the plotted points. This line will be a downward sloping line due to the negative relationship between ln(k) and 1/T. Find the slope of this line.
04

Calculate the activation energy

The slope of the graph is equal to -Ea/R. So, multiply the slope by -R (the gas constant). The value of R is 8.314 J/mol*K. The product will give the activation energy (Ea) in J/mol. Convert this energy into kJ/mol by dividing by 1000. That is the activation energy of the reaction.

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Most popular questions from this chapter

Sketch a potential energy versus reaction progress plot for the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) \quad \Delta H^{\circ}= \\ \quad-296 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{Cl}(g) \Delta H^{\circ}=243 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Consider the following elementary steps for a consecutive reaction: $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(A\).

The concentrations of enzymes in cells are usually quite small. What is the biological significance of this fact?

Cyclobutane decomposes to ethylene according to the equation $$ \mathrm{C}_{4} \mathrm{H}_{8}(g) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(g) $$ Determine the order of the reaction and the rate constant based on the following pressures, which were recorded when the reaction was carried out at \(430^{\circ} \mathrm{C}\) in a constant-volume vessel. $$ \begin{array}{rc} \hline \text { Time (s) } & P_{\mathrm{C}_{4} \mathrm{H}_{8}}(\mathrm{mmHg}) \\\ \hline 0 & 400 \\ 2,000 & 316 \\ 4,000 & 248 \\ 6,000 & 196 \\ 8,000 & 155 \\ 10,000 & 122 \\ \hline \end{array} $$

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ fructose \(\quad\) glucose This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \\ \hline \end{array} $$
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