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Chapter 14: Problem 33

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. $$ \begin{array}{lc} \mathrm{T}(\mathrm{K}) & \mathrm{k}\left(\mathrm{s}^{-1}\right) \\ \hline 273 & 7.87 \times 10^{3} \\ 298 & 3.46 \times 10^{5} \\ 318 & 4.98 \times 10^{6} \\ 338 & 4.87 \times 10^{7} \end{array} $$

Short Answer

Expert verified
Determine the activation energy by plotting ln(k) against 1/T using the provided data in the table. The negative slope from the graph multiplied by the gas constant (-m*R) gives the activation energy Ea for the reaction.

Step by step solution

01

Convert Given Equation Form

The first step is to convert the Arrhenius equation into a form that allows us to use linear regression to calculate the activation energy. By using the natural logarithm on both sides, the equation can be transformed into: \(ln(k) = ln(A) - Ea/R * 1/T \). Now, this equation resembles the formula of a straight line \(y = mx + c\), where the slope \(m\) represents \(-Ea/R\). By plotting \(ln(k)\) against \(1/T\), the activation energy can be determined from the slope of the line.
02

Prepare Your Data for Plotting

Now, prepare your data. For each pair of T and k-values, calculate the 1/T and ln(k) value. For example, using the first temperature value (273K), 1/T would be approximately 0.00367 and using the first k value we obtain ln(k)=8.97.
03

Plot the Data

Next, plot ln(k) on the y-axis and 1/T on the x-axis. You should observe a negative linear trend, which complies with the transformed Arrhenius equation.
04

Determine the Slope of the Line and Calculate Activation Energy

Find the slope of the line which is the ratio of the vertical change (y-axis change) to the horizontal change (x-axis change). A crucial point to remember is that the slope should be negative as per the equation. Assuming that the slope is m, the activation energy can be calculated by rearranging the previous formula: \(Ea = -m * R\). R is the universal gas constant, which is approximately 8.31 J/(mol*K) in SI units.

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Most popular questions from this chapter

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?

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