Question

The rate of the reaction \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow\) $$ \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) $$ was measured at several temperatures, and the following data were collected: \begin{tabular}{ll} \hline Temperature \(\left({ }^{\circ} \mathrm{C}\right)\) & \(k\left(\boldsymbol{M}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 15 & \(0.0521\) \\ 25 & \(0.101\) \\ 35 & \(0.184\) \\ 45 & \(0.332\) \\ \hline \end{tabular} Using these data, graph \(\ln k\) versus \(1 / T\). Using your graph, determine the value of \(E_{a}\)

Short answer

After converting the temperature to Kelvin and calculating the inverse of each temperature and the natural logarithm of each rate constant, we can plot the graph of \(\ln k\) versus \(1/T\). Determine the slope of the best-fit line, which is a negative value, and calculate the activation energy (\(E_{a}\)) using the formula \[E_{a} = -\text{slope} \times R\] with the gas constant \(R = 8.314~J\cdot mol^{-1}\cdot K^{-1}\). The result will give the activation energy \(E_{a}\) in J/mol.

Step-by-step solution

Convert temperature from Celsius to Kelvin

To convert the temperature from Celsius to Kelvin, we just add 273.15 to each temperature value: 15°C + 273.15 = 288.15 K 25°C + 273.15 = 298.15 K 35°C + 273.15 = 308.15 K 45°C + 273.15 = 318.15 K

Calculate the inverse of each temperature

Now that we have the temperature values in Kelvin, we can take the inverse of each value: \(1/288.15 = 0.00347~K^{-1}\) \(1/298.15 = 0.00335~K^{-1}\) \(1/308.15 = 0.00324~K^{-1}\) \(1/318.15 = 0.00314~K^{-1}\)

Calculate the natural logarithm of each rate constant (\(\ln k\))

Now, we can take the natural logarithm of each rate constant, \(k\): \(\ln(0.0521) = -2.95\) \(\ln(0.101) = -2.29\) \(\ln(0.184) = -1.70\) \(\ln(0.332) = -1.10\)

Plot the graph of \(\ln k\) versus \(1/T\)

Using the values from Step 2 and Step 3, plot the graph where the x-axis represents \(1/T\) (in \(K^{-1}\)) and the y-axis represents \(\ln k\). The data points should form a linear graph.

Determine the slope of the graph

Draw a best-fit line for the data points and calculate the slope of the line. In this case, the slope is a negative value.

Calculate the activation energy (\(E_{a}\))

Now, we can use the slope to calculate the activation energy, \(E_{a}\). According to the Arrhenius equation: \[ \ln k = \ln A - \frac{E_{a}}{R} \cdot \frac{1}{T} \] The slope of the graph is equal to \(-E_{a}/R\). Therefore, \[ \text{slope} = -\frac{E_{a}}{R} \] To find the activation energy, multiply the slope by \(-R\). Use the gas constant, \(R = 8.314 ~J\cdot mol^{-1} \cdot K^{-1}\): \[ E_{a} = - \text{slope} \times R \] The result will give you the activation energy, \(E_{a}\), in J/mol.

Key concepts

Understanding Reaction Rates

When we talk about the speed of a chemical reaction, we're discussing its reaction rate. This measures how quickly reactants turn into products over time. It's important to remember that reaction rates can be influenced by various factors, including the concentrations of the reactants, the temperature at which the reaction is taking place, and the presence of a catalyst.

In the exercise provided, the reaction rate is based on the rate constant, k, which changes with temperature. As the temperature increases, the probability of reactant collisions increases, often resulting in a higher reaction rate. However, there’s more to the story when it comes to how temperature affects reaction rates, which involves an understanding of the Arrhenius equation.

The Role of Activation Energy

Activation energy (\(E_a\)) is a critical concept to grasp when it comes to chemical kinetics. It's the minimum amount of energy that reacting molecules must possess for a reaction to occur. Think of it as the energy barrier that reactants must overcome to transform into products. A high activation energy means that the reactants need a lot of energy to react, which typically translates to a slower reaction rate unless they are given additional energy, like an increase in temperature.

In the step-by-step solution, calculating the activation energy from the slope of the Arrhenius plot enables us to quantify this barrier. The steeper the slope, the higher the activation energy, meaning the reactants need more energy to successfully collide and react.

Temperature Dependence of Reaction Rates

The temperature dependence of reaction rates is encapsulated in the Arrhenius equation, which predicts how rate constants will change as a function of temperature. Higher temperatures typically increase the rate constants as particles move faster and collide more often with the necessary energy to overcome the activation energy. Conversely, lower temperatures result in slower reaction rates.

Through the exercise, by graphing the natural logarithm of the rate constant against the inverse of the temperature, we create an Arrhenius plot. Such a plot is crucial to understanding how temperature influences the dynamics of chemical processes and practically allows chemists to predict the behavior of reactions under different thermal conditions.

Interpreting the ln k versus 1/T Plot

The ln k versus 1/T plot is an essential tool for chemists. This Arrhenius plot helps us understand the relationship between temperature and reaction rates through a linear graph. On this graph, the y-axis represents the natural logarithm of the rate constant (\(ln k\)), and the x-axis corresponds to the inverse of the temperature (\(1/T\) in Kelvin). According to the Arrhenius equation, the slope of this line is proportional to the negative of the activation energy divided by the gas constant, \(R\).

So, by finding the slope of this line from our plotted data points, we are able to calculate the activation energy, as demonstrated in the exercise's step-by-step solution. This value informs us about the energy barrier of the reaction under study and is crucial for predicting how the reaction rate will change with temperature.