Question

The compound \(\mathrm{AX}_{2}\) decomposes according to the equation \(2 \mathrm{AX}_{2}(g) \rightarrow 2 \mathrm{AX}(g)+\mathrm{X}_{2}(g) .\) In one experiment, \(\left[\mathrm{AX}_{2}\right]\) was measured at various times and these data were obtained: $$ \begin{array}{cc} \text { Time (s) } & {\left[A X_{2}\right](\mathrm{mol} / \mathrm{L})} \\ \hline 0.0 & 0.0500 \\ 2.0 & 0.0448 \\ 6.0 & 0.0300 \\ 8.0 & 0.0249 \\ 10.0 & 0.0209 \\ 20.0 & 0.0088 \end{array} $$ (a) Find the average rate over the entire experiment. (b) Is the initial rate higher or lower than the rate in part (a)? Use graphical methods to estimate the initial rate.

Short answer

The average rate is 0.00206 mol/L/s. The initial rate is 0.0026 mol/L/s and it is higher than the average rate.

Step-by-step solution

- Understand the Problem

Look at the given chemical reaction: \(2 \, \mathrm{AX}_{2}(g) \rightarrow 2 \, \mathrm{AX}(g) + \mathrm{X}_{2}(g)\). The data provided shows how the concentration of \(\left[\mathrm{AX}_{2}\right]\) changes over time during the decomposition reaction.

- Calculate the Average Rate

The average rate of reaction is the change in concentration of \(\left[\mathrm{AX}_{2}\right]\) over the total time: \[\text{Average Rate} = -\frac{\Delta [\mathrm{AX}_{2}]}{\Delta t} = -\frac{[\mathrm{AX}_{2}(20)] - [\mathrm{AX}_{2}(0)]}{20 - 0} = -\frac{0.0088 - 0.0500}{20.0 - 0.0}\] Calculate this value.

- Calculation

Substitute the values into the average rate formula: \[ = -\frac{0.0088 - 0.0500}{20.0} = -\frac{-0.0412}{20.0} = 0.00206 \, \text{mol/L/s}\]

- Determine the Initial Rate

To determine if the initial rate is higher or lower, consider the slope of the graph at time \(t = 0\). Use the first two data points, \((0.0, 0.0500)\) and \((2.0, 0.0448)\), to estimate the initial rate. Calculate the rate: \[\text{Initial Rate} = -\frac{[\mathrm{AX}_{2}(2.0)] - [\mathrm{AX}_{2}(0.0)]}{2.0 - 0.0} = -\frac{0.0448 - 0.0500}{2.0 - 0.0}\] Calculate this value.

- Calculation of Initial Rate

Substitute the values into the initial rate formula: \[ = -\frac{0.0448 - 0.0500}{2.0} = -\frac{-0.0052}{2.0} = 0.0026 \, \text{mol/L/s}\]

- Compare Initial Rate with Average Rate

Compare the initial rate of 0.0026 mol/L/s with the average rate of 0.00206 mol/L/s. The initial rate is higher.

Key concepts

Rate of Reaction

The rate of reaction is a fundamental concept in chemical kinetics. It measures how quickly the reactants are converted to products over a certain period. Factors that can influence the rate of reaction include the concentration of reactants, temperature, surface area, and the presence of catalysts.

In a chemical reaction, the rate can be expressed using the change in concentration of a reactant or product per unit of time. For the decomposition of \(\text{AX}_{2}\), the rate of reaction can be defined as:

\[ \text{Rate} = - \frac{\text{d}[\text{AX}_{2}]}{\text{d}t} \]

This equation indicates how the concentration of \([ \text{AX}_{2}]\) decreases with respect to time. The negative sign shows that \([ \text{AX}_{2}]\) is decreasing.

Decomposition Reaction

In the provided exercise, the reaction of \(\text{AX}_{2}\) is a decomposition reaction. Decomposition reactions involve a single compound breaking down into two or more simpler substances.

For example, the decomposition of \(\text{AX}_{2}\) can be written as:

\[ 2 \text{AX}_{2}(g) \rightarrow 2 \text{AX}(g) + \text{X}_{2}(g) \]

Here, two molecules of \(\text{AX}_{2}\) decompose to form two molecules of \(\text{AX}\) and one molecule of \(\text{X}_{2}\). These reactions are commonly affected by temperature and the presence of catalysts.

Reaction Rate Calculation

To calculate the average reaction rate, we need to find the change in concentration of a reactant over the change in time.

For the decomposition of \(\text{AX}_{2}\), given the data, the average rate over the entire experiment can be expressed as:

\[ \text{Average Rate} = -\frac{\big[ \text{AX}_{2}(20)\big] - \big[ \text{AX}_{2}(0)\big]}{20 - 0} \]

Substituting the provided values:

\[ = -\frac{0.0088 - 0.0500}{20} = 0.00206 \text{ mol/L/s} \]

Therefore, the average rate of reaction is 0.00206 mol/L/s. This method of calculation demonstrates how the concentration of \(\text{AX}_{2}\) decreases over the total experiment duration.

Initial Rate Determination

The initial rate of reaction is typically higher and represents the rate at the beginning of the reaction. This is often obtained from the slope of the concentration versus time graph at \(t = 0\).

Using the first two data points \((0.0, 0.0500)\) and \((2.0, 0.0448)\), the initial rate is calculated as follows:

\[ \text{Initial Rate} = -\frac{[\text{AX}_{2}(2.0)] - [\text{AX}_{2}(0.0)]}{2.0 - 0.0} \]

Substituting the values:

\[ -\frac{0.0448 - 0.0500}{2.0} = 0.0026 \text{ mol/L/s} \]

Comparing the initial rate (0.0026 mol/L/s) to the average rate (0.00206 mol/L/s), we find that the initial rate is indeed higher. This happens because, at the beginning, reactant molecules frequently collide more often, causing the reaction to proceed faster.