Question

Experimental data are listed here for the reaction \(A \rightarrow 2 B.\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{B}](\mathrm{mol} / \mathrm{L})} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data; connect the points with a smooth line; and calculate the rate of change of [B] for each 10 -second interval from 0.0 to 40.0 seconds. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of [A] related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to 20.0 seconds. (c) What is the instantaneous rate, \(\Delta[\mathrm{B}] / \Delta \mathrm{t},\) when \([\mathrm{B}]=0.750 \mathrm{mol} / \mathrm{L} ?\)

Short answer

(a) Rate decreases. (b) \( \frac{\Delta[A]}{\Delta t} = -\frac{1}{2} \times \frac{\Delta[B]}{\Delta t} \); Rate of \([A]\) is \(-0.0123 \ \text{mol/L/s}\). (c) Instantaneous rate is \(0.0159 \ \text{mol/L/s}\).

Step-by-step solution

Graph the Data

Plot the concentration of \( [B] \) versus time. Use time (in seconds) on the x-axis and \([B]\) (in mol/L) on the y-axis. Connect the data points with a smooth line to visualize the change over time.

Calculate Rate of Change of [B]

For each 10-second interval, calculate the rate of change using \( \text{Rate} = \frac{\Delta[B]}{\Delta t} \). Use the formula for each interval: - From 0 to 10 s: \( \text{Rate} = \frac{0.326 - 0.000}{10} = 0.0326 \ \text{mol/L/s}\) - From 10 to 20 s: \( \text{Rate} = \frac{0.572 - 0.326}{10} = 0.0246 \ \text{mol/L/s}\) - From 20 to 30 s: \( \text{Rate} = \frac{0.750 - 0.572}{10} = 0.0178 \ \text{mol/L/s} \) - From 30 to 40 s: \( \text{Rate} = \frac{0.890 - 0.750}{10} = 0.0140 \ \text{mol/L/s} \) Notice that the rate decreases over time, likely due to the consumption of reactants.

Relate Rates of [A] and [B]

According to the stoichiometry \( A \rightarrow 2B \), for every mole of \(A\) consumed, 2 moles of \(B\) are produced. Thus, \( \frac{\Delta[A]}{\Delta t} = -\frac{1}{2} \times \frac{\Delta[B]}{\Delta t} \).

Calculate Rate of Change of [A] (10-20 s)

Using the relationship: \( \text{Rate of change of } [A] = -\frac{1}{2} \times 0.0246 = -0.0123 \ \text{mol/L/s} \).

Calculate Instantaneous Rate at [B] = 0.750

Find the time when \([B] = 0.750 \ ext{mol/L}\), which occurs at 30 s. Use the average rate before and after 30 s to estimate: \( \text{Instantaneous Rate} \approx \frac{0.0178 + 0.0140}{2} = 0.0159 \, \text{mol/L/s} \).

Key concepts

Rate of Change

The rate of change in reaction kinetics is a measure of how quickly concentrations of reactants or products vary over time. In the given exercise, we are focusing on the change in concentration of product B over different time intervals as the reaction proceeds. By using the formula \(\text{Rate} = \frac{\Delta [B]}{\Delta t}\), where \(\Delta [B]\) represents the change in concentration of B and \(\Delta t\) is the change in time, we can ascertain how fast the product B is forming.
For example:
\(\cdot\) From 0 to 10 seconds, the rate is \(0.0326 \text{ mol/L/s}\).
\(\cdot\) From 10 to 20 seconds, the rate decreases to \(0.0246 \text{ mol/L/s}\).
This decrease suggests a slower formation of B as time goes on, which is often due to the consumption of reactants A, leading to a diminishing reaction rate. The concept of rate of change is crucial in analyzing reaction dynamics over specific time periods.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. In this exercise, the reaction is \(A \rightarrow 2B\), meaning in stoichiometric terms, one mole of A produces two moles of B.
From this relationship, we can derive that the rate of consumption of A is half the rate of formation of B, given by the equation:

• \(\frac{\Delta[A]}{\Delta t} = -\frac{1}{2} \times \frac{\Delta[B]}{\Delta t}\)

The negative sign signifies that the concentration of A decreases over time, as it is being consumed. Hence, if B's formation rate is \(0.0246 \text{ mol/L/s}\) from 10 to 20 seconds, then A's consumption rate within this interval is \(-0.0123 \text{ mol/L/s}\). This relationship highlights the stoichiometric ratio between A and B, emphasizing how stoichiometry is vital for predicting the quantities in reactions.

Instantaneous Rate

The instantaneous rate of a reaction is the rate at a particular moment in time, rather than over a span. It's often calculated at specific moments to understand the speed of reaction at that instant. In our example, when \(\mathrm{B} = 0.750 \text{ mol/L}\), we're looking for the rate at this specific concentration.
Given that this occurs at 30 seconds, the instantaneous rate can be approximated by averaging the rates of change immediately before and after this time point:
\[\text{Instantaneous Rate} \approx \frac{0.0178 + 0.0140}{2} = 0.0159 \text{ mol/L/s}\]
This gives us an estimate of how quickly the concentration of B is changing precisely when it is 0.750 mol/L. Instantaneous rates are crucial for understanding reactions in real-time and can aid in determining the mechanics of complex reactions.