Problem 5 - Chemistry and Chemical Reactivity

Question

Experimental data are listed here for the reaction \(\mathrm{A} \longrightarrow 2 \mathrm{B}.\) $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\(\mathrm{s})\end{array} & \begin{array}{l}{[\mathrm{B}]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data, connect the points with a smooth line, and calculate the rate of change of [B] for each \(10-\) s interval from 0.0 to 40.0 s. 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 \([\mathrm{A}]\) 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 \(20.0 \mathrm{s}\) (c) What is the instantaneous rate when \([\mathrm{B}]=0.750\) \(\mathrm{mol} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

(a) Rate decreases due to reduced reactant concentration. (b) Rate of [A] = -0.5 × Rate of [B]; from 10 to 20 s, it’s -0.0123 mol/L/s. (c) Instantaneous rate is 0.0159 mol/L/s.

Step 1: Tabulate Time Intervals

The data gives the concentration of \( [\mathrm{B}] \) at specific times: 0, 10, 20, 30, and 40 seconds. We will compute the rate of change of concentration over these intervals by calculating the difference in concentration divided by the time interval.

Step 2: Calculate the Rate of Change of [B]

For the interval 0 to 10 s: \( \text{Rate} = \frac{0.326 - 0.000}{10} = 0.0326 \) mol/L/s.\For the interval 10 to 20 s: \( \text{Rate} = \frac{0.572 - 0.326}{10} = 0.0246 \) mol/L/s.\For the interval 20 to 30 s: \( \text{Rate} = \frac{0.750 - 0.572}{10} = 0.0178 \) mol/L/s.\For the interval 30 to 40 s: \( \text{Rate} = \frac{0.890 - 0.750}{10} = 0.0140 \) mol/L/s.

Step 3: Analyze the Rate of Change

The rate of change of \([\mathrm{B}]\) decreases over time, indicating that the reaction is slowing down. This might occur because the concentration of reactant A decreases over time, thus slowing the production of B.

Step 4: Relation of Rate of [A] and [B]

According to the stoichiometry of the reaction \( \mathrm{A} \rightarrow 2\mathrm{B} \), the rate of change of \([\mathrm{A}]\) is half of the negative rate of change of \([\mathrm{B}]\) because 1 mole of \(\mathrm{A}\) produces 2 moles of \(\mathrm{B}\).

Step 5: Calculate Rate of Change of [A] from 10 to 20 s

Using the rate of \( [\mathrm{B}] \) from 10 to 20 s (0.0246 mol/L/s), the rate of change of \( [\mathrm{A}] \) is \(-0.5 \times 0.0246 = -0.0123 \) mol/L/s.

Step 6: Determine Instantaneous Rate When [B] = 0.750 mol/L

The instantaneous rate at \([\mathrm{B}] = 0.750\) mol/L is estimated using the average rate of surrounding intervals (20-30 s and 30-40 s). The rate is \(\frac{0.0178 + 0.0140}{2} = 0.0159\) mol/L/s.

Key Concepts

Rate of Change

In chemical reactions, the rate of change is a critical concept that helps us understand how fast a reaction proceeds. It measures the speed at which the concentration of a reactant or product changes over time. To calculate this, we look at the difference in concentration over a specific time interval. For instance, in our exercise, we calculated the rate of change for substance B by comparing its concentration at the start and end of each 10-second interval. The formula used is:\[ \text{Rate of Change} = \frac{\Delta [B]}{\Delta t} \]where \( \Delta [B] \) is the difference in concentration of B and \( \Delta t \) is the time duration. The rate is expressed in mol/L/s.As observed from the solution, the rate decreased across intervals—0.0326, 0.0246, 0.0178, and 0.0140 mol/L/s respectively—indicating the reaction is slowing. This could be due to the consumption of reactant A, reducing the formation of B. Understanding how rates change over time is fundamental in chemical kinetics.

Concentration

Concentration refers to the amount of a substance in a given volume and is usually expressed in mol/L. In the case of our exercise, it shows how much of the substance B is present at different times during the reaction. By knowing the concentration of a reactant or product at various points, we gain insights into how the reaction progresses.For the reaction \( A \rightarrow 2B \), initially, the concentration of B is zero. Over time, as A is converted to B, the concentration of B increases. This accumulation of B is monitored over successive 10-second intervals to observe the change and compute the rate of reaction.Changes in concentration tell us not just about the progress of the reaction but also about how fast it's occurring at any moment. In real-life chemical kinetics applications, such as manufacturing or environmental chemistry, these changes are key to optimizing and controlling processes.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that examines the speed and mechanisms of chemical reactions. Understanding reaction rates is a significant part of chemical kinetics, which involves examining how fast reactants turn into products. We uncovered aspects of chemical kinetics by calculating the change in concentration of B over time.Determining the rate involves linking it with the stoichiometry of the reaction. For our equation \( A \rightarrow 2B \), the stoichiometry tells us that 1 mole of A produces 2 moles of B. Thus, the rate of decrease of A is half the rate of increase of B. This relationship was fundamental in calculating the rate of change of A from the calculated rate of B.Moreover, instantaneous rates, like the one calculated when B concentration was 0.750 mol/L, allow us to determine the exact rate at a particular concentration, using surrounding interval rates. Chemical kinetics informs various fields, from designing chemical reactors to environmental impact studies, making it a cornerstone of modern chemistry.