Problem 35 - Chemistry The Central Science In Si Units

Question

The following data were measured for the reaction \(\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g):\) \begin{tabular}{lccc} \hline Experiment & {\(\left[\mathrm{BF}_{3}\right](M)\)} & {\(\left[\mathrm{NH}_{3}\right](M)\)} & Initial Rate \((M / \mathrm{s})\) \\ \hline 1 & 0.250 & 0.250 & 0.2130 \\ 2 & 0.250 & 0.125 & 0.1065 \\ 3 & 0.200 & 0.100 & 0.0682 \\ 4 & 0.350 & 0.100 & 0.1193 \\ 5 & 0.175 & 0.100 & 0.0596 \\ \hline \end{tabular} (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

Step-by-Step Solution

Answer

(a) Rate = k[BF₃][NH₃], (b) Second order, (c) k = 3.408 M⁻¹s⁻¹, (d) Rate = 0.1704 M/s.

Step 1: Determine Reaction Orders

To determine the orders with respect to each reactant, we observe how the rate changes as the concentration of one reactant changes while keeping the other constant.- **For NH₃ (g):** From Experiments 1 and 2, \[ \frac{\text{Rate}_1}{\text{Rate}_2} = \frac{0.2130}{0.1065} = 2 \] When \([NH₃]\) is halved from 0.250 M to 0.125 M, and the concentration of \([BF₃]\) remains 0.250 M, the initial rate is halved, indicating a first-order reaction with respect to NH₃.- **For BF₃ (g):** From Experiments 3 and 5, \[ \frac{\text{Rate}_3}{\text{Rate}_5} = \frac{0.0682}{0.0596} \approx 1.144 \] When \([BF₃]\) is increased from 0.175 M to 0.200 M, \([NH₃]\) remains 0.100 M, resulting in the ratio change indicating a first-order reaction with respect to BF₃. The slight discrepancy is due to experimental error.

Step 2: Write the Rate Law

Based on the observations from Step 1, the experimental rate law can be written as:\[\text{Rate} = k[BF_3]^1[NH_3]^1\]Thus, the reaction is first-order with respect to both \([BF₃]\) and \([NH₃]\).

Step 3: Determine Overall Order of Reaction

The overall order of a reaction is the sum of the exponents in the rate law.For this reaction:\[\text{Overall order} = 1 + 1 = 2\]Thus, the reaction is second order overall.

Step 4: Calculate the Rate Constant

Using the data from Experiment 1:\[0.2130 = k (0.250)(0.250)\]Solving for \(k\):\[k = \frac{0.2130}{0.250 \times 0.250} = 3.408 \, M^{-1}s^{-1}\]Thus, the rate constant \(k\) is 3.408 M⁻¹s⁻¹.

Step 5: Calculate the Rate for Given Concentrations

Using the rate law determined, calculate the rate for \([BF₃] = 0.100 \text{ M}\) and \([NH₃] = 0.500 \text{ M}\):\[\text{Rate} = 3.408 \times 0.100 \times 0.500\]\[\text{Rate} = 0.1704 \, M/s\]Thus, the rate when \([BF₃] = 0.100 \text{ M}\) and \([NH₃] = 0.500 \text{ M}\) is 0.1704 M/s.

Key Concepts

Rate Law

The rate law is a crucial component of chemical kinetics, as it describes the relationship between the rate of a chemical reaction and the concentration of its reactants. For our specific reaction involving \(\mathrm{BF}_3\) and \(\mathrm{NH}_3\), we determine the rate law by observing how the initial rate of reaction changes with varying concentrations of reactants. The general form of a rate law can be written as:

• \(\text{Rate} = k [A]^m [B]^n\)

Here, \(k\) is the rate constant, and \(m\) and \(n\) are the reaction orders concerning reactants \(\mathrm{A}\) and \(\mathrm{B}\), respectively. For our reaction, using experimental data, we found that both \([\mathrm{BF}_3]\) and \([\mathrm{NH}_3]\) are first-order. Thus, the rate law becomes:

• \(\text{Rate} = k [BF_3]^1 [NH_3]^1\)

This means the reaction rate is directly proportional to the concentration of each reactant.

Reaction Order

The reaction order provides insight into the influence of each reactant on the reaction rate. In simple terms, it tells us whether the rate will change significantly if we alter the concentration of a reactant. Reaction orders for each reactant are determined experimentally. In our example, both \(\mathrm{BF}_3\) and \(\mathrm{NH}_3\) exhibit first-order behavior. This is deduced by observing that:

• For \(\mathrm{NH}_3\): When the concentration is halved, the rate is also halved, indicating first-order dependence.
• For \(\mathrm{BF}_3\): Changing its concentration shows consistent rate changes proportional to its concentration, confirming a first-order dependence.

Understanding these orders helps us predict how the reaction will respond to changes in concentration.

Rate Constant

The rate constant \(k\) is a pivotal factor in the rate law equation. It is influenced by reaction conditions such as temperature and provides information about the speed of a reaction. In the given problem, using data from a specific experimental run, we calculate \(k\) using the formula:

• \[ k = \frac{\text{Rate}}{ [BF_3]^1 [NH_3]^1 } \]

From experiment 1, we found:

• \[ 0.2130 = k \times 0.250 \times 0.250 \]

Solving this gives us:

• \[ k = 3.408 \, M^{-1}s^{-1} \]

The units of \(M^{-1}s^{-1}\) indicate this is a second-order overall reaction, providing another layer of verification for our findings.

Overall Reaction Order

The overall reaction order is the sum of the individual orders of each reactant in the rate law expression. It gives us an overarching view of how the concentrations of reactants collectively affect the reaction rate. In our given example, we sum the exponents of \([BF_3]\) and \([NH_3]\) from the rate law:

• \([BF_3]^1 and [NH_3]^1\)

Thus, the overall order is calculated as:

• First-order in \(BF_3\) + First-order in \(NH_3\) = Second-order overall.

This understanding not only aids in predicting the reaction dynamics but also in determining the appropriate units for the rate constant \(k\). With second-order reactions, like this one, the rate equation simplifies to a form dependent on the product of two reactant concentrations, emphasizing the role of each reactant in the reaction process.