Problem 37 - Chemistry The Central Science In Si Units

Question

Consider the gas-phase reaction between nitric oxide and bromine at \(273^{\circ} \mathrm{C}: 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{NOBr}(g)\). The following data for the initial rate of appearance of NOBr were obtained: \begin{tabular}{lccc} \hline Experiment & {\([\mathrm{NO}](M)\)} & {\(\left[\mathrm{Br}_{2}\right](M)\)} & Initial Rate \((M / \mathrm{s})\) \\ \hline 1 & 0.10 & 0.20 & 24 \\ 2 & 0.25 & 0.20 & 150 \\ 3 & 0.10 & 0.50 & 60 \\ 4 & 0.35 & 0.50 & 735 \\ \hline \end{tabular} (a) Determine the rate law, (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. \((\mathbf{c})\) How is the rate of appearance of \(\mathrm{NOBr}\) related to the rate of disappearance of \(\mathrm{Br}_{2} ?\) (d) What is the rate of disappearance of \(\mathrm{Br}_{2}\) when \([\mathrm{NO}]=0.075 \mathrm{M}\) and \(\left[\mathrm{Br}_{2}\right]=0.25 \mathrm{M} ?\)

Step-by-Step Solution

Answer

(a) Rate = k[NO]^2[Br2]; (b) k = 12000 M^{-2}s^{-1}; (c) Rate of NOBr = -2*rate of Br2; (d) 8.44 M/s.

Step 1: Determine the Order with Respect to NO

To determine the order of reaction with respect to NO, compare experiments 1 and 2. In these experiments, the concentration of \( \mathrm{Br}_2 \) is constant while the concentration of \( \mathrm{NO} \) changes.\[\frac{{\text{rate}_2}}{{\text{rate}_1}} = \left(\frac{{[\mathrm{NO}]_2}}{{[\mathrm{NO}]_1}}\right)^m\]Substituting the values,\[\frac{150}{24} = \left(\frac{0.25}{0.10}\right)^m \6.25 = (2.5)^m\]Taking the logarithm: \[m \log(2.5) = \log(6.25) \m = \frac{\log(6.25)}{\log(2.5)} = 2\]Thus, the reaction is second order with respect to NO.

Step 2: Determine the Order with Respect to Br2

To determine the order of reaction with respect to \( \mathrm{Br}_2 \), compare experiments 1 and 3. In these experiments, the concentration of \( \mathrm{NO} \) is constant while the concentration of \( \mathrm{Br}_2 \) changes.\[\frac{\text{rate}_3}{\text{rate}_1} = \left(\frac{[\mathrm{Br}_2]_3}{[\mathrm{Br}_2]_1}\right)^n\]Substituting the values,\[\frac{60}{24} = \left(\frac{0.50}{0.20}\right)^n \2.5 = (2.5)^n\]Taking the logarithm: \[n \log(2.5) = \log(2.5) = \frac{\log(2.5)}{\log(2.5)} = 1\]Thus, the reaction is first order with respect to \( \mathrm{Br}_2 \).

Step 3: Write the Rate Law

Based on the orders determined:The rate law for the reaction is \[\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Br}_2]^1 \]where \( k \) is the rate constant.

Step 4: Calculate the Average Rate Constant

Using the rate law, calculate \( k \) for each experiment, and then average them.**Experiment 1:**\[24 = k(0.10)^2(0.20)\]\[k = \frac{24}{0.10^2 \times 0.20} = 12000 \, \text{M}^{-2}\text{s}^{-1}\]**Experiment 2:**\[150 = k(0.25)^2(0.20)\]\[k = \frac{150}{0.25^2 \times 0.20} = 12000 \, \text{M}^{-2}\text{s}^{-1}\]**Experiment 3:**\[60 = k(0.10)^2(0.50)\]\[k = \frac{60}{0.10^2 \times 0.50} = 12000 \, \text{M}^{-2}\text{s}^{-1}\]**Experiment 4:**\[735 = k(0.35)^2(0.50)\]\[k = \frac{735}{0.35^2 \times 0.50} = 12000 \, \text{M}^{-2}\text{s}^{-1}\]The average \( k \) value is \( 12000 \, \text{M}^{-2}\text{s}^{-1} \) as all are the same.

Step 5: Relate the Rate of NOBr and Br2

From the balanced equation, \( 2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \rightarrow 2 \mathrm{NOBr}(g) \), the stoichiometry shows that the rate of appearance of NOBr is twice the rate of disappearance of \( \mathrm{Br}_2 \).\[\text{Rate of NOBr appearance} = -2 \times \frac{\Delta[\mathrm{Br}_2]}{\Delta t}\]

Step 6: Calculate Rate of Disappearance of Br2

To find the rate of disappearance of \( \mathrm{Br}_2 \), use the rate law with \([\mathrm{NO}] = 0.075 \, \text{M}\) and \([\mathrm{Br}_2] = 0.25 \, \text{M}\):\[\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Br}_2] = 12000 \cdot (0.075)^2 \cdot (0.25)\]\[\text{Rate} = 12000 \cdot 0.005625 \cdot 0.25 = 16.875 \, \text{M/s}\]Thus, the rate of disappearance of \( \mathrm{Br}_2 \) is half this rate since the stoichiometry is \( 2:1 \). \[\text{Rate of disappearance of } \mathrm{Br}_2 = \frac{16.875}{2} = 8.4375 \, \text{M/s}\]

Key Concepts

Rate Law

Rate law is a mathematical equation that describes the relationship between the concentration of reactants and the rate of a chemical reaction. It's like the recipe that tells you how the reaction progresses. The rate law is generally expressed as:

• Rate = k [A]^x [B]^y

Here,

• k is the rate constant, a unique number for each reaction at a given temperature.
• [A] and [B] are the concentrations of the reactants.
• x and y represent the reaction orders for each reactant.

The rate law helps predict how adjusting concentrations can change the reaction rate. Understanding the rate law enables us to control reactions better, a crucial aspect in industrial and lab settings.

Reaction Order

Reaction order tells us how sensitive the reaction rate is to changes in reactant concentration. For each reactant in a reaction, we have an order. It's the power raised on each concentration term in the rate law expression:

• For example, if a reaction with respect to Nitric Oxide (NO) is determined to be second order, its concentration is squared in the rate equation.

Knowing the order means knowing how the rate will change:

• If first-order, doubling the concentration will double the rate.
• If second-order, doubling the concentration increases the rate by four times.

Different reactions have different orders for each reactant, and often, the only way to find these is through experimental data.

Rate Constant

The rate constant, denoted as k, is an important factor in the rate law equation. It isn't just a random number but depends on factors like temperature and the presence of catalysts.

• The rate constant unit varies depending on the overall order of the reaction.
• In this exercise, k was found to be 12000 M^-2s^-1, typical for a third-order reaction.

Understanding k is crucial because:

• It reflects how fast or slow a reaction happens.
• A larger k means a faster reaction, while a smaller k means it’s slower.
• It stays consistent if temperature and pressure remain unchanged, providing a reliable measure for reaction rate prediction.

Stoichiometry

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Think of it like a balanced scale; it tells us how many molecules or moles of each substance are needed. Stoichiometry gives us the proportions seen in the chemical equation:

• For this reaction, 2 moles of NO react with 1 mole of Br₂ to form 2 moles of NOBr.

Understanding stoichiometry allows us to:

• Predict how much of a reactant is needed to produce a desired amount of product.
• Calculate the rate of disappearance of reactants and appearance of products.
• Determine yield and efficiency of chemical processes.

So, when studying how NOBr's appearance is related to Br₂'s disappearance, stoichiometry provides the exact ratios you need.