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Chapter 35: Problem 7

The reaction rate as a function of initial reactant pressures was investigated for the reaction \(2 \mathrm{NO}(g)+\) \(2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g),\) and the following data were obtained: $$\begin{array}{cccc}\text { Run } & P_{o} \mathrm{H}_{2}(\mathrm{kPa}) & P_{o} \mathrm{NO}(\mathrm{kPa}) & \text { Rate }\left(\mathrm{kPa} \mathrm{s}^{-1}\right) \\\\\hline 1 & 53.3 & 40.0 & 0.137 \\\2 & 53.3 & 20.3 & 0.033 \\\3 & 38.5 & 53.3 & 0.213 \\\4 & 19.6 & 53.3 & 0.105\end{array}$$ What is the rate law expression for this reaction?

Short Answer

Expert verified
The rate law expression for this reaction is: Rate = k * [NO]虏 * [H鈧俔

Step by step solution

01

Examine the change on one reactant's pressure while keeping the other constant

Let's examine Runs 1 and 2. In these runs, the initial pressure of H鈧� gas is kept constant while the initial pressure of NO gas changes. From Run 1 to Run 2: \(P_{o} \mathrm{H}_{2}\) is constant. \(P_{o} \mathrm{NO}\) changes from 40.0 kPa to 20.3 kPa (approximately a factor of 0.5) Rate changes from 0.137 kPa s鈦宦� to 0.033 kPa s鈦宦� (approximately a factor of 0.24)
02

Determine the effect of the change in NO pressure on the reaction rate

Divide the rate from Run 2 by the rate from Run 1 and compare it to the change in NO pressure: \[\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.033}{0.137} \approx 0.24 \] Now divide the \(\mathrm{NO}\) pressure in Run 2 by the \(\mathrm{NO}\) pressure in Run 1: \[\frac{P_{o}\mathrm{NO}_2}{P_{o}\mathrm{NO}_1} \approx 0.5\] To find the order of the reaction with respect to NO, we can set up the following equation: \[\left(\frac{P_{o}\mathrm{NO}_2}{P_{o}\mathrm{NO}_1}\right)^{x} = \frac{\text{Rate}_2}{\text{Rate}_1}\] Solving for x: \[0.5^{x} = 0.24\] \[x \approx 2\] So the order of the reaction with respect to NO is approximately 2.
03

Examine the change on one reactant's pressure while again keeping the other constant

Let's now examine Runs 3 and 4. In these runs, the initial pressure of NO gas is kept constant while the initial pressure of H鈧� gas changes. From Run 3 to Run 4: \(P_{o} \mathrm{NO}\) is constant. \(P_{o} \mathrm{H}_{2}\) changes from 38.5 kPa to 19.6 kPa (approximately a factor of 0.51) Rate changes from 0.213 kPa s鈦宦� to 0.105 kPa s鈦宦� (approximately a factor of 0.49)
04

Determine the effect of the change in H鈧� pressure on the reaction rate

Divide the rate from Run 4 by the rate from Run 3 and compare it to the change in H鈧� pressure: \[\frac{\text{Rate}_4}{\text{Rate}_3} = \frac{0.105}{0.213} \approx 0.49 \] Now divide the \(\mathrm{H}_{2}\) pressure in Run 4 by the \(\mathrm{H}_{2}\) pressure in Run 3: \[\frac{P_{o}\mathrm{H}_{2}_4}{P_{o}\mathrm{H}_{2}_3} \approx 0.51\] To find the order of the reaction with respect to H鈧�, set up the following equation: \[\left(\frac{P_{o}\mathrm{H}_{2}_4}{P_{o}\mathrm{H}_{2}_3}\right)^{y} = \frac{\text{Rate}_4}{\text{Rate}_3}\] Solving for y: \[0.51^{y} = 0.49\] \[y \approx 1\] So the order of the reaction with respect to H鈧� is approximately 1.
05

Write the rate law expression

Based on our analysis on steps 2 and 4, the order of the reaction with respect to NO = 2, and the order of the reaction with respect to H鈧� = 1. Thus, the rate law expression is given by: Rate = k * [NO]虏 * [H鈧俔 where k is the rate constant and [NO] and [H鈧俔 represent the concentrations of NO and H鈧�, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics is the study of the rates at which chemical reactions proceed and the factors that affect these rates. It's a crucial area of chemistry because it helps us understand reaction mechanisms and the behavior of reactants over time. For example, when examining the reaction where nitrogen monoxide (NO) and hydrogen (H鈧�) gas react to form nitrogen (N鈧�) and water (H鈧侽), kinetics tells us not just that the products will form, but how quickly they'll form under certain conditions.

The rate of a reaction can be affected by a variety of factors, such as the concentration of reactants, temperature, and the presence of a catalyst. In the provided exercise, we see the initial pressures of the reactants used to study the reaction rate, which is common practice in gaseous reactions where pressure is directly proportional to concentration.
Rate Constant
The rate constant, symbolized by the letter 'k', is a proportionality factor in the rate law of a chemical reaction. It is unique for each reaction at a given temperature, and provides a quantitative measure of how fast a reaction occurs. In mathematical terms, the rate constant translates the relationship between reactant concentrations and the rate at which products are formed.

To find the rate constant, one usually needs to know the rate law expression and the concentrations of the reactants along with the actual rate of the reaction. Although we don't compute the actual value of 'k' in the provided exercise, understanding its conceptual role is important for analyzing reaction rates. It represents the intrinsic reaction speed and is influenced predominantly by the nature of the reactants and the temperature at which the reaction is taking place.
Reaction Order
The reaction order indicates the dependency of the reaction rate on the concentration of each reactant. It tells us how the rate of the reaction will change when the concentration of one of the reactants changes. In mathematical terms, each reactant's concentration is raised to a power called the 'order' with respect to that reactant. The overall reaction order is the sum of these individual orders.

In the exercise, we established the order of the reaction with respect to NO is 2 and with respect to H鈧� is 1 鈥� implying a second-order dependency on NO and first-order on H鈧�. This information is critical because it allows us to predict how the reaction rate will change under different conditions, making it a cornerstone concept in chemical kinetics.
Initial Reactant Pressures
For reactions involving gases, the role of pressure is vital. The initial pressures of reactants can be directly related to their molar concentrations through the ideal gas law. Thus, by varying the initial pressures of reactants and observing the changes in reaction rate, one can determine the order of the reaction with respect to each gas. This approach is evident in the exercise, where the pressure variation of NO and H鈧� influences the reaction rate and subsequently reveals the reaction orders.

In practical terms, knowing how the initial pressures (or concentrations) impact the rate can help control a reaction鈥攚hether to hasten the process in industrial applications or to slow it down for safety reasons. Understanding this relationship is fundamental for scientists and engineers when scaling reactions from laboratory to industrial scales.

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Most popular questions from this chapter

For the sequential reaction \(\mathrm{A} \stackrel{k_{1}}{\rightarrow} \mathrm{B} \stackrel{k_{B}}{\rightarrow} \mathrm{C},\) the rate constants are \(k_{A}=5 \times 10^{6} \mathrm{s}^{-1}\) and \(k_{B}=3 \times 10^{6} \mathrm{s}^{-1}\) Determine the time at which \([B]\) is at a maximum.

Consider the following reaction involving bromophenol blue (BPB) and \(\mathrm{OH}^{-}: \mathrm{HBPB}(a q)+\mathrm{OH}^{-}(a q) \rightarrow\) \(\mathrm{BPB}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) .\) The concentration of \(\mathrm{BPB}\) can be monitored by following the absorption of this species and using the Beer-Lambert law. In this law, absorption \(A\) and concentration are linearly related. a. Express the reaction rate in terms of the change in absorbance as a function of time. b. Let \(A_{o}\) be the absorbance due to HBPB at the beginning of the reaction. Assuming that the reaction is first order with respect to both reactants, how is the absorbance of HBPB expected to change with time? c. Given your answer to part (b), what plot would you construct to determine the rate constant for the reaction?

A technique for radioactively labeling proteins is electrophilic radioiodination in which an aromatic substitution of \(^{131}\) I onto a tyrosine residue is performed as follows: Using the activity of \(^{131} \mathrm{I}\), one can measure protein lifetimes in a variety of biological processes. 131 I undergoes beta decay with a half- life of 8.02 days. Initially a protein labeled with \(^{131}\) I has a specific activity of \(1.0 \mu \mathrm{Ci}\), which corresponds to 37,000 decay events every second. The protein is suspended in aqueous solution and exposed to oxygen for 5 days. After isolating the protein from solution, the protein sample is found to have a specific activity of \(0.32 \mu \mathrm{Ci}\). Is oxygen reacting with the tyrosine residues of the protein, resulting in the loss of \(^{131}\) I?

The growth of a bacterial colony can be modeled as a first-order process in which the probability of cell division is linear with respect to time such that \(d N / N=\zeta d t,\) where \(d N\) is the number of cells that divide in the time interval \(d t\) and \(\zeta\) is a constant. a. Use the preceding expression to show that the number of cells in the colony is given by \(N=N_{0} e^{\zeta t},\) where \(N\) is the number of cells in the colony and \(N_{0}\) is the number of cells present at \(t=0\) b. The generation time is the amount of time it takes for the number of cells to double. Using the answer to part (a), derive an expression for the generation time. c. In milk at \(37^{\circ} \mathrm{C}\), the bacterium Lactobacillus acidophilus has a generation time of about 75 min. Construct a plot of the acidophilus concentration as a function of time for time intervals of \(15,30,45,60,90,120,\) and 150 min after a colony of size \(N_{0}\) is introduced to a container of milk.

In the limit where the diffusion coefficients and radii of two reactants are equivalent, demonstrate that the rate constant for a diffusion controlled reaction can be written as $$k_{d}=\frac{8 R T}{3 \eta}$$
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