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Chapter 11: Problem 31

The reaction $$ 2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ is found to be first-order in \(\mathrm{H}_{2}(\mathrm{~g}) .\) Which rate equation cannot be correct? (a) Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\) (b) Rate \(=k\left[\mathrm{H}_{2}\right]\) $$ \text { (c) Rate }=k\left[\mathrm{NOl}^{2}\left[\mathrm{H}_{3}\right]^{2}\right. $$

Short Answer

Expert verified
Option (c) is incorrect.

Step by step solution

01

Understanding Reaction Order

The problem states that the reaction is first-order with respect to \(\mathrm{H}_2\). This means the rate of the reaction is directly proportional to the concentration of \(\mathrm{H}_2\). The power of the concentration term for \(\mathrm{H}_2\) in the rate equation should be 1.
02

Analyzing Rate Equations

Let's analyze each given rate equation to determine if they align with the first-order condition for \(\mathrm{H}_2\):(a) Rate \(=k[\mathrm{NO}]^{2}[\mathrm{H}_{2}]\) - The order with respect to \(\mathrm{H}_2\) is 1, which is consistent.(b) Rate \(=k[\mathrm{H}_{2}]\) - This equation is also consistent as it has a power of 1 for \(\mathrm{H}_2\).(c) Rate \(=k[\mathrm{NO}]^{2}[\mathrm{H}_{3}]^{2}\) - This equation mistakenly uses \(\mathrm{H}_3\) instead of \(\mathrm{H}_2\) and shows a second-order term for \(\mathrm{H}_2\), which is incorrect.
03

Identifying the Incorrect Equation

Since the reaction is first-order in \(\mathrm{H}_2\), any rate equation suggesting otherwise is incorrect. Option (c) has the mistake of indicating a second-order dependency for \(\mathrm{H}_2\) with a term \([\mathrm{H}_{2}]^2\), and the chemical species are incorrect due to \(\mathrm{H}_3\). Thus, option (c) cannot be correct.

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Key Concepts

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

Reaction Order
In the realm of chemical kinetics, the term **reaction order** is crucial in understanding how different substances influence the rate of a chemical reaction. This concept reflects how the concentration of one or more reactants affects the rate at which the product is formed. More specifically, it relates to the exponent used in the rate equation for a particular reactant.
  • If the order with respect to a reactant is 1, it implies that the rate of reaction is directly proportional to the concentration of that reactant.
  • For a zero-order reaction, the rate is unaffected by changes in concentration.
  • If the order is 2, the rate is proportional to the square of the reactant's concentration.
To determine the reaction order, we often look at experimental data or observe how variations in concentration affect the reaction rate. Knowing the order helps in predictive modeling of how fast a reaction proceeds under certain conditions. Understanding reaction order is essential as it allows chemists to manipulate reaction conditions effectively to increase yield or speed.
Rate Equation
The **rate equation** is an invaluable tool in chemistry, providing a mathematical description of the relationship between the reaction rate and the concentrations of reactants. Each component of the equation corresponds to a reagent and its respective reaction order: \[ ext{Rate} = k[A]^m[B]^n\] In this general form:
  • \(k\) is the rate constant specific to a reaction at a given temperature.
  • \([A]\) and \([B]\) are the concentrations of the reactants.
  • \(m\) and \(n\) are the orders of the reaction with respect to reactants \(A\) and \(B\), respectively.
The sum of \(m\) and \(n\) gives the overall order of the reaction.For instance, if a reaction is first-order with respect to a particular reactant, the rate equation will show an exponent of 1 for that reactant. Accurately writing the rate equation ensures proper analysis of reaction dynamics and understanding of how concentration changes impact rate.
First-order Reactions
When we discuss **first-order reactions**, we are focusing on reactions where the rate is directly proportional to the concentration of a single reactant. This type of reaction is common and easier to predict and analyze compared to higher-order reactions.The rate law for a first-order reaction can be expressed as:\[ ext{Rate} = k[A]\]In this expression:- \(A\) is the reactant whose concentration directly influences the rate.- \(k\) is the rate constant.A characteristic feature of first-order reactions is their exponential nature. As the reaction progresses, the concentration of the reactant decreases exponentially, which can be represented by the formula:\[[A] = [A]_0e^{-kt}\]This formula helps chemists determine how the concentration changes with time, aiding in reaction monitoring and optimization.Such simplicity in kinetics makes first-order reactions particularly favorable for initial studies in chemical kinetics, helping students and professionals alike gain insight into reaction mechanisms.

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Most popular questions from this chapter

Suppose a chemical reaction has an activation energy of \(76 \mathrm{~kJ} / \mathrm{mol}\), as in the example in Figure \(11.12 .\) Calculate by what factor the rate of the reaction at \(50 .{ }^{\circ} \mathrm{C}\) is increased over its rate at \(25^{\circ} \mathrm{C}\).

A reaction has the experimental rate law, Rate = \(k[\mathrm{~A}]^{2}[\mathrm{~B}]\). If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

Experiments show that the reaction of nitrogen dioxide with fluorine $$ 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{FNO}_{2}(\mathrm{~g}) $$ has the rate law $$ \text { Rate }=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{F}_{2}\right] $$ and the reaction is thought to occur in two steps: $$ \begin{array}{l} \text { Step } 1: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \\ \text { Step } 2: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g}) \end{array} $$ (a) Show that the sum of this sequence of reactions gives the balanced equation for the overall reaction. (b) Which step is rate-determining?

Using data given in the table for the reaction \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}\) calculate the average rate of reaction during each of these intervals: \(\begin{array}{ll}0.50 \mathrm{~h} . & \text { (b) } 0.50 \text { to } 1.0 \mathrm{~h} .\end{array}\) 0.00 to (a) 1.0 to (c) \(2.0 \mathrm{~h}\) (d) 2.0 to \(3.0 \mathrm{~h}\). \(4.0 \mathrm{~h}\) (f) 4.0 to \(5.0 \mathrm{~h}\). (e) 3.0 to \begin{tabular}{cccc} \hline Time (h) & {\(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{mol} / \mathrm{L})\)} & Time \((\mathrm{h})\) & {\(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{mol} / \mathrm{L})\)} \\ \hline 0.00 & 0.849 & 3.00 & 0.352 \\ 0.50 & 0.733 & 4.00 & 0.262 \\ 1.00 & 0.633 & 5.00 & 0.196 \\ 2.00 & 0.472 & & \\ \hline \end{tabular}

Hydrogenation reactions-processes in which \(\mathrm{H}_{2}\) is added to a molecule-are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Tell why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal that has the same mass.
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