How much and in what direction will each of the following affect the rate of the reaction: \(\mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)\) if the rate law for the reaction is rate \(=k\left[\mathrm{NO}_{2}\right]^{2} ?\) (a) Decreasing the pressure of \(\mathrm{NO}_{2}\) from 0.50 atm to 0.250 atm. (b) Increasing the concentration of CO from \(0.01 M\) to \(0.03 M\).

Short Answer

Expert verified
Decreasing the pressure of NO2 from 0.50 atm to 0.250 atm will decrease the reaction rate by a factor of four. Increasing the concentration of CO from 0.01 M to 0.03 M will have no effect on the reaction rate according to the given rate law.

Step by step solution

01

Understanding the Rate Law

The rate law given is rate = k[NO2]^2, which means that the rate of the reaction depends on the concentration of NO2 squared. The reaction rate does not depend directly on the concentration of CO for this rate law.
02

Analyzing the Effect of Decreasing NO2 Pressure

Since the rate is proportional to the square of the NO2 concentration, decreasing the pressure of NO2 from 0.50 atm to 0.250 atm will decrease the concentration of NO2. Because pressure is proportional to concentration, the concentration of NO2 is halved. This means that the rate of the reaction will decrease by a factor of four (since \(0.250^2 = (1/2)^2 = 1/4 \)).
03

Analyzing the Effect of Increasing CO Concentration

Increasing the concentration of CO from 0.01 M to 0.03 M will not affect the reaction rate according to the rate law provided. The rate law only includes [NO2] and has no term for [CO]. Thus, changing the concentration of CO has no effect on the rate.

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Key Concepts

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

Reaction Rate Law
The reaction rate law is a mathematical equation that expresses the speed of a chemical reaction as a function of the concentration of its reactants. In the context of the given exercise, the rate law for the reaction \(\mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)\) is defined as rate = \(k[\mathrm{NO}_{2}]^{2}\). This equation shows that the reaction rate is directly proportional to the square of the concentration of nitrogen dioxide \(\mathrm{NO}_{2}\).

Understanding the relationship between reaction rate and reactant concentration is fundamental to controlling chemical reactions. For instance, with this rate law, if the concentration of \(\mathrm{NO}_2\) doubles, the reaction rate will increase by a factor of four, because the rate is dependent on \(\mathrm{NO}_2\) concentration squared. In this exercise, it's crucial to note that the concentration of carbon monoxide (CO) is irrelevant to the rate at which the reaction occurs, as it does not appear in the rate law.
Concentration Dependence
The dependence of reaction rate on reactant concentration is highlighted by the rate law equation. In the provided example, only the concentration of \(\mathrm{NO}_2\) appears in the rate equation, indicating that the concentration of carbon monoxide (CO) does not impact the rate. It’s an important point of clarity for students who might mistakenly believe that changes in CO concentration would affect the rate.

To elaborate, the impact on reaction rate is only through the reactants included in the rate law. Learning to identify which reactant's concentration will change the rate is critical in predicting how a reaction proceeds under various conditions. In classroom or laboratory settings, understanding concentration dependence helps in the design of experiments to test reaction kinetics. This concept also has practical implications in industrial processes where controlling reactant concentrations can be crucial for efficiency and safety.
Pressure and Reaction Rate
Pressure is another factor that can influence the reaction rate, particularly in reactions involving gases, as in the given exercise. The relationship between pressure and reaction rate can be somewhat indirect, as pressure affects gas concentration and, in turn, concentration influences reaction rate.

In a closed system, an increase in pressure typically leads to an increase in gas concentration, thereby increasing the reaction rate according to the rate law. Conversely, decreasing the pressure reduces the concentration of the gas reactants, slowing the reaction rate. For the reaction between \(\mathrm{CO}\) and \(\mathrm{NO}_2\), decreasing \(\mathrm{NO}_2\) pressure from 0.50 atm to 0.250 atm results in halving its concentration, which then reduces the reaction rate by a factor of four, since rate is proportional to \(\mathrm{NO}_2\) concentration squared. This principle is vital for students to grasp, as it links the physical conditions of a reaction system with the kinetic behavior described by the rate law.

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Most popular questions from this chapter

Ozone decomposes to oxygen according to the equation \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) .\) Write the equation that relates the rate expressions for this reaction in terms of the disappearance of \(\mathrm{O}_{3}\) and the formation of oxygen.

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: \(\mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data: $$\begin{array}{|c|c|c|c|} \hline \text { \([\mathrm{NO}](M)\) }& 0.30 & 0.60 & 0.60 \\ \hline \text { \(\left[\mathrm{H}_{2}\right](M)\) } & 0.35 & 0.35 & 0.70 \\\ \hline \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\right)\) } & 2.835 \times 10^{-3} & 1.134 \times 10^{-2} & 2.268 \times 10^{-2} \\ \hline \end{array}$$

From the following data, determine the rate law, the rate constant, and the order with respect to \(A\) for the reaction \(A \longrightarrow 2 C\). $$\begin{array}{|c|c|c|c|} \hline \text { \([A](M)\) } & 1.33 \times 10^{-2} & 2.66 \times 10^{-2} & 3.99 \times 10^{-2} \\ \hline \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{h}^{-1}\right)\) } & 3.80 \times 10^{-7} & 1.52 \times 10^{-6} & 3.42 \times 10^{-6} \\ \hline \end{array}$$

What is the half-life for the first-order decay of phosphorus-32? $$\left(\frac{14}{6} \mathrm{C} \longrightarrow_{7}^{14} \mathrm{N}+\mathrm{e}^{-}\right)$$ The rate constant for the decay is \(4.85 \times 10^{-2}\) day \(^{-1}\).

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