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Express the rate of the following reaction equation in terms of the rate of concentration change for each of the three species involved: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) $$

Short Answer

Expert verified
Rate = \(-\frac{1}{2} \frac{d[\text{SO}_2]}{dt} = -\frac{d[\text{O}_2]}{dt} = \frac{1}{2} \frac{d[\text{SO}_3]}{dt}\).

Step by step solution

01

Understand the Reaction Equation

The given chemical reaction is \(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\). It involves three species: sulfur dioxide (\(\text{SO}_2\)), oxygen (\(\text{O}_2\)), and sulfur trioxide (\(\text{SO}_3\)). The coefficients indicate the stoichiometric ratios of the reactants and products.
02

Define Rate of Reaction

The rate of a reaction can be expressed as the change in concentration of any of the reactants or products per unit time. By convention, the rate is positive for products and negative for reactants. For our reaction, the rate can be expressed as the negative change in concentration of \(\text{SO}_2\) or \(\text{O}_2\) per unit time, or the positive change in concentration of \(\text{SO}_3\) per unit time.
03

Relate Rates for Each Species Involved

Using the stoichiometry of the reaction equation, we find that: \[-\frac{1}{2} \frac{d[\text{SO}_2]}{dt} = -\frac{d[\text{O}_2]}{dt} = \frac{1}{2} \frac{d[\text{SO}_3]}{dt}\] This represents the rate at which \(\text{SO}_2\) is consumed, \(\text{O}_2\) is consumed, and \(\text{SO}_3\) is produced. Notice that the rate of disappearance of \(\text{SO}_2\) and appearance of \(\text{SO}_3\) are both half of the rate of \(\text{O}_2\) due to their stoichiometric coefficients.
04

Write the General Rate Equation

We can now express the rate of the reaction in terms of the concentration changes of the individual species: \[Rate = -\frac{1}{2} \frac{d[\text{SO}_2]}{dt} = -\frac{d[\text{O}_2]}{dt} = \frac{1}{2} \frac{d[\text{SO}_3]}{dt}\] This equation ensures that the rate is consistent for both the disappearance of reactants and the formation of the product.

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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 a branch of chemistry that studies the rate at which reactions occur and the factors affecting these rates. Understanding the speed of a chemical reaction is crucial for predicting how a reaction progresses over time. With reaction rates, we can better understand the dynamics of reactants turning into products and control these processes in industrial and laboratory settings.

To determine the rate of a chemical reaction, we look at how quickly the concentrations of the reactants decrease or how rapidly the products increase. This is often influenced by factors such as temperature, catalysts, and concentration of the reactants. In our exercise, the reaction between sulfur dioxide (\(\text{SO}_2\)) and oxygen (\(\text{O}_2\)) to form sulfur trioxide (\(\text{SO}_3\)) provides a great example of how chemical kinetics can be applied.
  • An increase in temperature typically leads to faster reaction rates because the molecules move more energetically, increasing the likelihood of collisions.
  • Catalysts can speed up reactions without being consumed, often by providing an alternative pathway with a lower activation energy.
  • The concentration of reactants influences the reaction rate as higher concentrations lead to more collisions.
Kinetic studies help in designing better industrial processes and understanding biological pathways.
Stoichiometry
Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between the substances involved in reactions. It uses the balanced chemical equation to determine the ratios of reactants and products. For instance, in our reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\), the stoichiometric coefficients are crucial.

These coefficients (2:1:2 in this case) imply that two moles of \(\text{SO}_2\) react with one mole of \(\text{O}_2\) to produce two moles of \(\text{SO}_3\). This is vital because it tells us how much of each species is involved and allows us to calculate changes in concentration with accuracy.
  • Every coefficient in a balanced equation can be interpreted as a part of a whole, simplifying predictions about the amount of products produced or reactants consumed.
  • It ensures that atoms are conserved in a reaction, following the law of conservation of mass.
Understanding stoichiometry is essential for computing theoretical yields, deciding on reaction conditions, and analyzing data from experiments.
Rate of Reaction Calculation
Calculating the rate of reaction involves measuring how the concentration of a substance changes over time. In our example reaction, there are different expressions for the rate of reaction, depending on whether we're monitoring reactants or products.

For example, in the reaction \(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\), the rate can be calculated using the concentration change of:\(-\frac{1}{2} \frac{d[\text{SO}_2]}{dt}\) for \(\text{SO}_2\),\(-\frac{d[\text{O}_2]}{dt}\) for \(\text{O}_2}\), and \(\frac{1}{2} \frac{d[\text{SO}_3]}{dt}\) for \(\text{SO}_3\).
  • Reactant disappearance rates are negative as the reactants are consumed, while product formation rates are positive as products are generated.
  • In reactions with more than one reactant or product, rates are standardized by dividing by the stoichiometric coefficient to yield a consistent rate.
Applying these principles, chemists can determine how long a reaction might take to reach completion and monitor its progress under different conditions.

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Most popular questions from this chapter

The rate constant for the first-order reaction described by the equation CC1CC1 \((g)\) C=CC \((g)\) cyclopropane \(\quad\) propene at \(500^{\circ} \mathrm{C}\) is \(5.5 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life of \(\mathrm{cy}\) clopropane at \(500^{\circ} \mathrm{C}\). Given an initial cyclopropane concentration of \(1.00 \times 10^{-\mathrm{s}} \mathrm{M}\) at \(500^{\circ} \mathrm{C}\), calculate the concentration of cyclopropane that remains after \(2.0\) hours.

The radioisotope argon- 41 is used to measure the rate of the flow of gases from smokestacks. It is a \(\gamma\) -emitter with a half-life of \(109.2\) minutes. Calculate the fraction of an argon- 41 sample that remains after one day.

Potassium- 40 decays by the two different paths, shown below. $$ \begin{aligned} &{ }_{19}^{40} \mathrm{~K} \rightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e} \quad(89.3 \%) \\ &{ }_{19}^{40} \mathrm{~K} \stackrel{\mathrm{EC}}{\longrightarrow}{ }_{18}^{41} \mathrm{Ar} \quad(10.7 \%) \end{aligned} $$ where EC stands for electron capture. The overall half-life for the decay of \({ }_{19}^{40} \mathrm{~K}\) is \(1.248 \times 10^{9}\) years. The potassium- 40 to argon- 40 reaction is thought to be the source of argon in the earth's atmosphere. Potassium-argon dating is used in geology and archaeology to date sedimentary rocks. Estimate the age of sedimentary rocks with a \({ }^{40} \mathrm{Ar}\) -to- \({ }^{-4} \mathrm{~K}\) ratio of \(0.0102\).

Azomethane, \(\mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{5}(g)\), decomposes according to the equation $$ \mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{3}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}(g)+\mathrm{N}_{2}(g) $$ Given that the decomposition is a first-order process with \(k=4.0 \times 10^{-4} \mathrm{~s}^{-1}\) at \(300^{\circ} \mathrm{C}\), calculate the fraction of azomethane that remains after \(1.0\) hour.

The decomposition of \(\mathrm{NO}_{2}(g)\) as described by the equation $$ 2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is second order in \(\mathrm{NO}_{2}(g)\). The rate constant for the reaction at \(300^{\circ} \mathrm{C}\) is \(0.54 \mathrm{M}^{-1 . \mathrm{s}^{-1}} .\) If \(\left[\mathrm{NO}_{2}\right]_{0}=1.25 \mathrm{M}\) what is the value of \(\left[\mathrm{NO}_{2}\right]\) after the reaction has run for \(2.0\) minutes?

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