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Express the rate of the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) $$ in terms of (a) \(\Delta\left[\mathrm{N}_{2} \mathrm{O}\right]\) (b) \(\Delta\left[\mathrm{O}_{2}\right]\)

Short Answer

Expert verified
Question: Express the rate of the given reaction in terms of the change in concentration of N2O and O2. Reaction: \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) Answer: The rate of the reaction can be expressed in terms of the change in concentration of N2O as \(-\cfrac{1}{2} \cfrac{\Delta[\mathrm{N}_{2} \mathrm{O}]}{\Delta t}\) and in terms of the change in concentration of O2 as \(\cfrac{\Delta[\mathrm{O}_{2}]}{\Delta t}\).

Step by step solution

01

Determine the stoichiometric coefficient of N2O

Since there are 2 moles of N2O reacting, the stoichiometric coefficient of N2O is 2.
02

Express rate in terms of N2O

Using the information from Step 1, we can express the rate of the reaction in terms of the change in concentration of N2O: Rate = \(-\cfrac{1}{2} \cfrac{\Delta[\mathrm{N}_{2} \mathrm{O}]}{\Delta t}\) (b) Express the rate in terms of \(\Delta\left[\mathrm{O}_{2}\right]\):
03

Determine the stoichiometric coefficient of O2

From the given reaction, there is 1 mole of O2 produced, so the stoichiometric coefficient of O2 is 1.
04

Express rate in terms of O2

Using the information from Step 1, we can express the rate of the reaction in terms of the change in concentration of O2: Rate = \(\cfrac{\Delta[\mathrm{O}_{2}]}{\Delta t}\) So, the rate of the given reaction can be expressed in terms of the change in concentration of N2O as \(-\cfrac{1}{2} \cfrac{\Delta[\mathrm{N}_{2} \mathrm{O}]}{\Delta t}\) and in terms of the change in concentration of O2 as \(\cfrac{\Delta[\mathrm{O}_{2}]}{\Delta t}\).

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Key Concepts

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

Stoichiometry
Understanding stoichiometry is crucial when exploring chemical reactions. Stoichiometry refers to the quantitative relationship between the reactants and products in a chemical reaction. It's derived from the Greek words 'stoicheion' (element) and 'metron' (measure), and it helps us calculate the amounts of substances consumed and produced in a reaction. In the given exercise, 2 moles of dinitrogen monoxide () react to produce 2 moles of nitrogen () and 1 mole of oxygen gas ().

These numbers before each chemical species are known as stoichiometric coefficients and they indicate the proportions in which molecules react or form. The coefficients tell us, for example, that for every 2 molecules of decomposed, 2 molecules of and 1 molecule of are formed. Stoichiometry is pivotal for learning how to balance chemical equations and to understand the underlying mole concept—a fundamental building block in chemistry.
Reaction Rate Expression
The reaction rate expression lets us quantitatively describe the speed of a chemical reaction by relating the rate with the concentration changes of reactants or products over time. For the reaction given in the exercise, you can generate a rate expression using the stoichiometry of the reaction. When we say the rate of the reaction in terms of , we incorporate not just the change in concentration, but also the stoichiometric coefficient. That's why the rate is expressed as , where the negative sign indicates that the concentration of diminishes over time, and the factor of is because we are accounting for 2 moles of being consumed for every 1 mole of produced. The importance of a precise reaction rate expression lies in its power to predict how fast a product forms or a reactant is consumed, information that's vital in industries and research.
Concentration Change
With chemical reactions, we often want to know how the concentration of a substance changes as the reaction proceeds. This is what we refer to when we talk about concentration change in a reaction's context. The rate of the reaction in terms of the change in concentration of the gaseous oxygen can be expressed as . This denotes the amount of produced per unit of time. Monitoring concentration changes allows chemists to understand reaction kinetics - the study of the rates of chemical processes. Identifying these changes facilitates the control of reactions in industrial processes and helps to determine the best conditions for a reaction to occur. In laboratory settings, figuring out how concentration affects reaction rate can be crucial for experimental design and for scaling reactions from the lab to production.

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Most popular questions from this chapter

In a solution at a constant \(\mathrm{H}^{+}\) concentration, iodide ions react with hydrogen peroxide to produce iodine. $$ \mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)+\frac{1}{2} \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \frac{1}{2} \mathrm{I}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} $$ The reaction rate can be followed by monitoring the appearance of \(\mathrm{I}_{2}\). The following data are obtained: $$ \begin{array}{ccc} \hline\left[\mathrm{I}^{-}\right] & {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ \hline 0.015 & 0.030 & 0.0022 \\ 0.035 & 0.030 & 0.0052 \\ 0.055 & 0.030 & 0.0082 \\ 0.035 & 0.050 & 0.0087 \\ \hline \end{array} $$ (a) Write the rate expression for the reaction. (b) Calculate \(k\). (c) What is the rate of the reaction when \(25.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of \(\mathrm{KI}\) is added to \(25.0 \mathrm{~mL}\) of a \(10.0 \%\) by mass solution of \(\mathrm{H}_{2} \mathrm{O}_{2}(d=1.00 \mathrm{~g} / \mathrm{mL}) ?\) Assume volumes are additive.

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Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a \(2.50-\mathrm{L}\) flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes?

Consider the following hypothetical reaction: $$ \mathrm{X}(g) \longrightarrow \mathrm{Y}(g) $$ A \(200.0-\mathrm{mL}\) flask is filled with 0.120 moles of \(\mathrm{X}\). The disappearance of \(\mathrm{X}\) is monitored at timed intervals. Assume that temperature and volume are kept constant. The data obtained are shown in the table below. $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 20 & 40 & 60 & 80 \\ \text { moles of X } & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline \end{array} $$ (a) Make a similar table for the appearance of \(\mathrm{Y}\). (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two 20 -minute intervals. (c) What is the average rate of appearance of Y between the 20 - and 60 -minute intervals?

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