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Chapter 7: Problem 25

For the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\), rate and rate constant are \(1.02 \times 10^{-4}\) and \(3.4 \times 10^{-5} \mathrm{sec}^{-1}\) respectively then concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at that time will be a. \(1.732\) b. 3 c. \(3.4 \times 10^{5}\) d. \(1.02 \times 10^{-4}\)

Short Answer

Expert verified
The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at that time is 3.

Step by step solution

01

Understand the Relation

The rate of the reaction is given as \(1.02 \times 10^{-4}\). The rate formula for a reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(\text{Rate} = k \times [\mathrm{N}_2 \mathrm{O}_5]\), where \(k\) is the rate constant.
02

Insert Known Values

We know the rate \( = 1.02 \times 10^{-4}\) and the rate constant \(k = 3.4 \times 10^{-5} \mathrm{sec}^{-1}\). Substituting these into the rate equation gives \(1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [\mathrm{N}_2 \mathrm{O}_5]\).
03

Solve for the N2O5 Concentration

To find \([\mathrm{N}_2 \mathrm{O}_5]\), divide both sides by the rate constant: \([\mathrm{N}_2 \mathrm{O}_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}\). Calculating this gives \([\mathrm{N}_2 \mathrm{O}_5] = 3\).

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Key Concepts

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

Rate Constant
The rate constant, often denoted as "k", plays a crucial role in determining how quickly a chemical reaction occurs. It's a fundamental parameter in the rate equation, which describes the relationship between the rate of the reaction and the concentration of the reactants. In the given reaction, the rate constant was provided as \(3.4 \times 10^{-5} \text{ sec}^{-1}\). This unit indicates that it's a first-order reaction with respect to \(\mathrm{N}_2\mathrm{O}_5\).
  • The rate constant remains unaffected by the concentrations of reactants or products.
  • It can vary with temperature, meaning higher temperatures generally increase the rate constant, thus speeding up the reaction.
  • Understanding the rate constant helps predict how fast a reaction proceeds under given conditions.
The rate constant is unique for each reaction and can be influenced by catalysts as well.
This is because a catalyst can lower the activation energy, indirectly affecting the rate constant.
Concentration Calculation
Calculating the concentration is often a fundamental step in determining the progress or state of a chemical reaction. In our reaction, we needed to find the concentration of \(\mathrm{N}_2\mathrm{O}_5\) that corresponds to a specified rate of the reaction.
  • The rate equation used was \(\text{Rate} = k \times [\mathrm{N}_2 \mathrm{O}_5]\), where the values for rate and rate constant were known.
  • To isolate \([\mathrm{N}_2 \mathrm{O}_5]\), the rate equation was rearranged to \([\mathrm{N}_2 \mathrm{O}_5] = \frac{\text{Rate}}{k}\).
  • By substituting the given rate \(1.02 \times 10^{-4}\) and the rate constant \(3.4 \times 10^{-5} \text{ sec}^{-1}\), we calculated \([\mathrm{N}_2 \mathrm{O}_5] \) to be 3.
This calculation tells us that, at the time considered, three moles per unit volume of \(\mathrm{N}_2\mathrm{O}_5\) are undergoing the reaction.Having accurate concentration calculations can help gauge the efficiency and progress of the reaction.
N2O5 Decomposition
The decomposition of dinitrogen pentoxide (\(\mathrm{N}_2\mathrm{O}_5\)) is an example of a chemical reaction where a single reactant breaks down into simpler products.
  • The balanced chemical equation for this process is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2}\).
  • This reaction produces nitrogen dioxide (\(\mathrm{NO}_2\)) and oxygen (\(\mathrm{O}_2\)).
  • The stoichiometric coefficients in the balanced equation indicate that two moles of \(\mathrm{N}_2\mathrm{O}_5\) decompose to form four moles of \(\mathrm{NO}_2\) and one mole of \(\mathrm{O}_2\).
The decomposition of \(\mathrm{N}_2\mathrm{O}_5\) is crucial in studying kinetics because it is simple and provides clear stoichiometric relationships.Additionally, understanding its decomposition helps in atmospheric chemistry, where \(\mathrm{NO}_2\) contributes to smog and air pollution control.

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Most popular questions from this chapter

\text { For a first order reaction, }a. The degree of dissociation is equal to \(\left(1-\mathrm{e}^{-\mathrm{k} 1}\right)\) b. The pre-exponential factor in the Arrhenius equation has the dimensions of time \(\mathrm{T}^{-1}\). c. The time taken for the completion of \(75 \%\) reaction is thrice the \(t \frac{1}{2}\) of the reaction. d. both (a) and (b)

Consider a reaction \(\mathrm{aG}+\mathrm{bH} \rightarrow\) Products. When concentration of both the reactants \(\mathrm{G}\) and \(\mathrm{H}\) is doubled, the rate increases by eight times. However when concentration of \(\mathrm{G}\) is doubled keeping the concentration of \(\mathrm{H}\) fixed, the rate is doubled. The overall order of the reaction is a. 0 b. 1 c. 2 d. 3

Which of the following expressions is/are not correct? a. \(\log \mathrm{k}=\log \mathrm{A}-\frac{\mathrm{Ea}}{2.303 \mathrm{RT}}\) b. \(\operatorname{In} \mathrm{A}=\operatorname{In} \mathrm{k}+\frac{\mathrm{Ea}}{\mathrm{RT}}\). C. \(\mathrm{k} \mathrm{Ae}^{-\mathrm{RT} / \mathrm{Fa}}\) d. \(\operatorname{In} \mathrm{k}=\operatorname{In} \mathrm{A}+\mathrm{Ea} / \mathrm{RT}\)

The calculation of the Arrhenius factor is based on the a. Idea that the reactant species must come together, leading to the formation of the transition state which then transforms into the products b. Idea that, for a reaction to take place, the reactant species must come together c. Calculation of the order of thereaction d. Calculation of the molecularity of the reaction

The rate law has the form; Rate \(=\mathrm{k}[\mathrm{A}][\mathrm{B}]^{3 / 2}\), can the reaction be an elementary process? a. yes b. no c. may be yes or no d. can not be predicted
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