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For the following reaction at a particular temperature which takes place as- follows \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) \(2 \mathrm{NO}_{2}+1 / 2 \mathrm{O} 2 \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}\) The value of activation energies are \(E_{1}\) and \(E_{2}\) respectively then a. \(\mathrm{E}_{1}>\mathrm{E}_{2}\) b. \(\mathrm{E}_{1}=2 \mathrm{E}_{2}\) c. \(2 \mathrm{E}_{1}=\mathrm{E}_{2}\) d. \(\mathrm{E}_{1}<\mathrm{E}_{2}\)

Short Answer

Expert verified
d. \(E_1 < E_2\)

Step by step solution

01

Understand the Reactions

The given reactions are: 1) Forward Reaction: \(2 \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) and 2) Reverse Reaction: \(2 \mathrm{NO}_{2}+1/2 \mathrm{O}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}\). We are comparing the activation energies for these reactions, denoted as \(E_1\) for the forward reaction and \(E_2\) for the reverse reaction.
02

Analyze Reaction Direction

In chemical reactions, the forward reaction of breaking down a compound (\(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposing into \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\)) typically has a smaller activation energy compared to the reverse reaction where products recombine to form the original compound. Generally, building a more complex compound (as in the reverse reaction) requires more energy.
03

Compare Activation Energies

Since forming \(\mathrm{N}_{2} \mathrm{O}_{5}\) from \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) is more energy-intensive than decomposing \(\mathrm{N}_{2} \mathrm{O}_{5}\) to \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\), \(E_2\) (reverse reaction) is greater than \(E_1\) (forward reaction). Thus, we conclude that \(E_1 < E_2\).

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where substances called reactants transform into different substances known as products. In the given exercise, the reactants are substances like \(\mathrm{N}_2\mathrm{O}_5\) and the products result after decomposition, such as \(\mathrm{NO}_2\) and \(\mathrm{O}_2\).

During a chemical reaction, bonds between atoms are broken and new bonds are formed, leading to a change in the molecular structure of the involved chemicals.

Some key points about chemical reactions are:
  • They can be fast, like explosion-type reactions, or slow, like rust formation.
  • Catalysts are often used to speed up reactions without being consumed in the process.
  • Temperature, pressure, concentration, and the presence of catalysts can influence the rate and outcome of a reaction.
Understanding these factors helps in predicting how a reaction will proceed and what conditions are optimal for it.
Forward and Reverse Reactions
Forward and reverse reactions refer to the direction in which chemical reactions proceed. A forward reaction is when reactants are converted to products, as in the decomposition of \(\mathrm{N}_2\mathrm{O}_5\) into \(\mathrm{NO}_2\) and \(\mathrm{O}_2\). The reverse reaction, conversely, takes the products and converts them back into the original reactants, as when \(\mathrm{NO}_2\) and \(\mathrm{O}_2\) recombine to form \(\mathrm{N}_2\mathrm{O}_5\).

Some critical points about these reactions:
  • In a closed system, forward and reverse reactions can reach a state of equilibrium where the rate of the forward reaction equals the rate of the reverse reaction.
  • The dynamic balance in equilibrium doesn’t mean that the concentrations of reactants and products are equal, but rather that their ratios remain constant.
  • The concept of equilibrium is essential in understanding how reactions occur in real-world scenarios, such as in industrial chemical processes.
Energy Comparison in Reactions
The activation energy of a reaction is the minimum energy required for the reactants to undergo a transformation into products. This energy barrier must be overcome for a reaction to proceed. In the given reactions, we compare the activation energies \(E_1\) for the forward reaction and \(E_2\) for the reverse reaction.

Key aspects regarding the energy comparison include:
  • The forward reaction \(2\mathrm{N}_2\mathrm{O}_5 \rightarrow 4\mathrm{NO}_2+\mathrm{O}_2\) generally has a lower activation energy \(E_1\) compared to the reverse reaction \(2\mathrm{NO}_2+1/2\mathrm{O}_2 \rightarrow \mathrm{N}_2\mathrm{O}_5\) where \(E_2\) is higher.
  • Typically, decomposition reactions (breaking complex molecules into simpler ones) have lower activation energies than synthesis reactions (forming complex molecules from simpler ones).
  • The relationship \(E_1
Understanding these energy requirements helps in designing processes that optimize energy use in industrial applications.

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Most popular questions from this chapter

A mechanism for a naturally occurring reaction that destroys ozone is: Step I: \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{HO}(\mathrm{g}) \rightarrow \mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) Step II: \(\mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{HO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) Which species is a catalyst? a. \(\mathrm{O}\) b. \(\mathrm{O}_{3}\) c. \(\mathrm{HO}_{2}\) d. \(\mathrm{HO}\)

Which of the following are not the permitted values of molecularity? a. 0 b. 2 c. 5 d. 1

The rate constant of a reaction is given by In \(\mathrm{k}\left(\mathrm{sec}^{-1}\right)\) \(=14.34-\left(1.25 \times 10^{4}\right) / \mathrm{T}\) What will be the energy of activation? a. \(24.83 \mathrm{kcal} \mathrm{mol}^{-1}\) b. \(49.66 \mathrm{kcal} \mathrm{mol}^{-1}\) c. \(12.42 \mathrm{kcal} / \mathrm{mol}\) d. none

From the following data for the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) \(\begin{array}{llll}{[\mathrm{A}]} & {[\mathrm{B}]} & \text { initial rate } & \left.(\mathrm{mol}]^{-1} \mathrm{~s}^{-1}\right) \\ \mathrm{mol} 1^{-1} & \mathrm{~mol} & \mathrm{l}^{-1} 300 \mathrm{~K} & 320 \mathrm{~K} \\ 2.5 \times 10^{-4} & 3.0 \times 10^{-5} & 5.0 \times 10^{-4} & 2.0 \times 10^{-3} \\\ 5.0 \times 10^{-4} & 6.0 \times 10^{-5} & 4.0 \times 10^{-3} & \- \\ 1.0 \times 10^{-3} & 6.0 \times 10^{-5} & 1.6 \times 10^{-2} & -\end{array}\) Calculate the rate of the equation. a. \(\mathrm{r}=\mathrm{k}[\mathrm{B}]^{1}\) b. \(\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}\) c. \(r=k[A]^{2}[B]^{1}\) d. \(\mathrm{r}=\mathrm{k}[\mathrm{A}][\mathrm{B}]\)

The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at \(295 \mathrm{~K}\) What is the rate law? $$ \begin{array}{lll} \hline\left[\mathrm{Cl}_{2}\right] & {[\mathrm{NO}]} & \begin{array}{l} \text { Initial rate } \\ \left(\mathrm{mol}^{-1} \mathrm{~s}^{-1}\right) \end{array} \\ \hline 0.05 & 0.05 & 1 \times 10^{-3} \\ 0.15 & 0.05 & 3 \times 10^{-3} \\ 0.05 & 0.15 & 9 \times 10^{-3} \\ \hline \end{array} $$ a. \(\mathrm{r}=\mathrm{k}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) b. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}[\mathrm{NO}]^{2}\) c. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) d. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}\)

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