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The overall equation and rate law for the gas-phase decomposition of dinitrogen pentoxide are \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad\) rate \(=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Which of the following can be considered valid mechanisms for the reaction? I One-step collision II \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}_{3}(g)+2 \mathrm{NO}_{2}(g) \quad[\) slow \(]\) \(2 \mathrm{NO}_{3}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+2 \mathrm{O}(g)\) [fast] \(2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) [fast] III \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g)\) [fast] \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 3 \mathrm{NO}_{2}(g)+\mathrm{O}(g) \quad\) [slow] \(\mathrm{NO}_{3}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad[\) fast \(]\) \(\mathrm{IV} 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{3}(g)+3 \mathrm{O}(g) \quad[\) fast \(]\) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) [slow] \(2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g) \quad[\) fast \(]\) \(\mathrm{V} \quad 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{N}_{4} \mathrm{O}_{10}(g)\) [slow] \(\mathrm{N}_{4} \mathrm{O}_{10}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) [fast]

Short Answer

Expert verified
Only Mechanism III is valid.

Step by step solution

01

Understand the Given Rate Law

The rate law given is: \(\text{rate} = k\left[\text{N}_2 \text{O}_5 \right]\). This implies that the reaction is first-order with respect to \(\text{N}_2 \text{O}_5 \).
02

Evaluate Mechanism I: One-step Collision

For a one-step mechanism to be valid, the rate law derived from the stoichiometry must match the experimental rate law. However, for a single-step, a bimolecular collision would imply a second-order rate law, which contradicts the given first-order rate law. Hence, Mechanism I is not valid.
03

Evaluate Mechanism II

The first step: \(2 \text{N}_2 \text{O}_5 (g) \longrightarrow 2 \text{NO}_3 (g) + 2 \text{NO}_2 (g) \ [\text{slow}] \) suggests that the rate-determining step involves \( \left[ \text{N}_2 \text{O}_5 \right]^2 \), which implies a second-order rate law. This does not match the first-order rate law. Therefore, Mechanism II is not valid.
04

Evaluate Mechanism III

The second step: \( \text{NO}_2 (g) + \text{N}_2 \text{O}_5 (g) \longrightarrow 3 \text{NO}_2 (g) + \text{O} (g) \ [\text{slow}] \) matches the given rate law. The rate-determining step involves one molecule of \( \text{N}_2 \text{O}_5 \), giving \( \text{rate} = k \left[\text{N}_2 \text{O}_5 \right] \). Hence, Mechanism III is valid.
05

Evaluate Mechanism IV

The first step: \(2 \text{N}_2 \text{O}_5 (g) \rightleftharpoons 2 \text{NO}_2 (g) + \text{N}_2 \text{O}_3 (g) + 3 \text{O} (g) \ [\text{fast}] \), followed by the slow step involving \( \text{N}_2 \text{O}_3 \), would imply a rate law involving more than one molecule of \( \text{N}_2 \text{O}_5 \), which would not match the first-order rate law. Therefore, Mechanism IV is not valid.
06

Evaluate Mechanism V

The first step: \(2 \text{N}_2 \text{O}_5 (g) \longrightarrow \text{N}_4 \text{O}_10 (g) \ [\text{slow}] \), involves two molecules of \( \text{N}_2 \text{O}_5 \), implying a rate law of the form \( \text{rate} = k \left[\text{N}_2 \text{O}_5 \right]^2 \). This does not match the given first-order rate law. Thus, Mechanism V is not valid.

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Key Concepts

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

reaction mechanisms
Reaction mechanisms detail the step-by-step sequence through which reactants are transformed into products. For the decomposition of dinitrogen pentoxide (N2O5), the overall reaction is:
2 N2O5(g) → 4 NO2(g) + O2(g).
A mechanism must align with the observed rate law. A reaction can occur in a single step or multiple steps. These steps can be classified as 'fast' or 'slow' where the slow step is often the rate-determining step, influencing the overall reaction rate. In our case, a valid mechanism must produce the given rate law: rate = k[ N2 O5].
  • Mechanism I, a one-step collision, assumes a single-step reaction. This would result in a second-order rate law (rate = k[ N2 O5] 2), which does not match our first-order rate law.
  • Mechanism II suggests multiple steps, with a slow initial step involving two N2 O5 molecules. This again implies a second-order rate law, making it invalid.
  • Mechanism III proposes a fast equilibrium followed by a slow step. This slow step matches the observed rate, making this mechanism valid.
  • Mechanisms IV and V both imply rate laws that do not align with the given first-order rate law, thus they are invalid.
rate law
Rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For a reaction: aA + bB → products, the rate law could be written as: rate = k[A]^m [B]^n. Here, 'k' is the rate constant, and 'm' and 'n' are the reaction orders with respect to each reactant, determined experimentally.
For the decomposition of N2O5, the given rate law is rate = k[ N2 O5]. This signifies the reaction is first-order with respect to N2O5, meaning the rate depends linearly on the concentration of N2O5.
Let’s evaluate:
  • Mechanism I suggests a one-step collision, implying a bimolecular process, leading to a second-order rate law which contradicts the given first-order law.
  • Mechanism II, with a slow step involving 2 N2 O5 molecules, suggests a second-order rate law, disqualifying it.
  • Mechanism III has a slow step involving one N2 O5 molecule, correctly producing a first-order rate law, thus it fits our requirement.
  • Mechanism IV and V also do not match the first-order rate law because they involve steps suggesting higher-order rate laws.
kinetics
Kinetics studies the speed of chemical reactions and the factors affecting this speed. For the decomposition of N2O5, the kinetics is governed by the rate law:
rate = k[ N2 O5].
  • The observed first-order kinetics implies that the reaction rate depends directly on the concentration of N2O5.
  • Total reaction time is influenced by the slowest step in the reaction mechanism, the rate-determining step.
Mechanisms:
  • In Mechanism I, a single-step bimolecular collision does not align with the kinetic data, as it suggests a second-order process.
  • Mechanism II’s initial slow step also predicts a second-order rate, conflicting with our first-order observation.
  • Mechanism III accurately reflects the kinetic data: the slow step dictates a first-order dependence on N2O5, validating its position as the correct mechanism.
  • Mechanisms IV and V also fail to align with the kinetic data due to suggested higher-order dependence on N2O5.
To conclude, understanding the rate-determining step in kinetics is essential for establishing accurate rate laws and validating reaction mechanisms.

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Most popular questions from this chapter

16.89 A slightly bruised apple will rot extensively in about 4 days at room temperature \(\left(20^{\circ} \mathrm{C}\right)\). If it is kept in the refrigerator at \(0^{\circ} \mathrm{C}\), the same extent of rotting takes about 16 days. What is the activation energy for the rotting reaction?

Give the individual reaction orders for all substances and the overall reaction order from this rate law: $$ \text { Rate }=k\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{Cl}_{2}\right] $$

Heat transfer to and from a reaction flask is often a critical factor in controlling reaction rate. The heat transferred \((q)\) depends on a heat transfer coefficient \((h)\) for the flask material, the temperature difference \((\Delta T)\) across the flask wall, and the commonly "wetted" area (A) of the flask and bath: \(q=h A \Delta T\). When an exothermic reaction is run at a given \(T,\) there is a bath temperature at which the reaction can no longer be controlled, and the reaction "runs away" suddenly. A similar problem is often seen when a reaction is "scaled up" from, say, a half-filled small flask to a half-filled large flask. Explain these behaviors.

In a first-order decomposition reaction, \(50.0 \%\) of a compound decomposes in \(10.5 \mathrm{~min}\). (a) What is the rate constant of the reaction? (b) How long does it take for \(75.0 \%\) of the compound to decompose?

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ 2 \mathrm{D}(g)+3 \mathrm{E}(g)+\mathrm{F}(g) \longrightarrow 2 \mathrm{G}(g)+\mathrm{H}(g) $$ When [D] is decreasing at \(0.1 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\), how fast is [H] increasing?

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