/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 At low temperatures, the rate la... [FREE SOLUTION] | 91影视

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At low temperatures, the rate law for the reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)$$ is as follows: rate \(=\) constant \(\times\left[\mathrm{NO}_{2}\right]^{2}\). Which of the following mechanisms is consistent with the rate law? (a) \(\mathrm{CO}+\mathrm{NO}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}\) (b) \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4} \quad\) (fast) \(\mathrm{N}_{2} \mathrm{O}_{4}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{NO} \quad\) (slow) (c) \(2 \mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\) NO \(\quad\) (slow) \(\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} \quad\) (fast) (d) \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2} \quad\) (slow) \(\mathrm{O}_{2}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2} \quad\) (fast)

Short Answer

Expert verified
a) CO + NO鈧 鈫 CO鈧 + NO b) 2 NO鈧 鈬 N鈧侽鈧 (fast); N鈧侽鈧 + 2 CO 鈫 2 CO鈧 + 2 NO (slow) c) 2 NO鈧 鈫 NO鈧 + NO (slow); NO鈧 + CO 鈫 NO鈧 + CO鈧 (fast) d) 2 NO鈧 鈫 2 NO + O鈧 (slow); O鈧 + 2 CO 鈫 2 CO鈧 (fast) Answer: b) 2 NO鈧 鈬 N鈧侽鈧 (fast); N鈧侽鈧 + 2 CO 鈫 2 CO鈧 + 2 NO (slow)

Step by step solution

01

Understand the rate law

The given rate law is: rate = constant 脳 [NO鈧俔虏. This means that the rate of the reaction depends on the concentration of NO鈧, and it has an order of 2 with respect to NO鈧. This information will help us determine which mechanism matches this rate law.
02

Evaluate mechanism (a)

Mechanism (a) states that CO + NO鈧 鈫 CO鈧 + NO. However, this mechanism does not involve the square of the concentration of NO鈧, so it does not match the given rate law.
03

Evaluate mechanism (b)

Mechanism (b) has two steps: 1. 2 NO鈧 鈬 N鈧侽鈧 (fast) 2. N鈧侽鈧 + 2 CO 鈫 2 CO鈧 + 2 NO (slow) Since the slow step is what determines the overall reaction rate, in this mechanism, the reaction is first-order with respect to N鈧侽鈧 and first-order with respect to CO. However, since N鈧侽鈧 is formed from 2 NO鈧 molecules, if we substitute it with [NO鈧俔虏, this mechanism matches the given rate law.
04

Evaluate mechanism (c)

Mechanism (c) has two steps: 1. 2 NO鈧 鈫 NO鈧 + NO (slow) 2. NO鈧 + CO 鈫 NO鈧 + CO鈧 (fast) In this mechanism, the rate law is first-order with respect to NO鈧 and first-order with respect to CO. As the NO鈧 intermediate is not related to [NO鈧俔虏, this mechanism does not match the given rate law.
05

Evaluate mechanism (d)

Mechanism (d) has two steps: 1. 2 NO鈧 鈫 2 NO + O鈧 (slow) 2. O鈧 + 2 CO 鈫 2 CO鈧 (fast) In this mechanism, the rate law is first-order with respect to O鈧 and first-order with respect to CO. As the O鈧 intermediate is not related to [NO鈧俔虏, this mechanism does not match the given rate law.
06

Conclusion

From our analysis of the four different mechanisms, we determined that mechanism (b) matches the given rate law as it results in the expression: rate = constant 脳 [NO鈧俔虏. Therefore, the correct answer is (b): 2 NO鈧 鈬 N鈧侽鈧 (fast) N鈧侽鈧 + 2 CO 鈫 2 CO鈧 + 2 NO (slow)

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Key Concepts

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

Reaction Mechanism
In chemical kinetics, understanding the reaction mechanism gives us insight into how reactants transform into products. A reaction mechanism represents a detailed step-by-step pathway that a reaction follows. It is crucial to distinguish between the individual steps, which help us understand how the reacting molecules interact and change over time. In the context of the exercise, multiple proposed mechanisms were examined to see which aligns with the given rate law of the reaction.
  • A good reaction mechanism should be able to explain the sequence of elementary steps and predict the reaction rate.
  • Each step in a mechanism is an elementary reaction and is characterized by its own stepwise rate law.
  • The slowest step, often called the rate-determining step, largely controls the overall reaction rate.
Mechanism (b) was identified as consistent with the rate law given because it properly combines the elementary steps to reflect how the concentration of the reactants affects the rate.
Rate Law
The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. Derived from experimental data, it takes the form: rate = constant 脳 [Reactant1]鈦 脳 [Reactant2]岬...
  • These exponents represent the power to which the concentration of each reactant must be raised and are not necessarily the coefficients in the balanced equation.
  • In the provided problem, the rate law is given as rate = constant 脳 [NO鈧俔虏, indicating it is second-order with respect to NO鈧.
  • The rate constant is specific to a particular reaction at a certain temperature and must be evaluated empirically.
The rate law is crucial because it helps predict how changes in reactant concentration affect the speed of the reaction and aids in understanding the reaction mechanism itself.
Reaction Order
The reaction order tells us how the concentration of a reactant affects the rate of reaction and is an important part of a rate law. It is the sum of the powers of the concentration terms in the rate equation.
  • For example, if the rate law is rate = constant 脳 [A]虏 脳 [B]鹿, the reaction order is 2+1=3 (third order).
  • Reaction order can be an integer or a fraction, depending on the reaction type.
  • It offers insights into the reaction mechanism; for instance, a second-order reaction may suggest that two molecules must collide to proceed.
In this problem, the reaction is second-order relative to NO鈧, meaning its concentration squared directly affects the rate. Understanding this concept helps determine which reaction mechanism is correct, as seen with mechanism (b), where the overall order was a good fit for the observed rate law.
Intermediates
Intermediates play a key role in multi-step reactions by connecting sequential steps, appearing as products in one step and reactants in another. Although they are crucial for the reaction mechanism, intermediates do not appear in the overall balanced chemical equation.
  • They provide a snapshot of the molecular interaction happening during a reaction.
  • Intermediates are often highly reactive and short-lived, making their direct study difficult.
  • A correct reaction mechanism should predict the formation and consumption of intermediates accurately.
In our analysis, any proposed mechanism had to account for potential intermediates that matched the overall rate law. In mechanism (b), N鈧侽鈧 acted as an intermediate, consistent with the observed rate law when accounting for the formation's dependence on NO鈧. This understanding helps validate the mechanism as aligning well with the empirical data.

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Most popular questions from this chapter

The activation energy for the reaction involved in the souring of raw milk is \(75 \mathrm{~kJ}\). Milk will sour in about eight hours at \(21^{\circ} \mathrm{C}\left(70^{\circ} \mathrm{F}=\right.\) room temperature). How long will raw milk last in a refrigerator maintained at \(5^{\circ} \mathrm{C}\) ? Assume the rate constant to be inversely related to souring time.

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The equation for the iodination of acetone in acidic solution is $$\mathrm{CH}_{3} \mathrm{COCH}_{3}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)$$ The rate of the reaction is found to be dependent not only on the concentration of the reactants but also on the hydrogen ion concentration. Hence the rate expression of this reaction is $$\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]^{m}\left[\mathrm{I}_{2}\right]^{n}\left[\mathrm{H}^{+}\right]^{p}$$ The rate is obtained by following the disappearance of iodine using starch as an indicator. The following data are obtained: $$ \begin{array}{cccc} \hline\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right] & \left.\mathrm{[H}^{+}\right] & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 0.80 & 0.20 & 0.001 & 4.2 \times 10^{-6} \\ 1.6 & 0.20 & 0.001 & 8.2 \times 10^{-6} \\ 0.80 & 0.40 & 0.001 & 8.7 \times 10^{-6} \\ 0.80 & 0.20 & 0.0005 & 4.3 \times 10^{-6} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\). (d) What is the rate of the reaction when \(\left[\mathrm{H}^{+}\right]=0.933 M\) and \(\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]=3\left[\mathrm{H}^{+}\right]=10\left[\mathrm{I}^{-}\right] ?\)

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