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Chapter 11: Problem 79

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 + NO₂ → CO₂ + NO b) 2 NO₂ ⇌ N₂O₄ (fast); N₂O₄ + 2 CO → 2 CO₂ + 2 NO (slow) c) 2 NO₂ → NO₃ + NO (slow); NO₃ + CO → NO₂ + CO₂ (fast) d) 2 NO₂ → 2 NO + O₂ (slow); O₂ + 2 CO → 2 CO₂ (fast) Answer: b) 2 NO₂ ⇌ N₂O₄ (fast); N₂O₄ + 2 CO → 2 CO₂ + 2 NO (slow)

Step by step solution

01

Understand the rate law

The given rate law is: rate = constant × [NO₂]². This means that the rate of the reaction depends on the concentration of NO₂, and it has an order of 2 with respect to NO₂. This information will help us determine which mechanism matches this rate law.
02

Evaluate mechanism (a)

Mechanism (a) states that CO + NO₂ → CO₂ + NO. However, this mechanism does not involve the square of the concentration of NO₂, so it does not match the given rate law.
03

Evaluate mechanism (b)

Mechanism (b) has two steps: 1. 2 NO₂ ⇌ N₂O₄ (fast) 2. N₂O₄ + 2 CO → 2 CO₂ + 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 N₂O₄ and first-order with respect to CO. However, since N₂O₄ is formed from 2 NO₂ molecules, if we substitute it with [NO₂]², this mechanism matches the given rate law.
04

Evaluate mechanism (c)

Mechanism (c) has two steps: 1. 2 NO₂ → NO₃ + NO (slow) 2. NO₃ + CO → NO₂ + CO₂ (fast) In this mechanism, the rate law is first-order with respect to NO₃ and first-order with respect to CO. As the NO₃ intermediate is not related to [NO₂]², this mechanism does not match the given rate law.
05

Evaluate mechanism (d)

Mechanism (d) has two steps: 1. 2 NO₂ → 2 NO + O₂ (slow) 2. O₂ + 2 CO → 2 CO₂ (fast) In this mechanism, the rate law is first-order with respect to O₂ and first-order with respect to CO. As the O₂ intermediate is not related to [NO₂]², 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 × [NO₂]². Therefore, the correct answer is (b): 2 NO₂ ⇌ N₂O₄ (fast) N₂O₄ + 2 CO → 2 CO₂ + 2 NO (slow)

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Most popular questions from this chapter

The decomposition of \(\mathrm{Y}\) is a zero-order reaction. Its half-life at \(25^{\circ} \mathrm{C}\) and \(0.188 M\) is 315 minutes. (a) What is the rate constant for the decomposition of Y? (b) How long will it take to decompose a \(0.219 \mathrm{M}\) solution of \(\mathrm{Y}\) ? (c) What is the rate of the decomposition of \(0.188 \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) ? (d) Does the rate change when the concentration of \(\mathrm{Y}\) is increased to \(0.289 \mathrm{M}\) ? If so, what is the new rate?

The decomposition of dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{3}\right)\) to methane, carbon monoxide, and hydrogen gases is found to be first-order. At \(500^{\circ} \mathrm{C}\), a \(150.0\) -mg 35 sample of dimethyl ether is reduced to \(43.2 \mathrm{mg}\) after three quarters of an hour. Calculate (a) the rate constant. (b) the half-life at \(500^{\circ} \mathrm{C}\). (c) how long it will take to decompose \(95 \%\) of the dimethyl ether.

Iodine-131 is used to treat tumors in the thyroid. Its first-order half-life is \(8.1\) days. If a patient is given a sample containing \(5.00 \mathrm{mg}\) of \(\mathrm{I}-131\), how long will it take for \(25 \%\) of the isotope to remain in her system?

The decomposition of nitrosyl chloride $$\mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ is a second-order reaction. If it takes \(0.20 \mathrm{~min}\) to decompose \(15 \%\) of a \(0.300 \mathrm{M}\) solution of nitrosyl chloride, what is \(k\) for the reaction?

For the zero-order decomposition of HI on a gold surface $$\mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{I}_{2}(g)$$ it takes \(16.0 \mathrm{~s}\) for the pressure of HI to drop from \(1.00\) atm to \(0.200 \mathrm{~atm}\). (a) What is the rate constant for the reaction? (b) How long will it take for the pressure to drop from \(0.150 \mathrm{~atm}\) to \(0.0432\) atm? (c) What is the half-life of \(\mathrm{HI}\) at a pressure of \(0.500 \mathrm{~atm}\) ?
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