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Chapter 14: Problem 29

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is $2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$. The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(55^{\circ} \mathrm{C}\) the rate constant is \(4.12 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.050 \mathrm{M} ?(\mathbf{c})$ What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is tripled to $0.150 \mathrm{M}$ ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is reduced by \(10 \%\) to \(0.045 \mathrm{M}\) ?

Short Answer

Expert verified
(a) The rate law for the reaction is: Rate = k [N₂O₅], where k = 4.12 x 10⁻³ s⁻¹ at 55°C. (b) The rate of reaction when [N₂O₅] = 0.050 M is 2.06 x 10⁻⁴ M/s. (c) When the concentration of N₂O₅ is tripled to 0.150 M, the rate of reaction increases to 6.18 x 10⁻⁴ M/s. (d) When the concentration of N₂O₅ is reduced by 10% to 0.045 M, the rate of reaction decreases to 1.85 x 10⁻⁴ M/s.

Step by step solution

01

(a) Write the rate law for the reaction

Since the reaction is first-order in N₂O₅, the rate law for the reaction is: Rate = k [N₂O₅] where Rate is the rate of the reaction, k is the rate constant (4.12 x 10⁻³ s⁻¹ at 55°C), and [N₂O₅] is the concentration of N₂O₅.
02

(b) Calculate the rate of reaction when [N₂O₅] = 0.050 M

Using the rate law, substitute the rate constant k and the concentration of N₂O₅: Rate = (4.12 x 10⁻³ s⁻¹) (0.050 M) Rate = 2.06 x 10⁻⁴ M/s
03

(c) What happens to the rate when the concentration of N₂O₅ is tripled to 0.150 M

Since the rate law is first-order in N₂O₅, if the concentration is tripled, the rate of reaction will also triple. Use the rate law to calculate the new rate: Rate = (4.12 x 10⁻³ s⁻¹) (0.150 M) Rate = 6.18 x 10⁻⁴ M/s The rate of reaction increases to 6.18 x 10⁻⁴ M/s.
04

(d) What happens to the rate when the concentration of N₂O₅ is reduced by 10% to 0.045 M

When the concentration of N₂O₅ is reduced by 10% to 0.045 M, the rate of reaction will be 90% of the rate at 0.050 M concentration. Calculate the new rate using the rate law: Rate = (4.12 x 10⁻³ s⁻¹) (0.045 M) Rate = 1.85 x 10⁻⁴ M/s The rate of reaction decreases to 1.85 x 10⁻⁴ M/s when the concentration is reduced by 10% to 0.045 M.

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Most popular questions from this chapter

A certain enzyme catalyzes a biochemical reaction. In water, without the enzyme, the reaction proceeds with a rate constant of $6.50 \times 10^{-4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C} .$ In the presence of the enzyme in water, the reaction proceeds with a rate constant of $1.67 \times 10^{4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C}$. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

(a) A certain first-order reaction has a rate constant of $2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}\(. What is the value of \)k$ at \(60^{\circ} \mathrm{C}\) if $E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\( Another first-order reaction also has a rate constant of \)2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}$. What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if $E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{c})$ What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?

The reaction $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\( is second order in \)\mathrm{NO}$ and first order in \(\mathrm{O}_{2} .\) When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is $9.3 \times 10^{-5} \mathrm{M} / \mathrm{s} .(\mathbf{a})\( What is the rate of disappearance of \)\mathrm{O}_{2}$ at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8 ?\)
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