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Chapter 13: Problem 73

Hydrogen iodide decomposes slowly to \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) at \(600 \mathrm{~K}\). The reaction is second order in HI and the rate constant is \(9.7 \times 10^{-6} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Assume the initial concentration of HI is \(0.100 \mathrm{M}\). (a) What is its molarity after a reaction time of \(6.00\) days? (b) What is the time (in days) when the HI concentration reaches a value of \(0.085 \mathrm{M} ?\)

Short Answer

Expert verified
(a) Molarity after 6 days: 0.00508 M. (b) Time to reach 0.085 M: 2.10 days.

Step by step solution

01

Identify Known Values

For both parts (a) and (b), note that the reaction is second-order in HI with a rate constant \( k = 9.7 \times 10^{-6} \ \mathrm{M}^{-1} \mathrm{s}^{-1} \). The initial concentration \( [\mathrm{HI}]_0 = 0.100 \ \mathrm{M} \).
02

Apply Second-Order Reaction Formula

For a second-order reaction, use the formula: \[\frac{1}{[\mathrm{HI}]} - \frac{1}{[\mathrm{HI}]_0} = k \cdot t\]where \([\mathrm{HI}]\) is the concentration after time \( t \).
03

Convert Days to Seconds

Since the rate constant \( k \) is in seconds, convert 6 days into seconds for part (a): \[6 \times 24 \times 60 \times 60 = 518400 \ \mathrm{s}\]
04

Solve for Molarity After 6 Days (Part a)

Substitute known values for part (a):\[\frac{1}{[\mathrm{HI}]} - \frac{1}{0.100} = 9.7 \times 10^{-6} \times 518400\]Solve for \([\mathrm{HI}]\): \[\frac{1}{[\mathrm{HI}]} \approx 197 \]\([\mathrm{HI}] \approx 0.00508 \ \mathrm{M}\)
05

Set Up Equation for Part b

For part (b), when \([\mathrm{HI}] = 0.085 \ \mathrm{M}\), use the same formula:\[\frac{1}{0.085} - \frac{1}{0.100} = 9.7 \times 10^{-6} \cdot t\]
06

Solve for Time (Seconds)

Calculate time in seconds:\[\frac{1}{0.085} - 10 = 9.7 \times 10^{-6} \cdot t\]\[11.76 - 10 = 9.7 \times 10^{-6} \cdot t\]\[t \approx 181445 \ \mathrm{s}\]
07

Convert Time to Days

Convert seconds to days for part (b): \[\frac{181445}{24 \times 3600} \approx 2.10 \ \mathrm{days}\]

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Key Concepts

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

Rate Constant
The rate constant, often denoted by the symbol \( k \), is a crucial factor in chemical kinetics that quantifies the speed of a chemical reaction. It provides us with an understanding of how fast or slow a reaction proceeds at a given temperature. In the original exercise, the rate constant for the decomposition of hydrogen iodide (HI) is given as \( 9.7 \times 10^{-6} \, \mathrm{M}^{-1} \, \mathrm{s}^{-1} \). This signifies that the reaction rate depends heavily on the concentration of HI.
  • A higher rate constant implies a faster reaction.
  • The units of the rate constant differ based on the order of the reaction, providing insight into its kinetics. For a second-order reaction, it's expressed in \( \mathrm{M}^{-1} \, \mathrm{s}^{-1} \).
  • The rate constant \( k \) remains unchanged as long as the reaction conditions, such as temperature, remain constant.
The rate constant is therefore fundamental to predicting how long it will take for a reaction to reach a certain stage based on the concentration of reactants.
Reaction Kinetics
Reaction kinetics studies the rate at which reactions occur and how different factors affect these rates. One essential aspect is determining the reaction order, which in the case of hydrogen iodide is second order. This tells us that the rate of the reaction is proportional to the square of the concentration of HI. Here's how we use this information:
  • For a second-order reaction, the formula used is \( \frac{1}{[HI]} - \frac{1}{[HI]_0} = k \cdot t \). This helps calculate how concentration changes over time.
  • The kinetics framework allows us to analyze how quickly or slowly a reaction might proceed and offers with it strategies to optimize conditions for desired outcomes, whether in industry or lab studies.
  • Converting real-time measurements such as days to seconds is often necessary, aligning with the units of the rate constant to ensure accurate calculations.
In summary, reaction kinetics is a detailed assessment of a reaction's speed and the factors that may influence it, setting the stage for comprehensive analyses and predictions in various chemical processes.
Molarity Calculation
Molarity, defined as moles of solute per liter of solution, is a measure of concentration used throughout chemistry. Calculating molarity accurately enables chemists to understand how much of a reactant is available in the solution for reaction. In the scenario of our original exercise, determining molarity after a certain reaction time is key to understanding the reaction progress.
  • To find the molarity of HI after a given time, the second-order formula \( \frac{1}{[HI]} - \frac{1}{[HI]_0} = k \cdot t \) is rearranged and solved for \([HI]\).
  • The initial molarity \( [HI]_0 \) is given, and the kinetic equation allows us to solve for the remaining concentrations over specified times.
  • This calculated molarity provides insight into the consumption rate of reactants and can be used to gauge the extent of conversion to products.
Having a clear understanding of molarity and how to calculate it using reaction kinetics helps ensure precise and systematic monitoring of chemical reactions.

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Most popular questions from this chapter

The half-life for the first-order decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(1.3 \times 10^{-5} \mathrm{~s}\) $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ If \(\mathrm{N}_{2} \mathrm{O}_{4}\) is introduced into an evacuated flask at a pressure of \(17.0 \mathrm{~mm} \mathrm{Hg}\), how many seconds are required for the pressure of \(\mathrm{NO}_{2}\) to reach \(1.3 \mathrm{~mm} \mathrm{Hg} ?\)

If the rate of a reaction increases by a factor of \(2.5\) when the temperature is raised from \(20^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\), what is the value of the activation energy in \(\mathrm{kJ} / \mathrm{mol?}\) By what factor does the rate of this reaction increase when the temperature is raised from \(120^{\circ} \mathrm{C}\) to \(130^{\circ} \mathrm{C}\).

Bromomethane is converted to methanol in an alkaline solution. The reaction is first order in each reactant. $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$(a) Write the rate law. (b) How does the reaction rate change if the OH concentration is decreased by a factor of \(5 ?\) (c) What is the change in rate if the concentrations of both reactants are doubled?

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the hightemperature reaction $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How is the rate of consumption of \(\mathrm{O}_{2}\) related to the rate of consumption of \(\mathrm{NH}_{3} ?\) (b) How are the rates of formation of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) related to the rate of consumption of \(\mathrm{NH}_{3} ?\)

At high substrate concentrations, the rate of product formation is zeroth order in S. (a) By what factor does the rate of an enzyme-catalyzed reaction change when the substrate concentration is changed from \(2.8 \times 10^{-3} \mathrm{M}\) to \(4.8 \times 10^{-3} \mathrm{M} ?\) (b) Why is the enzyme-catalyzed reaction zero order in \(\mathrm{S}\) at high substrate concentrations?
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