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Hydrogen iodide decomposes when heated, forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g}) .\) The rate law for this reaction is \(-\Delta[\mathrm{HI}] / \Delta t=k[\mathrm{HI}]^{2} .\) At \(443^{\circ} \mathrm{C}, k=30 . \mathrm{L} / \mathrm{mol} \cdot\) min. If the initial \(\mathrm{HI}(\mathrm{g})\) concentration is \(3.5 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) what concentration of HI (g) will remain after \(10 .\) minutes?

Short Answer

Expert verified
The concentration of HI after 10 minutes is approximately 0.00305 mol/L.

Step by step solution

01

Understand the Rate Law

The rate law for the decomposition of HI is given by \(-\Delta[\mathrm{HI}] / \Delta t=k[\mathrm{HI}]^{2}\). This indicates it is a second-order reaction with respect to HI.
02

Identify the Integrated Rate Law for Second-Order Reactions

For a second-order reaction: \[\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\] where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
03

Substitute Known Values

Insert the known values into the integrated rate equation: \[\frac{1}{[HI]_t} = \frac{1}{3.5 \times 10^{-2}} + (30)(10)\]
04

Calculate the Concentration

First, calculate \(\frac{1}{3.5 \times 10^{-2}} = 28.57\). Then, multiply \(30\times 10 = 300\). Add these values: \[\frac{1}{[HI]_t} = 28.57 + 300 = 328.57\]
05

Determine the Final Concentration

Take the reciprocal to find \([HI]_t\): \([HI]_t = \frac{1}{328.57} \approx 0.00305 \text{ mol/L}\).

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Key Concepts

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

Rate law
The rate law is an equation that relates the reaction rate with the concentration of reactants. For a second-order reaction involving hydrogen iodide (HI), the rate law is expressed as \(-\Delta[\text{HI}] / \Delta t = k[\text{HI}]^2\). This equation helps us understand how the speed of the reaction changes as the concentration of HI changes.
In this format:
  • \(-\Delta[\text{HI}] / \Delta t\) represents the rate of the reaction, essentially how fast or slow the HI is being consumed over time.
  • \(k\) is the rate constant, which provides a relationship between the concentration of HI and the reaction rate. The value of \(k\) affects how swiftly the reaction progresses, but it does not change with varying concentrations.
  • \([\text{HI}]^2\) depicts the concentration of HI raised to the power of two. This indicates the reaction's dependence on the square of the HI concentration.
The factor of HI squared signifies that doubling the concentration of HI will quadruple the reaction rate, highlighting that this is a second-order reaction.
Integrated rate law
For a second-order reaction, the integrated rate law formulates how the concentration of reactants decrease over time. The equation for a second-order reaction given as: \[\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\]This equation is effective in predicting the concentration of a reactant, like HI, after some time.
Here:
  • \([A]_t\) is the concentration of HI at time \(t\).
  • \([A]_0\) is the initial concentration of HI at time zero.
  • \(k\) is the rate constant, the same as in the rate law equation.
  • \(t\) represents the elapsed time.
Utilizing this formula, you can rearrange and substitute known values to determine the concentration of HI at any point in time. The integrated rate law offers a way to measure how much HI remains after a specific duration, offering a predictive vision into the reaction process.
Decomposition reaction
A decomposition reaction involves breaking a compound into simpler products. For hydrogen iodide (HI), this process occurs through heat, producing hydrogen (H₁) and iodine (I₂). Such reactions are common in chemistry when a compound is unstable or external factors like heat influence it to decompose.
For HI's decomposition:
  • \(\text{HI} \rightarrow \text{H}_2 + \text{I}_2\)
When heated, HI molecules split apart into their elemental forms, H₁ and I₂ gas. Typically, decomposition reactions depend directly on external factors like temperature, which in this case, is the primary driver.
Understanding how these reactions proceed is crucial, especially when contemplating reaction rates and mechanisms. This insight will help foresee changes in concentration and estimate how long the decomposition takes under certain conditions.
Rate constant
The rate constant \(k\) is a significant component in both the rate law and integrated rate law equations. It essentially dictates how fast the decomposition reaction will proceed under given conditions.
For the decomposition of hydrogen iodide, \(k = 30\, \text{L/mol}\cdot \text{min}\) at 443°C. This value holds under specified experimental circumstances, meaning changes in temperature or pressure might alter \(k\).
  • \(k\) is usually determined experimentally and is unique for each reaction.
  • In equations, it serves as a proportionality factor linking concentration and reaction rate.
While the rate constant remains steady over a specific range of conditions, different reactions will have distinct \(k\) values. Being able to compute \(k\) is crucial, as it contributes vastly to predicting reaction behaviors and planning chemical processes effectively.

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Most popular questions from this chapter

The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After 27 minutes at \(27^{\circ} \mathrm{C},\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

Draw a reaction coordinate diagram for an exothermic reaction that occurs in a single step. Identify the activation energy and the net energy change for the reaction on this diagram. Draw a second diagram that represents the same reaction in the presence of a catalyst, assuming a single step reaction is involved here also. Identify the activation energy of this reaction and the energy change. Is the activation energy in the two drawings different? Does the energy evolved in the two reactions differ?

The dimerization of butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) to form 1,5-cyclooctadiene is a second-order process that occurs when the diene is heated. In an experiment, a sample of 0.0087 mol of \(\mathrm{C}_{4} \mathrm{H}_{6}\) was heated in a \(1.0-\mathrm{L}\) flask. After 600 seconds, \(21 \%\) of the butadiene had dimerized. Calculate the rate constant for this reaction.

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) is a first-order reaction. If \(2.56 \mathrm{mg}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is present initially and \(2.50 \mathrm{mg}\) is present after 4.26 minutes at \(55^{\circ} \mathrm{C},\) what is the value of the rate constant, \(k ?\)

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2},\) with a half-life of \(30 .\) minutes at room temperature: $$\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ If the partial pressure of HOF in a \(1.00-\mathrm{L}\). flask is initially \(1.00 \times 10^{2} \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C},\) what are the total pressure in the flask and the partial pressure of HOF after exactly 30 minutes? After 45 minutes?

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