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In a kinctics experiment, a chemist places crystals of iodine in a closed reaction vessel, introduces a given quantity of \(\mathrm{H}_{2}\) gas, and obtains data to calculate the rate of HI formation. In a second experiment, she uses the same amounts of iodine and hydrogen but first warms the flask to \(130^{\circ} \mathrm{C}\), a temperature above the sublimation point of iodine. In which of these experiments does the reaction proceed at a higher rate? Explain.

Short Answer

Expert verified
The reaction proceeds at a higher rate in the second experiment due to increased temperature and iodine being in the gas phase, which both enhance effective collisions.

Step by step solution

01

Understand the Problem

The task is to determine which of the two experiments proceeds at a higher rate of HI formation, one with iodine crystals and another with iodine gas at a higher temperature.
02

Consider the Effect of Temperature on Reaction Rate

The rate of a chemical reaction generally increases with temperature. According to the Arrhenius equation, higher temperatures increase the number of molecules that have sufficient energy to undergo the reaction.
03

Effect of Physical State

When iodine is in its gas phase (sublimated at 130°C), the molecules are more likely to collide with \(\text{H}_2\) molecules compared to when iodine is in its solid crystal form. This increases the effective collisions and hence the reaction rate.
04

Combine Insights

Combining the effects of higher temperature and the gas phase of iodine, the reaction is expected to proceed at a higher rate in the second experiment where iodine is sublimated to gas and the flask is warmed to 130°C.

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Key Concepts

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

kinetics experiment
In a kinetics experiment, chemists study the rate at which reactions occur. The reaction rate is determined by measuring the rate of formation of products or the rate of consumption of reactants.
For example, in your exercise, the chemist is examining the formation rate of hydrogen iodide (HI) from iodine crystals and hydrogen gas.
By analyzing reaction rates, chemists can infer the speed of chemical processes and understand the underlying mechanisms.
Important factors in a kinetics experiment include temperature, concentration of reactants, and physical state of reactants.
Arrhenius equation
The Arrhenius equation describes the relationship between the rate of a chemical reaction and temperature. It is given by:
\[ k = A e^{-E_a / RT} \]
Here, \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.
As temperature increases, the exponent \(-E_a / RT\) becomes less negative, leading to a higher value of \(k\).
This means that higher temperatures result in a faster reaction rate because more molecules have the required energy to overcome the activation energy barrier.
effect of temperature on reaction rate
Temperature has a significant effect on the rate of chemical reactions. Higher temperatures increase the kinetic energy of molecules, leading to more frequent and energetic collisions.
In the context of your exercise, when the flask is warmed to 130°C, the iodine sublimates and the reaction rate increases.
Here are some key points:
  • Increased molecular speed due to higher kinetic energy.
  • Higher probability of collisions between reactants.
  • Increased number of molecules with sufficient energy to react (as described by Arrhenius equation).
If you compare the two experiments, the reaction in the warmed flask happens faster.
physical state of reactants
The physical state of reactants can affect the rate of a chemical reaction. In general, reactions between gases or solutions occur faster than those involving solids.
In your exercise, iodine is in two different states: solid and gas.
  • Solid iodine has less surface area for collisions with hydrogen gas.
  • Gaseous iodine (obtained by warming the flask) mixes more thoroughly with hydrogen gas.
As a result, in the second experiment, where iodine is in the gas state, the reaction rate increases due to more effective collisions between reactants.
sublimation
Sublimation is the transition of a substance from the solid phase directly to the gas phase without passing through the liquid phase.
In your exercise, iodine undergoes sublimation when the flask is warmed to 130°C.
  • Sublimation increases the interaction between iodine and hydrogen gas.
  • This results in more frequent and effective collisions, accelerating the reaction.
Understanding sublimation helps explain why reactions involving gaseous iodine proceed faster compared to those with solid iodine.

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Most popular questions from this chapter

Reaction rate is expressed in terms of changes in concentration of reactants and products. Write a balanced equation for the reaction with this rate expression: $$ \text { Rate }=-\frac{\Delta\left[\mathrm{CH}_{4}\right]}{\Delta t}=-\frac{1}{2} \frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{H}, \mathrm{O}]}{\Delta t}=\frac{\Delta\left[\mathrm{CO}_{2}\right]}{\Delta t} $$

For the reaction \(4 \mathrm{~A}(g)+3 \mathrm{~B}(g) \longrightarrow 2 \mathrm{C}(g)\) the following data were obtained at constant temperature: $$ \begin{array}{cccc} \text { Experiment } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \end{array} & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 5.00 & 0.100 & 0.100 \\ 2 & 45.0 & 0.300 & 0.100 \\ 3 & 10.0 & 0.100 & 0.200 \\ 4 & 90.0 & 0.300 & 0.200 \end{array} $$ (a) What is the order with respect to each reactant? (b) Write the rate law. (c) Calculate \(k\) (using the data from Expt 1 ). (d) Using the value of \(k\) calculated in part (c), calculate the rate when \([\mathrm{A}]=\) \([\mathrm{B}]=0.400 \mathrm{~mol} / \mathrm{L}\).

Acetone is one of the most important solvents in organic chemistry, used to dissolve everything from fats and waxes to airplane glue and nail polish. At high temperatures, it decomposes in a first-order process to methane and ketene \(\left(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\right) .\) At \(600^{\circ} \mathrm{C},\) the rate constant is \(8.7 \times 10^{-3} \mathrm{~s}^{-1}\). (a) What is the half-life of the reaction? (b) How long does it take for \(40 . \%\) of a sample of acetone to decompose? (c) How long does it take for \(90 . \%\) of a sample of acetone to decompose?

The growth of Pseudomonas bacteria is modeled as a first-order process with \(k=0.035 \mathrm{~min}^{-1}\) at \(37^{\circ} \mathrm{C}\). The initial \(P\) seudomonas population density is \(1.0 \times 10^{3}\) cells/L. (a) What is the population density after \(2 \mathrm{~h}\) ? (b) What is the time required for the population to go from \(1.0 \times 10^{3}\) to \(2.0 \times 10^{3}\) cells/L?

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ \mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g}) $$ When [B] is decreasing at \(0.5 \mathrm{~mol} / \mathrm{L}\) -s, how fast is \([\mathrm{A}]\) decreasing?

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