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The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

Short Answer

Expert verified
The activation energy for the given reaction is approximately \(94.23 \, kJ/mol\).

Step by step solution

01

Write down the Arrhenius equation

We first write down the Arrhenius equation: \( k = Ae^{-Ea/RT}\).
02

Insert the given values and set up two equations

We now insert the given values into the Arrhenius equation. We get two equations: \(5.4*10^{-4} = Ae^{-Ea/(8.314*599)}\) and \(2.8*10^{-2} = Ae^{-Ea/(8.314*683)}\). Here we have used R = 8.314 J/(K*mol).
03

Divide the two equations

We divide the first equation by the second to get rid of A: \( \frac{5.4*10^{-4}}{2.8*10^{-2}} = \frac{e^{-Ea/(8.314*599)}}{e^{-Ea/(8.314*683)}} \)
04

Simplify and solve for Ea

This simplifies to \(0.0193 = e^{Ea/(8.314)*(1/683-1/599)}\). We then take the natural logarithm of both sides and solve for Ea: \(Ea= -8.314*ln(0.0193)/(1/683-1/599)\).
05

Calculate Ea

Plugging in the numbers gives us \(Ea=93830.77 \, J/mol \) or \(94.23\,kJ/mol\).

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Key Concepts

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

Arrhenius equation
The Arrhenius equation is a fundamental formula in chemical kinetics, illustrating how reaction rates change with temperature. It is expressed as \[ k = Ae^{-Ea/RT} \] where:
  • \( k \) is the rate constant of the reaction.
  • \( A \) is the pre-exponential factor, a constant for each chemical reaction.
  • \( E_a \) is the activation energy, the minimum energy required for a reaction to occur.
  • \( R \) is the universal gas constant (8.314 J/(mol·K)).
  • \( T \) is the temperature in Kelvin.
This equation helps us understand and calculate how sensitive a reaction is to changes in temperature. The exponential term \( e^{-Ea/RT} \) shows that even a small increase in temperature can significantly increase the rate constant if the activation energy is relatively high, as the fraction of molecules with enough energy to react increases.
Rate constant
The rate constant \( k \) is a crucial component in understanding chemical reactions. It is part of the kinetic rate law equation which dictates how fast a reaction proceeds: \[ ext{Rate} = k[ ext{Reactants}]^n \]where \( [ ext{Reactants}]^n \) represents the concentration of reactants raised to their respective order. The rate constant encapsulates all the variables influencing the speed of the reaction that aren't concentration-dependent, including:
  • Temperature: As seen in the Arrhenius equation, higher temperatures generally increase \( k \).
  • Nature of the reactants: Some substances react faster while others are more sluggish.
  • Presence of a catalyst: Catalysts lower the activation energy, thereby increasing \( k \).
Ultimately, \( k \) allows us to predict how quickly a reaction will occur under given conditions.
Temperature dependence of reaction rates
The rate of chemical reactions is highly dependent on temperature. Generally, increasing the temperature increases the reaction rate. This relationship is described quantitatively by the Arrhenius equation. Here's why temperature affects reactions:
  • Increased Molecular Motion: Higher temperatures give molecules more kinetic energy, causing them to move faster and collide more often.
  • Higher Energy Collisions: With increased kinetic energy, the likelihood of molecules overcoming the activation energy barrier increases, leading to more successful reactions.
  • Transition State Theory: The rate constant \( k \) is influenced by the frequency of collisions and the probability that the collisions will have enough energy to surpass the activation energy, both of which are enhanced by higher temperatures.
Understanding this concept is crucial for controlling and optimizing reactions in both industrial processes and laboratory settings.
Chemical kinetics
Chemical kinetics is the scientific discipline that studies the rates of chemical reactions, shedding light on how different conditions affect reaction speeds. By understanding kinetics, we can learn:
  • Reaction Mechanism: The step-by-step sequence of elementary reactions by which overall chemical change occurs.
  • Factors Affecting Rates: concentration of reactants, temperature, surface area, and the presence of catalysts.
  • Order of Reactions: describing how the rate depends on the concentration of reactants.
Chemical kinetics is not only theoretical but has practical applications in designing chemical processes, developing new materials, and understanding natural phenomena. Mastery of this subject allows scientists and engineers to predict the outcomes and speed of reactions, ultimately leading to innovations in technology and medicine.

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Most popular questions from this chapter

The following three-step mechanism has been proposed for the reaction of chlorine and chloroform. $$\begin{aligned} & \text { (1) } \quad \mathrm{Cl}_{2}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons_{k-1}} 2 \mathrm{Cl}(\mathrm{g})\\\ & \text { (2) } \quad \mathrm{Cl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(\mathrm{g})+\mathrm{CCl}_{3}(\mathrm{g})\\\ &\text { (3) } \quad \mathrm{CCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CCl}_{4}(\mathrm{g}) \end{aligned}$$ The numerical values of the rate constants for these steps are \(k_{1}=4.8 \times 10^{3} ; \quad k_{-1}=3.6 \times 10^{3} ; \quad k_{2}=1.3 \times 10^{-2} ; k_{3}=2.7 \times 10^{2} .\) Derive the rate law and the magnitude of \(k\) for the overall reaction.

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ Which of these sets of data corresponds to a (a) zero-order, (b) first-order, (c) second-order reaction?

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

We have used the terms order of a reaction and molecularity of an elementary process (that is, unimolecular, bimolecular). What is the relationship, if any, between these two terms?

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