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The rate constant of a first-order reaction is \(4.60 \times\) \(10^{-4} \mathrm{~s}^{-1}\) at \(350^{\circ} \mathrm{C}\). If the activation energy is 104 \(\mathrm{kJ} / \mathrm{mol}\), calculate the temperature at which its rate constant is \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The new temperature at which the rate constant of the reaction is \(8.80 × 10^{-4} s^{-1}\) is calculated by following the steps shown above using Arrhenius equation.

Step by step solution

01

Write down the known values

We are given three values: the first rate constant \(k_1 = 4.60 × 10^{-4} s^{-1}\), the activation energy \(E_a = 104 kJ/mol\) or \(104000 J/mol\) and the first temperature \(T_1 = 350°C = 350 + 273.15 = 623.15 K\), and we are trying to solve for the new temperature \(T_2\) at which the rate constant is \(k_2 = 8.80 × 10^{-4} s^{-1}\). The gas constant \(R = 8.314 JK^{-1}mol^{-1}\).
02

Solve for \(lnA\)

From Arrhenius' equation in logarithmic form, we can solve for \(lnA\) using the known values of \(k_1\), \(T_1\), \(R\) and \(E_a\). We have \(lnA = lnk_1 + \frac{E_{a}}{R}(\frac{1}{T_1})\). This gives us the pre-exponential factor, \(A\).
03

Substitute and Solve for \(T_2\)

Substitute the value of \(lnA\) into the logarithmic form of the Arrhenius equation for \(T_2\). Rearranging the equation gives us \(T_2 = \frac{1}{\frac{E_a}{R} - \frac{lnk_2}{lnA}}\) (after simplification). Substitute the respective values and solve for \(T_2\).

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Key Concepts

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

First-order Reactions
A first-order reaction is a type of chemical reaction where the reaction rate is directly proportional to the concentration of one reactant. This means that if you double the concentration of the reactant, the reaction rate doubles as well. Such reactions are described by a rate law of the form:
  • \( ext{Rate} = k[A] \)
Here, \( k \) is the rate constant, and \( [A] \) represents the concentration of the reactant.
First-order reactions are common in processes such as radioactive decay and simple chemical reactions. They are called 'first-order' because the rate depends linearly on the concentration of only one reactant.
Arrhenius Equation
The Arrhenius equation is a formula that describes how the rate constant \( k \) of a reaction depends on temperature and activation energy. It is expressed as:
  • \( k = Ae^{- rac{E_a}{RT}} \)
Where:
  • \( k \) is the rate constant,
  • \( A \) is the pre-exponential factor (frequency factor),
  • \( E_a \) is the activation energy in Joules per mole,
  • \( R \) is the universal gas constant (8.314 J/mol·K),
  • \( T \) is the temperature in Kelvin.
This equation shows that the rate constant increases with an increase in temperature or a decrease in activation energy. The Arrhenius equation helps understand how chemical reactions speed up as temperature rises.
Activation Energy
Activation energy is the minimum amount of energy required for a chemical reaction to occur. It acts as a barrier that reactants must overcome to transform into products. In the context of the Arrhenius equation, activation energy \( E_a \) has a critical role, as it affects the rate of the reaction.
Higher activation energy means that fewer molecules have sufficient energy to react at a given temperature, resulting in a slower reaction rate. Conversely, a lower activation energy means that more molecules can overcome the energy barrier, speeding up the reaction. Temperature impacts this as well, since increased temperature boosts the energy of molecules, allowing more of them to surpass the activation energy barrier.
Rate Constant
The rate constant \( k \) is a crucial factor that determines the speed of a chemical reaction. In the context of the Arrhenius equation, the rate constant changes with temperature and activation energy, reflecting the likelihood of reaction occurrences between molecules.
  • The larger the rate constant, the faster the reaction proceeds.
  • Rate constants can indicate the reaction mechanism and are unique to each reaction under specific conditions.
In the given exercise, the rate constant increases when the temperature changes, demonstrating the principle articulated by the Arrhenius equation. This is why calculating the rate constant at different temperatures is essential for studying reaction kinetics.

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Most popular questions from this chapter

Variation of the rate constant with temperature for the first-order reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(g)+\mathrm{O}_{2}(g) $$ is given in the following table. Determine graphically the activation energy for the reaction. $$ \begin{array}{lc} \mathrm{T}(\mathrm{K}) & \mathrm{k}\left(\mathrm{s}^{-1}\right) \\ \hline 273 & 7.87 \times 10^{3} \\ 298 & 3.46 \times 10^{5} \\ 318 & 4.98 \times 10^{6} \\ 338 & 4.87 \times 10^{7} \end{array} $$

In recent years ozone in the stratosphere has been depleted at an alarmingly fast rate by chlorofluorocarbons (CFCs). A CFC molecule such as \(\mathrm{CFCl}_{3}\) is first decomposed by UV radiation: $$ \mathrm{CFCl}_{3} \longrightarrow \mathrm{CFCl}_{2}+\mathrm{Cl} $$ The chlorine radical then reacts with ozone as follows: $$ \begin{array}{c} \mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{ClO}+\mathrm{O}_{2} \\ \mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2} \end{array} $$ (a) Write the overall reaction for the last two steps. (b) What are the roles of \(\mathrm{Cl}\) and \(\mathrm{ClO} ?\) (c) Why is the fluorine radical not important in this mechanism? (d) One suggestion to reduce the concentration of chlorine radicals is to add hydrocarbons such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) to the stratosphere. How will this Work?

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of the term "rate determining."

Briefly comment on the effect of a catalyst on each of the following: (a) activation energy, (b) reaction mechanism, (c) enthalpy of reaction, (d) rate of forward step, (e) rate of reverse step.

When a mixture of methane and bromine is exposed to light, the following reaction occurs slowly: $$ \mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g) $$ Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

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