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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 \(104 \mathrm{~kJ} /\) mol, calculate the temperature at which its rate constant is \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The temperature at which the rate constant is \(8.80 \times 10^{-4} s^{-1}\) is approximately \(533 °C\).

Step by step solution

01

Understand the Arrhenius equation

The Arrhenius equation is \(k=Ae^{-Ea/RT}\) where \(k\) is the rate constant, \(A\) is the frequency factor, \(Ea\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature. Taking the natural logarithm on both sides, it becomes \(ln k = ln A - Ea/RT\). We will make use of this transformed equation.
02

Plug in the given values

We know two rate constants correspond to two temperatures. We can express them as \(ln k_1 = ln A - Ea/R \times 1/T_1\) and \(ln k_2 = ln A - Ea/R \times 1/T_2\). Subtracting the two equations gives \(ln (k_2/k_1) = Ea/R \times (1/T_1 - 1/T_2)\). We have the values for \(k_1\), \(k_2\), \(R\), \(Ea\) and \(T_1\). The only unknown is \(T_2\), which is what we're asked to find.
03

Solve for T_2

Rearrange the equation to solve for \(T_2\). This gives \(T_2 = 1/ ((1/T_1) - (R/Ea) \times ln(k_2/k_1)) \). Remember to convert the activation energy to J/mol (1 kJ = 1000 J), the gas constant R to its appropriate value in J/(mol*K) (R = 8.314 J/(mol*K)), and T_1 to Kelvin (T_1= 350°C+273=623 K). Then plug in \(k_1 = 4.60 × 10^{-4} s^{-1}\), \(Ea = 104 × 10^3 J/mol\), \(R = 8.314 J/(mol*K)\), \(T_1 = 623 K\) and \(k_2 = 8.80 × 10^{-4} s^{-1}\) to calculate \(T_2\).
04

Convert to Degree Celsius

The solution from Step 3 will yield the temperature in Kelvin. We need to convert this to Degree Celsius by subtracting by 273. This gives the final answer.

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Key Concepts

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

Understanding the Rate Constant
The rate constant, represented by the symbol \(k\), is a crucial component in chemical kinetics. It indicates the speed of a chemical reaction. Specifically, the rate constant is a measure of how quickly a reaction proceeds, and it varies with conditions like temperature and the presence of a catalyst.
In the context of the Arrhenius equation, the rate constant is not fixed but instead is dependent on temperature. This relationship is expressed as \(k = Ae^{-Ea/RT}\), where \(A\) is the frequency factor, \(Ea\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
It's important to understand that a higher value of \(k\) implies a faster reaction, assuming the reactant concentrations are constant. In our exercise, we see two different rate constants, \(4.60 \times 10^{-4} \mathrm{~s}^{-1}\) at \(350^{\circ}\mathrm{C}\) (623 K), and a higher rate constant \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\) at another unknown temperature. This change highlights the temperature dependence of reaction rates.
What is Activation Energy?
Activation energy, symbolized as \(Ea\), is the minimum energy that reactant molecules need in order to undergo a chemical transformation to products. It can be thought of as the barrier that must be overcome for a reaction to proceed.
This concept is crucial in relation to reaction rates. The higher the activation energy, the slower the reaction will be at a given temperature because fewer molecules have the necessary energy to cross the barrier.
Within the Arrhenius equation, \(Ea\) plays a central role in determining how the rate constant \(k\) changes with temperature. This is because \(Ea\) is in the exponent of the equation \(k = Ae^{-Ea/RT}\), emphasizing its influence on \(k\). The exercise provided gives us an \(Ea\) of \(104\, \mathrm{kJ/mol}\), which directly affects how \(k\) changes as the temperature is adjusted. When calculating temperature effects, \(Ea\) is typically converted to Joules, as in our step-by-step solution, since the universal gas constant \(R\) is usually recorded in \(J/(mol \, K)\).
Temperature Dependence of Reaction Rates
Chemical reaction rates are highly dependent on temperature, which is captured eloquently in the Arrhenius equation. As temperature increases, the average kinetic energy of molecules increases, which leads to more collisions between molecules with sufficient energy to overcome the activation energy barrier.
The mathematical relationship, according to the Arrhenius equation, states that even a small increase in temperature can cause an exponential increase in the rate constant \(k\). This is manifested in the formula \(k = Ae^{-Ea/RT}\) where As temperature \(T\) increases, the term \(\frac{Ea}{RT}\) decreases, leading to an increase in \(k\).
In the exercise, we solve for a new temperature \(T_2\) using the equation \(\ln (k_2/k_1) = Ea/R \times (1/T_1 - 1/T_2)\). This relationship highlights how a change in the rate constant, from \(4.60 \times 10^{-4} \mathrm{~s}^{-1}\) to \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\), is predominantly driven by a change in temperature. Understanding this dependency is key to controlling reaction speeds in industrial and laboratory settings.

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Most popular questions from this chapter

Consider the reaction $$ \mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{Z} $$ From the following data, obtained at \(360 \mathrm{~K}\), (a) determine the order of the reaction, and (b) determine the initial rate of disappearance of \(\mathrm{X}\) when the concentration of \(\mathrm{X}\) is \(0.30 \mathrm{M}\) and that of Y is \(0.40 \mathrm{M}\). $$ \begin{array}{ccc} \hline \text { Initial Rate of } & & \\ \text { Disappearance of } \mathbf{X}(\boldsymbol{M} / \mathbf{s}) & {[\mathrm{X}](M)} & {[\mathrm{Y}](M)} \\ \hline 0.053 & 0.10 & 0.50 \\ 0.127 & 0.20 & 0.30 \\ 1.02 & 0.40 & 0.60 \\ 0.254 & 0.20 & 0.60 \\ 0.509 & 0.40 & 0.30 \\ \hline \end{array} $$

Assume that the formation of nitrogen dioxide: $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the condition described in (b), the half-life of the reaction has been estimated to be \(6.4 \times 10^{3}\) min. What would be the half-life if the initial concentration of NO were \(10 \mathrm{ppm} ?\)

The reaction of \(\mathrm{G}_{2}\) with \(\mathrm{E}_{2}\) to form \(2 \mathrm{EG}\) is exothermic, and the reaction of \(\mathrm{G}_{2}\) with \(\mathrm{X}_{2}\) to form \(2 \mathrm{XG}\) is endothermic. The activation energy of the exothermic reaction is greater than that of the endothermic reaction. Sketch the potential energy profile diagrams for these two reactions on the same graph.

List four factors that influence the rate of a reaction.

An instructor performed a lecture demonstration of the thermite reaction (see Example 6.10 ). He mixed aluminum with iron(III) oxide in a metal bucket placed on a block of ice. After the extremely exothermic reaction started, there was an enormous bang, which was not characteristic of thermite reactions. Give a plausible chemical explanation for the unexpected sound effect. The bucket was open to air.

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