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The gas-phase reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has an activation energy of \(103 \mathrm{kJ},\) and the rate constant is \(0.0900 \min ^{-1}\) at \(328.0 \mathrm{K}\). Find the rate constant at \(318.0 \mathrm{K}\)

Short Answer

Expert verified
At 318.0 K, the rate constant is approximately 0.00212 min⁻¹.

Step by step solution

01

Identify the given data

We are given: - Activation Energy \(E_a = 103 \text{ kJ/mol} = 103,000 \text{ J/mol}\),- Initial rate constant \(k_1 = 0.0900 \ \text{min}^{-1}\),- Initial temperature \(T_1 = 328.0 \ \text{K}\),- New temperature \(T_2 = 318.0 \ \text{K}\).We need to find the rate constant \(k_2\) at temperature \(T_2\).
02

Use the Arrhenius Equation

The Arrhenius equation relates the rate constants at two different temperatures to the activation energy. It is given by:\[k_2 = k_1 \cdot e^{\left(\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right)}\] where \(R = 8.314 \ \text{J/(mol K)}\) is the universal gas constant.
03

Calculate the exponent

Substitute the known values into the expression for the exponent:\[\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{103,000}{8.314}\left(\frac{1}{328.0} - \frac{1}{318.0}\right)\]Calculate the difference in the reciprocals of the temperatures and then the whole exponent.
04

Compute the value of the exponent

\[\left(\frac{1}{328.0} - \frac{1}{318.0}\right) = -0.00030208\]\[\frac{103,000}{8.314} \times (-0.00030208) = -3.749\]
05

Substitute back to find k2

Substitute this value back into the Arrhenius equation to find the new rate constant:\[k_2 = 0.0900 \cdot e^{-3.749}\]Calculate the value of \(e^{-3.749}\) and then evaluate \(k_2\).
06

Calculate the value for k2

\[ k_2 = 0.0900 \cdot 0.0235 = 0.00212 \ \text{min}^{-1}\] Therefore, the rate constant at \(318.0 \ \text{K}\) is approximately \(0.00212 \ \text{min}^{-1}\).

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Key Concepts

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

Rate Constant
The rate constant, symbolized as "k," plays a crucial role in chemical reactions. It indicates how fast a reaction proceeds at a given temperature. Mathematically, it appears in the rate equation, which links the reaction rate to the concentrations of reactants.

Generally, the higher the rate constant, the faster the reaction. However, this value is not absolute and varies with temperature. That's where the Arrhenius equation becomes particularly useful. It helps us understand how the rate constant adjusts when the temperature changes, as shown in the earlier exercise.

In the given problem, the initial rate constant is provided at a specific temperature, and the task is to find this constant at a different temperature using the Arrhenius equation. The equation underscores that even a small change in temperature can significantly alter the rate constant, impacting the speed of the reaction.
Activation Energy
Activation energy is the minimum energy required for a chemical reaction to occur. It serves as an energy barrier that reactants must overcome to transform into products. Higher activation energy means that reactants need more energy to initiate a reaction.

In the context of the exercise, the activation energy is given as 103 kJ/mol. This value means that the reactants need to overcome this energy barrier to form products. The concept of activation energy is crucial because it affects how easily reactants can be converted into products and thus influences the rate constant.

The Arrhenius equation clearly shows that the higher the activation energy, the lower the rate constant, assuming the temperature remains constant. This inverse relationship helps explain why reactions with high activation energy proceed more slowly than those with lower activation energies.
Temperature Dependence of Reaction Rates
Temperature has a profound effect on reaction rates. Typically, increasing the temperature increases the rate of a reaction. This happens because elevated temperatures provide reactant molecules with more kinetic energy, increasing their chances to overcome the activation energy barrier.

As depicted in the exercise, when temperature decreases, as from 328 K to 318 K, the rate constant also drops, indicating a slower reaction rate. This is described by the Arrhenius equation, which relates temperature changes to changes in the rate constant.

The reason for this dependency is that temperature affects the kinetic energy distribution of reactant molecules. At higher temperatures, a greater proportion of molecules have enough energy to overcome the activation energy barrier. Consequently, reactions typically go faster. Understanding this temperature dependence is essential when considering reaction environments and can guide chemists in optimizing conditions for desired reaction rates.

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Most popular questions from this chapter

If the rate constant for a reaction triples when the temperature rises from \(3.00 \times 10^{2} \mathrm{K}\) to \(3.10 \times 10^{2} \mathrm{K},\) what is the activation energy of the reaction?

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)

When heated, cyclopropane is converted to propene (see Example 15.5 ). Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=1.02 \times10^{-3} \mathrm{s}^{-1},\) respectively. Determine the activation energy, \(E_{\mathrm{a}}\) from these data.

Hundreds of different reactions can occur in the stratosphere, among them reactions that destroy the earth's ozone layer. The table below lists several (second-order) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant. $$\begin{array}{ll}\hline & \text { Rate Constant } \\\\\text { Reaction } & \left(298 \mathrm{K}, \mathrm{cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\right) \\\\\hline \text { (a) } \mathrm{Cl}+0_{3} \longrightarrow \mathrm{Cl} 0+0_{2} & 1.2 \times 10^{-11} \\\\\text {(b) } \mathrm{Cl}+\mathrm{CH}_{4} \longrightarrow\mathrm{HCl}+\mathrm{CH}_{3} & 1.0 \times 10^{-13} \\\\\text {(c) } \mathrm{Cl}+\mathrm{C}_{3} \mathrm{H}_{8} \longrightarrow\mathrm{HCl}+\mathrm{C}_{3} \mathrm{H}_{7} & 1.4 \times 10^{-10} \\\\\text {(d) } \mathrm{Cl}+\mathrm{CH}_{2} \mathrm{FCl} \longrightarrow\mathrm{HCl}+\mathrm{CHFCl} & 3.0 \times 10^{-18} \\\\\hline\end{array}$$ For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

The decomposition of HOF occurs at \(25^{\circ} \mathrm{C}\) $$2 \mathrm{HOF}(\mathrm{g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ Using the data in the table below, determine the rate law and then calculate the rate constant. $$\begin{array}{lc}\hline \begin{array}{l}{[\mathrm{HOF}]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} & \begin{array}{l}\text { Time } \\\\(\mathrm{min})\end{array} \\\\\hline 0.850 & 0 \\\0.810 & 2.00 \\\0.754 & 5.00 \\\0.526 & 20.0 \\\0.243 & 50.0 \\\\\hline\end{array}$$

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