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Chapter 11: Problem 70

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to \(\mathrm{NO}_{2}\) and \(\mathrm{NO}_{3}\) is a first-order gas-phase reaction. At \(25^{\circ} \mathrm{C}\), the reaction has a half-life of \(2.81\) s. At \(45^{\circ} \mathrm{C}\), the reaction has a half-life of \(0.313 \mathrm{~s}\). What is the activation energy of the reaction?

Short Answer

Expert verified
Answer: The activation energy of the reaction is approximately 60,535 J/mol.

Step by step solution

01

Write down the given information

We are given the following information: 1. The decomposition of N2O5 is a first-order reaction. 2. The half-life at 25°C is 2.81 seconds. 3. The half-life at 45°C is 0.313 seconds. Let's denote the half-life at 25°C as t₁ and the half-life at 45°C as t₂. t₁ = 2.81 s t₂ = 0.313 s T₁ = 25°C + 273.15 = 298.15 K T₂ = 45°C + 273.15 = 318.15 K
02

Find the rate constants (k)

The relationship between the half-life (t) and the rate constant (k) for a first-order reaction is given by: t = ln(2) / k We can use this equation to find the rate constants at both temperatures. k₁ = ln(2) / t₁ = ln(2) / 2.81 = 0.2468 s⁻¹ k₂ = ln(2) / t₂ = ln(2) / 0.313 = 2.2135 s⁻¹
03

Write down the Arrhenius equation

The Arrhenius equation is given by: k = Ae^(-Ea / RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.
04

Write down the ratio of the rate constants

Dividing the Arrhenius equations for both temperatures, we get: k₁ / k₂ = e^((Ea / R) * (1 / T₁ - 1 / T₂))
05

Solve for the activation energy (Ea)

Rearrange the equation obtained in Step 4 to solve for Ea: Ea = R * ln(k₁ / k₂) / (1 / T₂ - 1 / T₁) Plug in the values and calculate the activation energy: Ea = 8.314 J/mol·K * ln(0.2468 s⁻¹ / 2.2135 s⁻¹) / (1 / 318.15 K - 1 / 298.15 K) Ea ≈ 60535 J/mol This is the answer: The activation energy of the reaction is approximately 60,535 J/mol.

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Most popular questions from this chapter

The reaction $$\mathrm{ICl}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{I}_{2}(g)+\mathrm{HCl}(g)$$ is first-order in both reactants. The rate of the reaction is \(4.89 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) when the ICl concentration is \(0.100 M\) and that of the hydrogen gas is \(0.030 \mathrm{M}\) (a) What is the value of \(k\) ? (b) At what concentration of hydrogen is the rate \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and \([\mathrm{ICl}]=0.233 \mathrm{M?}\) (c) At what concentration of iodine chloride is the rate \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) if the hydrogen concentration is three times that of ICl?

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

The decomposition of nitrosyl chloride $$\mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ is a second-order reaction. If it takes \(0.20 \mathrm{~min}\) to decompose \(15 \%\) of a \(0.300 \mathrm{M}\) solution of nitrosyl chloride, what is \(k\) for the reaction?

The decomposition of phosphine, \(\mathrm{PH}_{3}\), to \(\mathrm{P}_{4}(g)\) and \(\mathrm{H}_{2}(g)\) is firstorder. Its rate constant at a certain temperature is \(1.1 \mathrm{~min}^{-1}\). (a) What is its half-life in seconds? (b) What percentage of phosphine is decomposed after \(1.25 \mathrm{~min}\) ? (c) How long will it take to decompose one fifth of the phosphine?

Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta \mathrm{t}\).)
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