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Chapter 13: Problem 41

For the reaction $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the frequency factor \(A\) is \(8.7 \times 10^{12} \mathrm{~s}^{-1}\) and the activation energy is \(63 \mathrm{~kJ} / \mathrm{mol} .\) What is the rate constant for the reaction at \(75^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The rate constant for the reaction at \(75^{\circ} \mathrm{C}\) is \(2.37 \times 10^2 s^{-1}\).

Step by step solution

01

Understand the Arrhenius equation

The rate constant of a reaction can be calculated using the Arrhenius equation: \[k = Ae^{-Ea/RT}\] where \(k\) is the rate constant, \(A\) is the frequency factor, \(Ea\) is the activation energy, \(R\) is the gas constant, and \(T\) is the absolute temperature (in Kelvin).
02

Convert activation energy

Firstly, the activation energy is given in kilojoules per mole, but in the Arrhenius equation it needs to be in Joules per mole. Convert \(63 kJ/mol\) into Joules by multiplying by 1000: \(Ea = 63000 J/mol\).
03

Convert temperature

The absolute temperature in Kelvin can be found by adding 273.15 to the temperature in degrees Celsius. So the temperature will be \(75°C + 273.15 = 348.15 K\).
04

Use the gas constant

The gas constant used in the Arrhenius equation is \(R = 8.314 J/(mol\cdot K)\). We use this specific value to match the units of activation energy (in Joules) and temperature (in Kelvin).
05

Calculate the rate constant

With all the necessary data, the rate constant can be calculated: \(k = 8.7 \times 10^{12} e^{(-63000/(8.314 \times 348.15))}\), which, when calculated results in \(k = 2.37 \times 10^2 s^{-1}\).

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Most popular questions from this chapter

Consider a car fitted with a catalytic converter. The first 5 minutes or so after it is started are the most polluting. Why?

The rate law for the reaction \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2} .\) Which of the following mechanisms can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{c} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N} \\ \mathrm{N}+\mathrm{NO} \longrightarrow \mathrm{N}_{2}+\mathrm{O} \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \end{array} $$ $$ \begin{aligned} &\text { Mechanism II }\\\ &\begin{array}{l} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} \end{aligned} $$ Mechanism III $$ \begin{aligned} 2 \mathrm{NO} & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} & \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

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.

The rate constants of some reactions double with every 10 -degree rise in temperature. Assume that a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

The integrated rate law for the zero-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-k t .\) (a) Sketch the follow- ing plots: (i) rate versus \([\mathrm{A}]_{t}\) and (ii) \([\mathrm{A}]_{t}\) versus \(t\). (b) Derive an expression for the half-life of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when \([\mathrm{A}]_{t}=0\)
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