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Chapter 14: Problem 45

The equation for the combustion of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \longrightarrow 4 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ Explain why it is unlikely that this equation also represents the elementary step for the reaction.

Short Answer

Expert verified
This equation is unlikely to represent an elementary step due to the improbability of nine reactant molecules colliding simultaneously to form the product molecules.

Step by step solution

01

Understand the Reaction

Firstly, you need to understand the given reaction. Ethane (\(C_{2}H_{6}\)) is combusted with oxygen (\(O_{2}\)) to produce carbon dioxide (\(CO_{2}\)) and water (\(H_{2}O\)). As per the reaction, 2 molecules of ethane and 7 molecules of oxygen are combining to get 4 molecules of carbon dioxide and 6 molecules of water.
02

Identify the number of molecules taking part in the reaction

Next, identify the total number of molecules involved in the reaction. Here, a total of 9 molecules (2 ethane and 7 oxygen) are combining to give 10 molecules (4 carbon dioxide and 6 water).
03

Analyze the likelihood of this being an elementary step

An elementary reaction involves a single molecular event happening, i.e., the reactant molecules colliding and immediately forming the product molecules. Since in our case there needs to be collision between 9 molecules simultaneously, it is statistically improbable. Hence, it is unlikely that this equation also represents the elementary step for the reaction.

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Most popular questions from this chapter

Polyethylene is used in many items such as water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules (the basic unit is called a monomer) together (see p. 369 ). The initiation step is $$ \mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot \quad \text { initiation } $$ The \(\mathrm{R}\) - species (called a radical) reacts with an ethylene molecule \((\mathrm{M})\) to generate another radical $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} \cdot $$ Reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: $$ \mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad \text { propagation } $$ This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine $$ \mathrm{M}^{\prime} \cdot+\mathrm{M}^{\prime \prime} \cdot \stackrel{k_{t}}{\longrightarrow} \mathrm{M}^{\prime}-\mathrm{M}^{\prime \prime} \quad \text { termination } $$ (a) The initiator used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]:\) $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} \cdot $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min} .\) (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\) or \(438 \mathrm{~min},\) at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the above polymerization process and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long high-molar-mass polyethylenes?

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) $$ \begin{array}{cc} {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{M})} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \end{array} $$ Determine graphically the rate law for the reaction and calculate the rate constant.

Consider a car fitted with a catalytic converter. The first 10 min or so after it is started are the most polluting. Why?

Consider the reaction $$ \mathrm{A} \longrightarrow \mathrm{B} $$ The rate of the reaction is \(1.6 \times 10^{-2} M / \mathrm{s}\) when the concentration of A is \(0.35 M\). Calculate the rate constant if the reaction is (a) first order in \(\mathrm{A},\) (b) second order in A.

Explain why termolecular reactions are rare.
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