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Chapter 14: Problem 26

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\). If the concentra- tion of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.025 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?(\mathbf{b})\) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is \(10 \mathrm{kPa}\) per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

Short Answer

Expert verified
CO2 and H2O increase by 0.050 M/s each. NH3 pressure increases by 20 kPa/hr; total pressure is constant.

Step by step solution

01

Understanding Combustion Reaction of Ethylene

The chemical reaction for the combustion of ethylene is \( \mathrm{C}_{2} \mathrm{H}_{4}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \). When ethylene is burned, CO2 and water are produced in a 2:2 molar ratio. We need to find the rate of increase in concentration for both \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) given the rate of decrease for \(\mathrm{C}_{2} \mathrm{H}_{4}\).
02

Rate of Change in Concentrations

Since one mole of \(\mathrm{C}_{2} \mathrm{H}_{4}\) produces two moles of \(\mathrm{CO}_{2}\) and two moles of \(\mathrm{H}_{2} \mathrm{O}\), the rate of change of concentration is proportional. Therefore, as the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) decreases by \(0.025 \text{ M/s}\), the concentration of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) increases by \(2 \times 0.025 = 0.050 \text{ M/s}\) each.
03

Understanding Hydrazine Reaction

The chemical reaction involving hydrazine is \( \mathrm{N}_{2} \mathrm{H}_{4}(g) + \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \). We are given that the partial pressure of \(\mathrm{N}_{2} \mathrm{H}_{4}\) decreases by \(10 \text{ kPa/hr}\), and we need to find how this change affects \(\mathrm{NH}_{3}\) and the total pressure.
04

Rate of Change in NH3 Partial Pressure and Total Pressure

Since one mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) forms two moles of \(\mathrm{NH}_{3}\), the rate of increase in \(\mathrm{NH}_{3}\) partial pressure is \(2 \times 10 = 20 \text{ kPa/hr}\). For the total pressure, the reaction consumes 1 mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) and 1 mole of \(\mathrm{H}_{2}\) and produces 2 moles of \(\mathrm{NH}_{3}\), leading to no net change in the number of moles. Thus, the total pressure remains constant.

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Key Concepts

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

Combustion Reactions
Combustion reactions involve a substance combining with oxygen to produce energy, light, and new products, often carbon dioxide and water. In the combustion of ethylene \((\mathrm{C}_{2} \mathrm{H}_{4})\), this process is shown by the reaction:- \(\mathrm{C}_{2} \mathrm{H}_{4}(g) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\)During combustion, the ethylene molecules break up and recombine with oxygen molecules to form carbon dioxide and water vapor. This not only releases energy but follows a set stoichiometric ratio. Each mole of ethylene reacts with three moles of oxygen, forming two moles each of carbon dioxide and water. This specific manner in which substances react is defined by stoichiometry, which ensures the conservation of mass.
Stoichiometry
Stoichiometry is the art of understanding and applying the quantitative relationships in chemical reactions. By following stoichiometry, we can predict how much of a product will form from a given amount of reactants.In our example, since one mole of \(\mathrm{C}_{2} \mathrm{H}_{4}\) produces two moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), there's a direct relationship between the rate of decrease in the reactant and the rate of increase in the products.- If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) decreases by \(0.025 \text{ M/s}\), then the concentration of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) increases by \(2 \times 0.025 = 0.050 \text{ M/s}\)This is a simple yet powerful concept in stoichiometry, reflecting the coefficients in the balanced chemical equation. These coefficients tell us the proportions in which substances react and are produced.
Gas Laws
Gas laws describe the behavior of gases in response to changes in temperature, volume, and pressure. These laws are essential when dealing with reactions that involve gases.In the hydrazine reaction:- \(\mathrm{N}_{2} \mathrm{H}_{4}(g) + \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)\)We observe changes in pressure 鈥� a form of gas behavior. Here's what happens:- The decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure is \(10 \text{ kPa/hr}\)- Each mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\) gives rise to two moles of \(\mathrm{NH}_{3}\), thus increasing its partial pressure by \(2 \times 10 = 20 \text{ kPa/hr}\)- The total pressure remains unchanged since the number of gas moles does not alter: two moles consumed and two moles produced.Understanding these gas laws allows us to predict and manipulate conditions in reactions that involve gaseous reagents or products.

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Most popular questions from this chapter

In solution, chemical species as simple as \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right]\) of a solution containing an acidcatalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\) concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: \begin{tabular}{ll} \hline Temperature \((\mathrm{K})\) & Rate Constant \(\left(\mathrm{s}^{-1}\right)\) \\ \hline 300 & \(3.2 \times 10^{-11}\) \\ 320 & \(1.0 \times 10^{-9}\) \\ 340 & \(3.0 \times 10^{-8}\) \\ 355 & \(2.4 \times 10^{-7}\) \\ \hline \end{tabular} From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\).

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}\), \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+\mathrm{Br}^{-}(a l c),\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M}\), the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)\) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

Enzymes are often described as following the two-step mechanism: $$ \begin{array}{l} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \text { (fast) } \\ \mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \text { (slow) } \end{array} $$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, \(\mathrm{ES}=\) enzyme- substrate complex, and \(\mathrm{P}=\) product. (a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of \(\mathrm{E}\) with \(\mathrm{I}\), an inhibitor.
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