Problem 26 - Chemistry The Central Science

Question

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g)\) \(\longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.036 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\) (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 74 torr per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

Step-by-Step Solution

Answer

The rates of change in the concentrations of CO₂ and H₂O are -0.072 M/s and -0.072 M/s, respectively. The rate of change of NH₃ partial pressure is -148 torr/h and the rate of change of total pressure in the vessel is -296 torr/h.

Step 1: Identify the Balanced Chemical Equation

We are given the balanced chemical equation for the combustion of ethylene: \[C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(g)\]

Step 2: Use Stoichiometry to Relate Rates

The stoichiometry of the reaction gives us the following relationships between the rates of change of concentration for each species: \[\frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}=-0.036 M/s=-1 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\] \[\frac{d[\mathrm{O}_{2}]}{dt}=-3 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\] \[\frac{d[\mathrm{CO}_{2}]}{dt}=2 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\] \[\frac{d[\mathrm{H}_{2}\mathrm{O}]}{dt}=2 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\]

Step 3: Calculate Rates of Change

Using the relationships from the previous step, we can find the rates of change of the concentrations of CO₂ and H₂O: \[\frac{d[\mathrm{CO}_2]}{dt} = 2 \cdot (-0.036\,\text{M}/\text{s}) = -0.072\,\text{M}/\text{s}\] \[\frac{d[\mathrm{H}_2\mathrm{O}]}{dt} = 2 \cdot (-0.036\,\text{M}/\text{s}) = -0.072\,\text{M}/\text{s}\] The rates of change in the concentrations of CO₂ and H₂O are -0.072 M/s and -0.072 M/s, respectively. #Problem (b)#

Step 1: Identify the Balanced Chemical Equation

We are given the balanced chemical equation for the given reaction: \[N_2H_4(g) + H_2(g) \longrightarrow 2NH_3(g)\]

Step 2: Use Stoichiometry to Relate Rates

The stoichiometry of the reaction gives us the following relationships between the rates of change of partial pressure for each species: \[\frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}=-74\,\text{torr}/\text{h}=-1 \cdot \frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}\] \[\frac{dP_{\mathrm{H}_{2}}}{dt}=-1 \cdot \frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}\] \[\frac{dP_{\mathrm{NH}_{3}}}{dt}=2 \cdot \frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}\]

Step 3: Calculate Rates of Change

Using the relationships from the previous step, we can find the rates of change of the partial pressures of each reaction component: \[\frac{dP_{\mathrm{NH}_3}}{dt} = 2 \cdot (-74\,\text{torr}/\text{h}) = -148\,\text{torr}/\text{h}\] The rate of change of NH₃ partial pressure is -148 torr/h.

Step 4: Calculate the Rate of Change of Total Pressure

The rate of change of total pressure is the sum of the rates of change for all gas species: \[\frac{dP_{\mathrm{total}}}{dt}=\frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}+\frac{dP_{\mathrm{H}_{2}}}{dt}+\frac{dP_{\mathrm{NH}_{3}}}{dt}\] Plug in the calculated rates of change: \[\frac{dP_{\mathrm{total}}}{dt} = -74\,\text{torr}/\text{h} + (-74\,\text{torr}/\text{h}) + (-148\,\text{torr}/\text{h}) = -296\,\text{torr}/\text{h}\] The rate of change of total pressure in the vessel is -296 torr/h.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand how reactants and products are related in a chemical reaction. In simple terms, stoichiometry involves using balanced chemical equations to determine the proportions of substances involved in a reaction.

For example, if we look at the combustion of ethylene with the equation: \[C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(g)\]we can see that one molecule of ethylene reacts with three molecules of oxygen to produce two molecules each of carbon dioxide and water.

In stoichiometry, these coefficients tell us the ratio in which the substances react. If you know the rate of change of one substance, you can calculate the rates for others using these ratios.

• If ethylene decreases at a rate of 0.036 M/s, then oxygen decreases at three times that rate, due to its coefficient of 3 in the equation.
• Similarly, carbon dioxide and water each increase at twice the rate of ethylene's decrease, thanks to their coefficients of 2.

Rate of Reaction

The rate of reaction refers to how quickly or slowly reactants are converted into products in a chemical reaction. It's crucial for understanding how a reaction progresses over time. In the case of ethylene combustion, the rate refers to the change in concentration of a reactant or product per unit time.

The rate of reaction can depend on several factors such as temperature, concentration of reactants, surface area, and the presence of catalysts.

In our examples:

• The rate of decrease in ethylene concentration is given as 0.036 M/s, which is directly used to find the rates of other substances via stoichiometry.
• For the reaction involving hydrazine, where the rate is given in torr/hour, pressure units are used instead of concentration due to the gaseous state of reactants and products.

This allows for conversion and calculation of rates for each substance involved based on their stoichiometric relationships.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, showing the substances involved and their changes during the process. A balanced chemical equation gives insight into the quantitative aspects of the reaction.

Writing balanced equations requires ensuring that the number of atoms for each element is the same on both sides of the equation, following the law of conservation of mass.

In the context of our exercise, the given reactions:

• Combustion of ethylene: \(C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g)\) shows how ethylene and oxygen are transformed into carbon dioxide and water.
• Decomposition of hydrazine: \(N_2H_4(g) + H_2(g) \rightarrow 2NH_3(g)\) illustrates the transformation of hydrazine and hydrogen into ammonia.

Understanding these equations lays the groundwork for calculating the change in concentration or pressure of reactants and products using stoichiometry.