Problem 22 - 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 \(0.025 \mathrm{M} / \mathrm{s}\), 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 63 torr \(/ \mathrm{h}\). 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 \(CO_2\) and \(H_2O\) are -0.050 M/s (both decreasing). The rate of change in NH₃ partial pressure is 126 torr/h (increasing). There is no rate of change in the total pressure in the vessel.

Step 1: Relate the rate of change of each species using stoichiometry

Since the stoichiometry of the reaction is \(1C_2H_4 : 3O_2 : /tinyoslatex[ 2CO_2]: 2H_2O\), we can use the stoichiometric coefficients to find the rates of change in the concentrations of \(CO_2\) and \(H_2O\). For every 1 mole of \(C_2H_4\) that reacts, 2 moles of \(CO_2\) and 2 moles of \(H_2O\) are produced. So, the rate of change of \(CO_2\) and \(H_2O\) relative to \(C_2H_4\) is 2:1 and 2:1, respectively.

Step 2: Calculate the rates of change of CO₂ and H₂O concentrations

The rate of change of the concentration of \(C_2H_4\) is -0.025 M/s. We found the rate of change of CO₂ and H₂O relative to C₂H₄ to be 2:1, and 2:1, respectively. The rates of change of CO₂ and H₂O concentrations are: \[\frac{d[CO_2]}{dt} = 2\frac{d[C_2H_4]}{dt}\] and \[\frac{d[H_2O]}{dt} = 2\frac{d[C_2H_4]}{dt}\] Substituting the known rate of change of C₂H₄ and solving: \[\frac{d[CO_2]}{dt} = 2(-0.025 \,\mathrm{M/s}) = -0.050 \,\mathrm{M/s}\] and \[\frac{d[H_2O]}{dt} = 2(-0.025 \,\mathrm{M/s}) = -0.050 \,\mathrm{M/s}\] So, the rates of change in the concentrations of CO₂ and H₂O are -0.050 M/s (both decreasing). (b) The given reaction is: \[N_2H_4(g) + H_2(g) \longrightarrow 2NH_3(g)\] We are given the rate of decrease in N₂H₄ partial pressure and asked to find the rate of change in NH₃ partial pressure and total pressure.

Step 1: Relate the rate of change of each species using stoichiometry

Since the stoichiometry of the reaction is 1 N₂H₄ : 1 H₂ : 2 NH₃, we can use the stoichiometric coefficients to find the rate of change in the partial pressure of NH₃ and the total pressure. For every mole of N₂H₄ that reacts, 2 moles of NH₃ are produced. So, the rate of change of NH₃ relative to N₂H₄ is 2:1.

Step 2: Calculate the rate of change of NH₃ partial pressure

The rate of decrease in N₂H₄ partial pressure is 63 torr/h. The rate of change of NH₃ relative to N₂H₄ is 2:1. The rate of change of NH₃ partial pressure is: \[\frac{dP(NH_3)}{dt} = 2\frac{dP(N_2H_4)}{dt}\] Substituting the known rate of change in N₂H₄ partial pressure and solving: \[\frac{dP(NH_3)}{dt} = 2(63 \,\mathrm{torr/h}) = 126 \,\mathrm{torr/h}\] So, the rate of change in NH₃ partial pressure is 126 torr/h (increasing).

Step 3: Calculate the rate of change of total pressure

The total pressure in the vessel is the sum of the individual partial pressures of all the species: \(P_{total} = P(N_2H_4) + P(H_2) + P(NH_3)\). Using the previously calculated rate of change of N₂H₄ (-63 torr/h) and NH₃ (126 torr/h) partial pressures, we can find the rate of change of the total pressure. Since there is no information given about the rate of change of H₂ partial pressure, we will assume that its rate of decrease is equal to the rate of decrease of N₂H₄ partial pressure, as both reactants have stoichiometric coefficients of 1. Thus, the rate of change of total pressure is: \[\frac{dP_{total}}{dt} = \frac{dP(N_2H_4)}{dt} + \frac{dP(H_2)}{dt} + \frac{dP(NH_3)}{dt}\] Substituting the known values and solving: \[\frac{dp_{total}}{dt} = -63 \,\mathrm{torr/h} - 63 \,\mathrm{torr/h} + 126 \,\mathrm{torr/h} = 0\] So, there is no rate of change in the total pressure in the vessel.

Key Concepts

Stoichiometry: The Heart of Reaction Calculations

Stoichiometry is the key principle that allows us to connect the quantities of reactants and products in a chemical reaction. It is like a recipe for a chemical reaction, indicating the amount of each substance involved. For any reaction, it's essential to understand the ratios of the reactants and products. These ratios are given by the coefficients in a balanced chemical equation.

In the case of the combustion of ethylene, the stoichiometric coefficients tell us that 1 molecule of ethylene (\(C_{2}H_{4}\)) reacts with 3 molecules of oxygen (\(O_{2}\)) to produce 2 molecules of carbon dioxide (\(CO_{2}\)) and 2 molecules of water (\(H_{2}O\)).

• This means for every 1 ethylene molecule consumed, 2 carbon dioxide and 2 water molecules are formed.
• The ratios are crucial because they allow us to calculate the rate of formation of products if we know the rate of consumption of reactants, or vice versa.

Analyzing these ratios helps us determine how quickly products appear or reactants disappear in a reaction, which is fundamental in predicting how long a reaction will take and the quantity of products formed.

Understanding Partial Pressure in Gas Reactions

In a gaseous reaction, each gas present in the container exerts its own partial pressure, which is the pressure it would exert if it occupied the entire volume alone at the same temperature. The concept of partial pressure is particularly useful because it allows chemists to apply the stoichiometry of a reaction to gas-phase reactions using pressure measurements.

For instance, in the decomposition of hydrazine (\(N_{2}H_{4}\)) to ammonia (\(NH_{3}\)), the decrease in the partial pressure of hydrazine directly correlates to the increase in the partial pressure of ammonia. Since 1 mole of \(N_{2}H_{4}\) produces 2 moles of \(NH_{3}\), the rate of increase in \(NH_{3}\)'s partial pressure is twice that of the decrease in \(N_{2}H_{4}\).

• This relationship helps track changes in gas concentrations and, consequently, rates of reactions, just by measuring pressure changes.
• Calculations involving partial pressures are a practical way to understand how different gases contribute to the total pressure in a reaction vessel.

Using partial pressures instead of concentrations is often preferred in gas reactions because it's easier to measure.

The Principle of Combustion Reactions

A combustion reaction is a high-energy chemical reaction where a substance combines with oxygen to release energy, often in the form of heat and light. These reactions are fundamental in applications ranging from engines to home heating systems.

In a typical combustion process, like that of ethylene (\(C_{2}H_{4}\)), the organic compound reacts with oxygen (\(O_{2}\)) to form more stable products such as carbon dioxide (\(CO_{2}\)) and water (\(H_{2}O\)).

• Combustion reactions are generally exothermic, meaning they release heat and energy, which is why they're harnessed for powering engines and generating electricity.
• The completeness of a combustion reaction depends on an adequate amount of oxygen; insufficient oxygen results in incomplete combustion, producing carbon monoxide and other pollutants instead of carbon dioxide.

Understanding combustion reactions helps in controlling emissions and improving the efficiency of energy systems by ensuring complete combustion.

Exploring Gas Laws in Chemical Reactions

Gas laws are fundamental principles that describe how gases behave under different conditions of pressure, volume, and temperature. These laws help predict the properties of gases in a reaction chamber and are crucial for understanding reactions involving gaseous compounds.

When gases take part in reactions, like the formation of ammonia from hydrazine and hydrogen, the behavior of gases can be described using laws such as the Ideal Gas Law. This law, expressed as \(PV = nRT\), relates the pressure (P), volume (V), and temperature (T) of a gas to the amount of gas present (n, in moles) and the ideal gas constant (R).

• Using these relationships, chemists can calculate how changing one condition, like temperature, affects the others, such as pressure.
• For example, in a sealed reaction vessel, the total pressure depends on both reactant and product gas partial pressures, demonstrating how reactions can alter pressure in practical scenarios.

Gas laws provide a framework for understanding the dynamics of gaseous systems, especially in industrial processes or laboratory settings where conditions need to be tightly controlled.