Question

(a) Consider the combustion of ethylene, ${\mathrm{C}}_2{\mathrm{H}}_4(g)+$ $3{\mathrm{O}}_2(g)⟶2{\mathrm{CO}}_2(g)+2{\mathrm{H}}_2\mathrm{O}(g).$ 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 ${\mathrm{H}}_2\mathrm{O}?(\mathbf{b})$ The rate of decrease in ${\mathrm{N}}_2{\mathrm{H}}_4$ partial pressure in a closed reaction vessel from the reaction ${\mathrm{N}}_2{\mathrm{H}}_4(g)+{\mathrm{H}}_2(g)⟶2{\mathrm{NH}}_3(g)$ is 74 torr per hour. What are the rates of change of ${\mathrm{NH}}_3$ partial pressure and total pressure in the vessel?

Short answer

The rates of change in the concentrations of CO₂ and H₂O are both -0.072 M/s. The rate of change of NH₃ partial pressure is -148 torr/h, and the rate of change of total pressure in the vessel is 0 torr/h.

Step-by-step solution

Write down the given rate of change of C₂H₄

The rate of decrease in the concentration of ethylene (C₂H₄) is given as: $\frac{d[{\mathrm{C}}_2{\mathrm{H}}_4]}{dt}=-0.036\mathrm{M}/\mathrm{s}$

Use stoichiometry to find the rates of change of CO₂ and H₂O

From the balanced reaction, we can see the stoichiometric ratios of the reactants and products: 1 mol C₂H₄ : 3 mol O₂ : 2 mol CO₂ : 2 mol H₂O Now, we can find the rates of change in the concentrations of CO₂ and H₂O: $\frac{d[{\mathrm{CO}}_2]}{dt}=2\times \frac{d[{\mathrm{C}}_2{\mathrm{H}}_4]}{dt}$ $\frac{d[{\mathrm{H}}_2\mathrm{O}]}{dt}=2\times \frac{d[{\mathrm{C}}_2{\mathrm{H}}_4]}{dt}$

Calculate the rates of change of CO₂ and H₂O

Substitute the given rate of decrease in the concentration of ethylene (C₂H₄) into the equations from step 2: $\frac{d[{\mathrm{CO}}_2]}{dt}=2\times (-0.036\mathrm{M}/\mathrm{s})=-0.072\mathrm{M}/\mathrm{s}$ $\frac{d[{\mathrm{H}}_2\mathrm{O}]}{dt}=2\times (-0.036\mathrm{M}/\mathrm{s})=-0.072\mathrm{M}/\mathrm{s}$ So, the rate of change in the concentrations of CO₂ and H₂O are -0.072 M/s. #b) Finding the rates of change of NH₃ partial pressure and total pressure in the vessel#

Write down the given rate of decrease of N₂H₄ partial pressure

The rate of decrease in the partial pressure of N₂H₄ is given as: $\frac{d{P}_{{\mathrm{N}}_2{\mathrm{H}}_4}}{dt}=-74\mathrm{torr}/\mathrm{h}$

Use stoichiometry to find the rate of change of NH₃ partial pressure

From the balanced reaction, we can see the stoichiometric ratios of the reactants and products: 1 mol N₂H₄ : 1 mol H₂ : 2 mol NH₃ Now, we can find the rate of change in the partial pressure of NH₃: $\frac{d{P}_{{\mathrm{NH}}_3}}{dt}=2\times \frac{d{P}_{{\mathrm{N}}_2{\mathrm{H}}_4}}{dt}$

Calculate the rate of change of NH₃ partial pressure

Substitute the given rate of decrease in the partial pressure of N₂H₄ into the equation from step 2: $\frac{d{P}_{{\mathrm{NH}}_3}}{dt}=2\times (-74\mathrm{torr}/\mathrm{h})=-148\mathrm{torr}/\mathrm{h}$ So, the rate of change of NH₃ partial pressure is -148 torr/h.

Calculate the rate of change of total pressure in the vessel

Since one mole of reactants produces one mole of products, there is no change in the total pressure in the vessel. Therefore, the rate of change of the total pressure in the vessel is: $\frac{d{P}_{\mathrm{total}}}{dt}=0\mathrm{torr}/\mathrm{h}$

Key concepts

Reaction Stoichiometry

Understanding reaction stoichiometry is pivotal for analyzing most chemical reactions, including the calculation of reaction rates. Stoichiometry involves the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's anchored in the balanced chemical equation which provides the mole ratios of the substances involved.

Consider the combustion of ethylene given in the exercise. The balanced equation, ${\mathrm{C}}_2{\mathrm{H}}_4(g)+3{\mathrm{O}}_2(g)⟶2{\mathrm{CO}}_2(g)+2{\mathrm{H}}_2\mathrm{O}(g)$,indicates that 1 mole of ethylene reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 2 moles of water. These ratios are used to determine how changes in the concentration of one species will affect the concentration of others. For example, if the concentration of ethylene decreases, the concentrations of carbon dioxide and water will increase, which can be expressed with specific rate relationships based on stoichiometry.

Ethylene Combustion

Ethylene combustion is a chemical reaction where ethylene (${\mathrm{C}}_2{\mathrm{H}}_4$) reacts with oxygen (${\mathrm{O}}_2$) to form carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$). This process is an example of a combustion reaction, a type of exothermic reaction that releases energy, mainly in the form of heat and sometimes light.

In such reactions, it's crucial to monitor the rate at which the reactants are consumed and the products are formed, which not only affects the energy release but also the control of the reaction process. The reaction rate can be manipulated by various factors such as concentration, temperature, and presence of a catalyst. For instance, as the exercise states, the rate at which the concentration of ethylene decreases is -0.036 M/s, which can be used to calculate the rate of formation of the products using the stoichiometry of the balanced equation.

Rate of Reaction

The rate of a reaction refers to the speed at which reactants are converted into products. This rate can be expressed in terms of the change in concentration of a reactant or product over a certain time period. In the exercise, the rate of ethylene concentration decrease is given, and using the stoichiometric coefficients from the balanced equation, the rates of formation for carbon dioxide and water can be deduced.

For a reaction like ethylene combustion, the rate will determine the amount of heat and light energy produced per unit of time, which is essential for safety and efficiency in industrial processes. Understanding the factors that influence the reaction rate, which include the nature of the reactants, surface area, temperature, concentration, and presence of catalysts, is fundamental for controlling and optimizing chemical reactions.

Partial Pressure Changes

In gas-phase reactions, partial pressure is a measure of the pressure contributed by a specific gas in a mixture of gases. Modification in partial pressure reflects changes in gas concentration. When chemical reactions occur in a closed system, like the decomposition of hydrazine (${\mathrm{N}}_2{\mathrm{H}}_4$), as described in the exercise, the rates of partial pressure changes for each gas can be calculated.

The exercise showcases how stoichiometry dictates the change in partial pressure of ammonia (${\mathrm{NH}}_3$) based on the change in partial pressure of hydrazine. Due to the stoichiometric relationship, a 74 torr/h decrease in hydrazine results in a 148 torr/h increase of ammonia. Nonetheless, since the reaction does not result in a net change in the amount of gas (molecules are neither created nor destroyed, only rearranged), the total pressure in the vessel remains unchanged, demonstrating another essential aspect of chemical reactions: conservation of mass in a closed system.