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 \(10 \mathrm{kPa}\) per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

Short answer

For part (a), the rates of change in concentrations of CO$_{2}$ and H$_{2}$O are both -0.05 M/s. For part (b), the rate of change of NH$_{3}$ partial pressure is -20 kPa/h and the rate of change of total pressure in the vessel is -40 kPa/h.

Step-by-step solution

Write down the balanced chemical equation

We are given the balanced chemical equation: \[C_{2}H_{4}(g) + 3O_{2}(g) \longrightarrow 2CO_{2}(g) + 2H_{2}O(g)\]

Analyze the stoichiometry

Notice that 1 mole of ethylene (C2H4) reacts with 3 moles of oxygen (O2) to produce 2 moles of carbon dioxide (CO2) and 2 moles of water (H2O).

Find the rate of change of concentration for CO2 and H2O

We are given the rate of change of concentration of C2H4 as: -0.025 M/s. According to stoichiometry, \[\frac{d[CO_2]}{dt} = 2 \times \frac{d[C_2 H_4]}{dt}\] Similarly, \[\frac{d[H_2O]}{dt} = 2 \times \frac{d[C_2 H_4]}{dt}\] Plugging in the given value of \(\frac{d[C_2 H_4]}{dt}\), we get: \[\frac{d[CO_2]}{dt} = 2 \times (-0.025) M/s = -0.05 M/s\] \[\frac{d[H_2O]}{dt} = 2 \times (-0.025) M/s = -0.05 M/s\] (b)

Write down the balanced chemical equation

We are given the balanced chemical equation: \[N_{2}H_{4}(g) + H_{2}(g) \longrightarrow 2NH_{3}(g)\]

Analyze the stoichiometry

Notice that 1 mole of nitrogen hydride (N2H4) reacts with 1 mole of hydrogen (H2) to produce 2 moles of ammonia (NH3).

Find the rate of change of NH3 partial pressure

We are given the rate of change of partial pressure of N2H4 as -10 kPa/h. According to stoichiometry, \[\frac{dP(NH_3)}{dt} = 2 \times \frac{dP(N_2 H_4)}{dt}\] Plugging in the given value of \(\frac{dP(N_2 H_4)}{dt}\), we get: \[\frac{dP(NH_3)}{dt} = 2 \times (-10) \,\text{kPa/h} = -20\, \text{kPa/h}\]

Find the rate of change of total pressure

Recall that the change in total pressure is given by the sum of the change of pressure of each individual gas: \[\frac{dP_{total}}{dt} = \frac{dP(N_2 H_4)}{dt} + \frac{dP(H_2)}{dt} + \frac{dP(NH_3)}{dt}\] Since the stoichiometry given in the balanced equation shows that 1 mole of N2H4 and 1 mole of H2 react to form 2 moles of NH3, the rate of change of partial pressure of H2 is equal to that of N2H4. Therefore, \[\frac{dP(H_2)}{dt} = -10\, \text{kPa/h}\] Now, we substitute these values into the equation for total pressure change: \[\frac{dP_{total}}{dt} = (-10) + (-10) + (-20)\, \text{kPa/h} = -40\, \text{kPa/h}\]

Key concepts

Reaction Rates

Understanding reaction rates is crucial in the study of chemical kinetics. Reaction rates tell us how fast a reaction proceeds. Specifically, it measures the change in concentration of a reactant or product over time. If you've ever noticed rust forming on iron or fruit ripening, you've witnessed reaction rates in action.
In the given problem, the reaction rate involves measuring how quickly ethylene (\(\mathrm{C}_{2}\mathrm{H}_{4}\) ) is consumed. With a given rate of \(-0.025 \, \mathrm{M/s}\), we determine how rapidly associated products, like \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\), are formed. Reaction rates can be influenced by various factors, including:

• Concentration of reactants
• Temperature
• Presence of catalysts
• Surface area of reactants

Learning these concepts enhances your predictive capability of reaction outcomes in different conditions.

Stoichiometry

Stoichiometry is an essential concept that helps us understand the quantitative relationships between reactants and products in a chemical reaction. It allows us to derive the conversion ratios, which are key in calculating how much of a substance is consumed or produced as the reaction proceeds.
In the provided reaction equations, stoichiometry tells us, for instance, that for every 1 mole of \(\mathrm{C}_{2}\mathrm{H}_{4}\) consumed, 2 moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) are produced. This 1:2 ratio aids in calculating the change in concentration for these products given the rate at which \(\mathrm{C}_{2}\mathrm{H}_{4}\) decreases.

• Consider relative amounts in a balanced equation e.g.,\(1\) \(\mathrm{C}_{2}\mathrm{H}_{4}\) reacts to produce \(2\) \(\mathrm{CO}_{2}\)
• Use ratios to find rates of change for other substances
• Aids in balancing reactions for accurate computations

Understanding stoichiometry is like having a recipe, ensuring you know the exact proportions needed for a chemical reaction.

Partial Pressure Changes

In a gas reaction, partial pressure is an important parameter to understand. Partial pressure refers to the pressure that a gas in a mixture would exert if it alone occupied the entire volume of the original mixture. This concept becomes particularly relevant in closed systems, where gas reactions are taking place.
For the reaction involving \(\mathrm{N}_{2}\mathrm{H}_{4}\) and \(\mathrm{H}_{2}\) forming \(\mathrm{NH}_{3}\), the rates of partial pressure changes can be calculated using stoichiometry. If the partial pressure of\(\mathrm{N}_{2}\mathrm{H}_{4}\)drops by 10 kPa per hour, stoichiometry helps predict the pressure change for \(\mathrm{NH}_{3}\) as -20 kPa/h.

• Partial pressures reflect mole ratios post-reaction
• Can affect reaction rates and equilibrium positions
• Useful for predicting outcomes in chemical engineering applications

Calculating partial pressures allows chemists to better anticipate the behavior of gases in reactions, both experimentally and industrially.