(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

01

Identify the Balanced Chemical Equation

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

02

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: \[\frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}=-0.036 M/s=-1 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\] \[\frac{d[\mathrm{O}_{2}]}{dt}=-3 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\] \[\frac{d[\mathrm{CO}_{2}]}{dt}=2 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\] \[\frac{d[\mathrm{H}_{2}\mathrm{O}]}{dt}=2 \cdot \frac{d[\mathrm{C}_{2}\mathrm{H}_{4}]}{dt}\]

03

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)#

04

Identify the Balanced Chemical Equation

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

05

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: \[\frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}=-74\,\text{torr}/\text{h}=-1 \cdot \frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}\] \[\frac{dP_{\mathrm{H}_{2}}}{dt}=-1 \cdot \frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}\] \[\frac{dP_{\mathrm{NH}_{3}}}{dt}=2 \cdot \frac{dP_{\mathrm{N}_{2}\mathrm{H}_{4}}}{dt}\]

06

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.

07

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.

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