(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?

Step by step solution

01

Write down the balanced chemical equation

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

02

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).

03

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)

04

Write down the balanced chemical equation

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

05

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).

06

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: \[\frac{dP(NH_3)}{dt} = 2 \times (-10) \,\text{kPa/h} = -20\, \text{kPa/h}\]

07

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: \[\frac{dP_{total}}{dt} = (-10) + (-10) + (-20)\, \text{kPa/h} = -40\, \text{kPa/h}\]

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