Question

(a) Consider the combustion of ethylene, given below.

C₂H₄(g) + 3 O₂(g) ? 2 CO₂ (g) + 2 H₂O (g)

If the concentration of C₂H₄ is decreasing at the rate of 0.26 M/s, what are the rates of change in the concentrations of CO₂ and H₂O?
CO₂

(b) The rate of decrease in N₂H₄ partial pressure in a closed reaction vessel from the following reaction is 41 torr/hr.

N₂H₄ (g) + H₂ (g) ? 2 NH₃ (g)

What are the rates of change of NH₃ partial pressure and total pressure in the vessel?
NH₃

Answer 1

(a)

The given reaction is

C₂H₄(g) + 3 O₂(g) ? 2 CO₂ (g) + 2 H₂O (g)

rate = -d[C2H4]/dt = -1/3d[O2]/dt = +1/2d[CO2]/dt = +1/2d[H2O]/dt

Given, -d[C2H4]/dt = 0.26 M/s

Rate of change in concentration of CO2 is

-d[C2H4]/dt = +1/2d[CO2]/dt

0.26 M/s = +1/2d[CO2]/dt

d[CO2]/dt = 2 x 0.26 = 0.52 M/s

Rate of change in concentration of H2O is

-d[C2H4]/dt = +1/2d[H2O]/dt

0.26 M/s = +1/2d[H2O]/dt

+1/2d[H2O]/dt =  2 x 0.26 = 0.52 M/s

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(b)

The given reaction is

N₂H₄ (g) + H₂ (g) ? 2 NH₃ (g)

rate = -d[N2H4]/dt ] = -d[H2]/dt = +1/2d[NH3]/dt

-d[N2H4]/dt ] = 41 torr/hr

The rate of change of NH3 partial pressure is

-d[N2H4]/dt ] = +1/2d[NH3]/dt

d[NH3]/dt = 2 x -d[N2H4]/dt ] = 2 x 41 torr/hr = 82 torr/hr