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Combustion of Ethylene: C2H4 + 3 O2 -> 2 CO2 + 2H20 a) If the concentration...

Combustion of Ethylene: C2H4 + 3 O2 -> 2 CO2 + 2H20

a) If the concentration of C2H4 is decreasing at the rate of 0.025 M/s what are the rates of change in the concentration of CO2 and H2O? Why do I have to divide 0.025 by two?

B) The rate of decrease in N2H4 partial pressure in a closed reaction vessel from the reaction N2H4 + H2 -> 2NH is 63 torr. What are the rates of change of NH3 partial pressure and total pressure in the vessel?

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Answer #1

a. No, you don't have to divide 0.025 by 2 but you have to multiply it by 2. That is because, as you can see from the reaction, the coefficient of CO2 and H2O is 2. So, for every mole of C2H4 consumed, 2 moles of CO2 and 2 moles of H2O are formed. So, the rates of increase in the concentrations of CO2 and H2O are twice as much as the rate of decrease in the concentration of C2H4.

So, the answer is 2 x 0.025 = 0.050 M/s

2. Similarly, the rate of change in the concentration of NH3

= 2 x rate of change of change in the concentration of N2H4

= 2 x 63

= 126 torr/s

As you can see from the reaction, the moles of gases on the reactant side is 2 (1 mole N2H4 and 1 mole H2). And on the product side, the moles of gases is 2 (2Mole NH3). So, the moles of gases consumed is equal to the moles of gases formed. So, the total pressure will remain constant. So, the rate of change of total pressure is 0.

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