[A5HLEY]

Hi all. I'm having trouble getting started on this problem. Any suggestions would be appreciated!'


The complete combustion of ethylene [C2H4(g)] produces CO2(g) and H2O(g). If the concentration of ethylene is decreasing at the rate of 0.324 M/s, what are the rates of change in the concentrations of CO2 and H2O?

[froster]

Aren't they equal? A mole of ethylene consumed equals a mole of carbon dioxide and a mole of water. So as fast as ethylene is being consumed should be as fast as the products are formed because of the 1:1 stoichiometry.

Am I right here? Sorry it's been a while... What's throwing me off is that the rate of formation just might be half instead of equal... 0.5(.324M/s)

[Borek]

Aren't they equal? A mole of ethylene consumed equals a mole of carbon dioxide and a mole of water. So as fast as ethylene is being consumed should be as fast as the products are formed because of the 1:1 stoichiometry.

You are close - it is all defined by the reaction stoichiometry, but you are wrong - it is not 1:1.

Ashley - start with balanced reaction equation.

[A5HLEY]

Aren't they equal? A mole of ethylene consumed equals a mole of carbon dioxide and a mole of water. So as fast as ethylene is being consumed should be as fast as the products are formed because of the 1:1 stoichiometry.

You are close - it is all defined by the reaction stoichiometry, but you are wrong - it is not 1:1.

Ashley - start with balanced reaction equation.

I think its 2 C2H4O + 5 O2 --> 4 CO2 + 4 H2O

[Borek]

Ethylene is not C₂H₄O.

[A5HLEY]

Oops! Wrong problem!

C2H4 + 3O2 --> 2CO2 + 2 H2O

Is that better?

[froster]

In that case wouldn't the rate of formation for each of the products be 2*0.324M/s?

[Borek]

OK, OK