7A.3

[Malia Shitabata 1F]

Ethene is a component of natural gas, and its combustion has been thoroughly studied. At a certain temperature and pressure, the unique rate of the combustion reaction C2H4(g)+3O2(g)->2CO2(g)+2H2O(g) is 0.44 mol/Ls. What is the rate at which oxygen reacts.

Why do you multiply 0.44 by 3 instead of leaving it as rate=0.44[O2]^3? Isn't 3 the order and not the concentration of O2?

[ShravanPatel2B]

you would multiply .44 by 3 because there are 3 moles of oxygen in the chemical equation

[Sean Cheah 1E]

0.44, in this case, is not your k value, but rather the actual observed rate so you cannot put it into a rate law the way that you did. The unique rate refers to the -1/a*(d[R]/dt) or 1/b*(d[P]/dt) for the general reaction aR --> bP. Thus, to get the actual rate at which a specific product is produced (d[P]/dt) or at which a specific reactant is consumed (d[R]/dt), simply multiply the unique rate by the appropriate coefficient.

[William Francis 2E]

Unique average reaction rate is described on page 590 of the textbook. Coefficients in a balanced chemical equation do not indicate the order of the reaction. The rate at which oxygen reacts is equal to the change in the concentration of oxygen divided by time regardless of coefficients. However, the coefficients come into play when comparing consumption or formation of different species in a reaction. The unique reaction rate is given as 0.44 mol/Ls. As described on page 590 of the textbook, this unique reaction rate can be set equal to (-1/3)(Δ[O₂]/Δt). Multiply each side of the equation by 3, and you've got the answer! The negative sign doesn't matter here in the equation by the way because it's just asking for the rate of reaction rather than a rate of formation or consumption.

[Julie_Reyes1B]

This is more of a test of whether we understand how coefficients are used in the unique reaction rate formula, which is the one that it like 1/a (dA/dt)= 1/b (dB/dt)...etc.

[Samuel Tzeng 1B]

0.44 is not the rate constant but the actual rate so you would multiply the number of oxygen's by 0.44