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Can you explain problem 11 on Spring 2009 Exam 2b

Methane (CH₄) reacts with oxygen to form carbon dioxide and water. under some set of conditions at some point in time, you find that 6.495g of methane react every minute in a 600.0 mL vessel.
a. what is the rate of methane consumption?
b. what is the rate of oxygen consumption?
c. what is the rate of carbon dioxide production?
d. what is the rate of water production?
e. what is the rate of the reaction?

Reaction rates are expressed as (change in concentration)/(change in time), so for this reaction under these conditions, the rate of methane consumption is:
{(6.495g CH₄)/(16.043g/mol)/(0.6000L)}/(1min) = 0.6748 M/min
For the rest of the rates, we'll need a balanced reaction:
CH₄(g) + 2 O₂(g) --> CO₂(g) + 2 H₂O(g)
Average rates are determined by the stoichiometry of the reaction. For every mol of CH₄ consumed, 2 mols of O₂ are consumed, so the rate of O₂ consumption must be twice as fast as the rate of CH₄ comsumption. 2x(0.6748 M/min)=1.350 M/min
Every mol of CH₄ that reacts produces 1 mol of CO₂, so the rate of CO₂ production is the same as the rate of CH₄ consumption, 0.6748 M/min.
Every mole of CH₄ that reacts produces 2 mols of H₂O, so the rate of H₂O production is twice the rate of CH₄ consumption, 1.350 M/min. {Which is the same as the rate of O₂ consumption. Note that things that have the same coefficient in the balanced reaction have the same rate.}
The rate of the reaction is also related to the stoichiometry of the reactants. Take the rate of consumption or production of anything in the reaction and divide it by its coefficient, to get the rate of the reaction, 0.6748 M/min.

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