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.
Other questions, let me know...
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