I'm confused on the following question and was wondering if someone could provide a mental walkthrough of how they got the answer. (The answer is given by the textbook: (a) Second order in NO, first order in O2 )
"When the NO concentration is doubled, the rate of the reaction 2NO (g) + O2 (g) ---> 2NO2 (g) increases by a factor of 4. When both the O2 and the NO concentrations are doubled, the rate increases by a factor of 8.What are (a) the reaction order with respect to each reactant,"
7A.3A Reaction Concentration
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Re: 7A.3A Reaction Concentration
When NO is doubled, the reaction rate increases by a factor of 4: the reaction order with respect to NO must be 2nd order (2^2 = 4)
When both NO and O2 are doubled the reaction rate increases by a factor of 8: we already know that NO is 2nd order, so its doubling alone results in the reaction order increasing by a factor of 4. Therefore, O2 must be first order, so that its doubling alone results in the reaction rate doubling. (Overall: 2^2 * 2^1 = 4 * 2 = 8)
When both NO and O2 are doubled the reaction rate increases by a factor of 8: we already know that NO is 2nd order, so its doubling alone results in the reaction order increasing by a factor of 4. Therefore, O2 must be first order, so that its doubling alone results in the reaction rate doubling. (Overall: 2^2 * 2^1 = 4 * 2 = 8)
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Re: 7A.3A Reaction Concentration
It helps me to isolate the reactant's effects on rate in an algebraic format.
2A = 4
2A*2B=8
Where A and B are the reactant's orders.
Therefore A must equal 2 and B must equal 1.
2A = 4
2A*2B=8
Where A and B are the reactant's orders.
Therefore A must equal 2 and B must equal 1.
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