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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Tiffany Hagan Tiffany Hagan
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11 years ago
When 340 mg of NO2 is confined to a 150 mL reaction vessel and heated to 300?C, it de- composes by a second-order process. In the rate law for the decomposition of NO2,

k= 0.54 L/mol x sec

What is the initial reaction rate? Answer in units of L/mol x sec

2. By what factor does the reaction rate change if the mass of NO2 present in the container is increased to 760 mg?
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#1
11 years ago
For 2nd order dissociation of NO2        Rate = K [NO2]^2
[NO2] initially is 340 mg /150 mL = 0.049 mol / L
So initial rate =  0.54  L/mol.sec x [0.049]^2 mol^2/L^2 = 1.3 x10^-3 mol/L.sec. (Note : your units are wrong)

if the concn. is increased to 760mg/150 mL then [NO2} = 0.11 mol/L
and initial rate = 0.54 x [0.11]^2 = 6.53 x10^-3 mol/L.sec. (about 5.026 times)
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