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what is the rate law of this?
the reaction 2NO+O2-->2NO2 is second order in NO and first order in O2. when [NO]=0.030 M and [O2]=0.040 M, the observe rate is 7.2 x 10^-5 M/s.
(a) what is the rate law?
(b) what is the value of the rate constant?
(c)what are the units of the rate constant?
(d)what would happen to the rate if the concentration of NO were decrease by a factor of 2?
1 Answer
- 8 years agoFavorite Answer
Rate=k[NO]^2[O2]^1
7.2x10^-5=k[0.03]^2[0.04]^1.
K=2litre^2mole^-2 s
Rate will increase four times