The reaction O_2(g) + 2 NO(g) rightarrow 2 NO_2(g) was studied at a certain temp

Question

The reaction O_2(g) + 2 NO(g) rightarrow 2 NO_2(g) was studied at a certain temperature with the following results: (a) What is the rate law for this reaction? Rate = k [O_2(g)] [NO(g)] Rate = k [O_2(g)]^2 [NO(g)] Rate = k [O_2(g)] [NO(g)]^2 Rate = k [O_2(g)]^2 [NO(g)]^2 Rate = k [O_2(g)] [NO(g)]^3 Rate = k [O_2(g)]^4 [NO(g)] (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of O_2(g) is 0.0214 M and that of NO(g) is 0.0444 M if the temperature is the same as that used to obtain the data shown above? M/s

Explanation / Answer

a) with respect to O2 reaction is first order

with respect to NO second order

so rate law as follows

rate = K [O2(g)][NO(g)]2

b) K = rate / [O2] [NO]2

K = 0.112 / [0.0205][0.0205]2  = 0.112 / 8.61 x 10-6

K = 1.3 x 103 M-2 s-1

c) rate = 1.3 x 103 [0.0214] [0.0444]2

rate = 5.48 x 10-2 M/s

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