Consider the following reaction: O2(g) + 2 NO(g) -> 2 NO2(g) (a) The rate law fo

Question

Consider the following reaction:

O2(g) + 2 NO(g) -> 2 NO2(g)

(a) The rate law for this reaction is first order in O2(g) and second order in NO(g). What is the rate law for this reaction?

Rate = k [O2(g)] [NO(g)]
Rate = k [O2(g)]2 [NO(g)]
Rate = k [O2(g)] [NO(g)]2
Rate = k [O2(g)]2 [NO(g)]2
Rate = k [O2(g)] [NO(g)]3
Rate = k [O2(g)]4 [NO(g)]

(b) If the rate constant for this reaction at a certain temperature is 10800, what is the reaction rate when [O2(g)] = 0.0212 M and [NO(g)] = 0.0291 M?

Rate = _ M/s.

(c) What is the reaction rate when the concentration of O2(g) is doubled, to 0.0424 M while the concentration of NO(g) is 0.0291 M?

Rate = _ M/s

Explanation / Answer

Arrhenius equation is k=Ae^(-Ea/RT) where k=rate constant A=frequency factor Ea=activation energy R=gas constant T=temperature 2.3x10^-12 L/mol-s=Ae^(-111x10^3 J/mol/(8.31 J/mol-K*273 K)) A=4.08299x10^9 L/mol-s Now find the rate constant, k at 519 K. k=4.08299x10^9 L/mol-s*e^(-111x10^3 J/mol/(8.31 J/mol-K *273 K)) =0.0271 L/mol-s reaction is second order so Rate=k[NO2]² Integrated rate law is 1/[At]=kt+(1/[Ai]) where [At]=amount at time t k=rate constant t=time [Ai]=initial amount 1/[78.57]=0.0271 L/mol-s*t+1/[100] t=0.1006 seconds

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