The reaction 2 NO2(g) + O3(g) N2O5(g) + O2(g) was studied at a certain temperatu

Question

The reaction 2 NO2(g) + O3(g) N2O5(g) + O2(g) was studied at a certain temperature with the following results: Experiment [NO2(g)] (M) [O3(g)] (M) Rate (M/s) 1 0.835 0.835 47900 2 0.835 1.67 95800 3 1.67 0.835 95800 4 1.67 1.67 1.92e+05 (a) What is the rate law for this reaction? A) Rate = k [NO2(g)] [O3(g)] B) Rate = k [NO2(g)]2 [O3(g)] C) Rate = k [NO2(g)] [O3(g)]2 D) Rate = k [NO2(g)]2 [O3(g)]2 E) Rate = k [NO2(g)] [O3(g)]3 F) Rate = k [NO2(g)]4 [O3(g)] (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of NO2(g) is 0.842 M and that of O3(g) is 1.82 M if the temperature is the same as that used to obtain the data shown above? ____M/s

Explanation / Answer

Doubling the concentration of B (trial 2 vs. trial 1) doubles the rate. R = k [B]^n 2R = k [2B]^n n must be 1 (lst order) Doubling the concentration of A (trial 3 vs. trial 1) quadruples the rate. R = k [A]^m 4R = k [2A]^m m must be 2 (2nd order) R = k [A]^2 [B]^1 [O]/[O]initial = exp(-kt) If you plot -ln[O] vs. t, you expect a straight line, and you get one with slope = 100.15. But remember, as you correctly point out, that the real rate equation is: Rate = k[NO2][O] The slope of 100.15 isn't k, it is actually k[NO2], since the slope came from psuedo-first-order reaction data with excess NO2. So, 100.15 = k[NO2] k = 100.15/(1.0 x 10^13) = 1.0 x 10^-11 M^-1 s^-1

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