The gas-phase decomposition of NOBr, 2 NOBr ----> 2 NO(g) + Br2(g), is studied a

Question

The gas-phase decomposition of NOBr, 2 NOBr ----> 2 NO(g) + Br2(g), is studied at a certain temperature, giving the following data:
Time (s) [NOBr] (M)
0.0e+00 0.100
5.0e+01 0.0160
1.0e+02 0.00870
1.5e+02 0.00597
2.0e+02 0.00455


Figure out whether this reaction is first-order or second-order with respect to the concentration of NOBr. Calculate the value of the rate constant. Pick the choice from below which gives the correct reaction order and value of the rate constant for this reaction.



a) This reaction is second-order with the rate constant k = 2.10e-01 M-1 s-1.
b) This reaction is first-order with the rate constant k = 5.25e-01 s-1.
c) This reaction is first-order with the rate constant k = 3.50e-01 s-1.
d) This reaction is second-order with the rate constant k = 1.05e+00 M-1 s-1.
e) This reaction is first-order with the rate constant k = 2.63e-01 s-1.

Explanation / Answer

the reaction is first order in O atoms because the rate data fits the first-order kinetic equation, which is: [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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