The decomposition of NOBr is studied manometrically because the number of moles
ID: 788021 • Letter: T
Question
The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish brown:
2 NOBr (g) ? 2 NO (g) + Br2 (g)
Use the data below to answer the following:
a) Determine the average rate over the entire experiment.
b) Determine the average rate between 2.00 and 4.00 s.
c) Use graphical methods to estimate the initial reaction rate.
d) Use graphical methods to estimate the rate at 7.00 s.
e) At what time does the instantaneous rate equal the average rate over the entire experiment?
Time (s)
[NOBr] (mol/L)
0.00
0.0100
2.00
0.0071
4.00
0.0055
6.00
0.0045
8.00
0.0038
10.00
0.0033
Time (s)
[NOBr] (mol/L)
0.00
0.0100
2.00
0.0071
4.00
0.0055
6.00
0.0045
8.00
0.0038
10.00
0.0033
Explanation / Answer
since K = products / reactants
the reverse is the inverse
when they ask for the K of the reverse reaction, it has all the same concentrations, but inversed:
2NOBr(g) <==> 2NO(g) Br2(g)
K1 = [NO]2 [Br2] / [NOBr]2
& the reverse reaction:
2NO(g) Br2(g) <==> 2NOBr(g)
K2 = [NOBr]2 / [NO]2 [Br2]
K2 is the inverse of K1
K2 = 1/K1
K2 = 1 / 1.3 times 10^{ - 2
K2 = 76.92
your answer is Kc = 77
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