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nekit [7.7K]
4 years ago
7

Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the followi

ng chemical equation. 2 NO2(g) → 2 NO(g) + O2(g). A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C. If the half-life and the rate constant at 300°C are 22 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?
Chemistry
1 answer:
givi [52]4 years ago
7 0
T(1/2)=1/(k[NO2])
[NO2]=1/(kt(1/2))
[NO2]=1/(0.54M-1 s-1*22s) seconds cancel
[NO2]=1/11.88=0.0842
inital mole= 0.0842*volume(2.5)=0.210 mols
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7.3\times 10^2\ days

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Using integrated rate law for first order kinetics as:

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\frac {[A_t]}{[A_0]} = 1 - 0.05 = 0.95

t = ?

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The balanced equation is given as;

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