2 N 2 O 5 ( g ) ? 4 NO 2 ( g ) + O 2 ( g ) (a) How many moles of N 2 O 5 will re
ID: 809878 • Letter: 2
Question
2 N2O5(g) ? 4 NO2(g) + O2(g)
(a) How many moles of N2O5 will remain after 1.5 min?
_______ mol
(b) How many minutes will it take for the quantity of N2O5 to drop to 0.005 mol?
_____ min
(c) What is the half-life of N2O5 at 70?C?
_____ min
Explanation / Answer
PART-1
1.5 min = 90 s
ln x / 0.0450 = - 6.82 x 10^-3 x 90
ln (x / 0.0450)=- 0.6138
e^-0.6138= x / 0.0450
0.54129=x / 0.0450
x = moles N2O5 = 0.02435
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PART-2
ln 0.005/ 0.0450= - 6.82 x 10^-3 t
-2.1972=-6.92 x 10^-3 t
t = 322.17 s => 5. min 22sec
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PART-3
t1/2 = ln 2 / 6.82 x 10^-3
= 106.6 s => 1.69 min
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