Decomposition of N2O5 is first order reaction is 2.4 hr at
STP. Starting with 10.8 gm of N2O5 how much
oxygen will be obtained after a period of 9.6 hr?
(1) 1.5 litre
(2) 3.36 litre
(3) 1.05 litre
(4) 0.07 litre
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The amount of oxygen obtained after a period of 9.6 hr is 0.07 L.
Explanation:
Given data:
Mass of N2O5 = 10.8 g
Period of first order reaction = 2.4 hr
One mole of N2O5 is present initially.
Therefore the half life time of N2O5 "T(1/2)" =2.4 = ln 2/K
K= ln 2/2.4
Now at T=9.6
K=ln(0.1/N)/9.6=ln 2/9.6
After solving the above equation it gives N=0.1/16
Now number of liter of O2 = N x 22.4/2
= 0.1/16 x 22.4/2
= 0.00625 x 11.2 = 0.07 L
Thus the amount of oxygen obtained after a period of 9.6 hr is 0.07 L.
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