Chemical Kinetics and Nuclear ChemistryHard
Question
The reaction 2N2O5(g) → 4NO2(g) + O2(g) follows first order kinetics. The pressure of a vessel containing only N2O5 was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. will be (Assume temperature remains constant) :
Options
A.125 mm Hg
B.106.25 mm Hg
C.116.25 mm Hg
D.150 mm Hg
Solution
2N2O5(g) → 4NO2(g) + O2(g)
t = 0 50 0 0
t = 30 50 - 2x 4x x
⇒ 87.5 = 50 + 3x
⇒ 3x = 37.5 ⇒ x = 12.5
⇒ PN2O5 after 30 min = 50 - 25 = 25
⇒ t1/2 = 30 min.
Hence after 60 min, (two half lives), PN2O5 remaining =
= 12.5 torr.
⇒ Hence decrease in PN2O5 = 50 - 12.5 = 37.5 torr.
⇒ PNO2 = 2 × 37.5 = 75 torr
PO2 =
= 18.75 torr
⇒ Ptotal = 12.5 + 75 + 18.75
= 106.25 torr.
t = 0 50 0 0
t = 30 50 - 2x 4x x
⇒ 87.5 = 50 + 3x
⇒ 3x = 37.5 ⇒ x = 12.5
⇒ PN2O5 after 30 min = 50 - 25 = 25
⇒ t1/2 = 30 min.
Hence after 60 min, (two half lives), PN2O5 remaining =
= 12.5 torr.⇒ Hence decrease in PN2O5 = 50 - 12.5 = 37.5 torr.
⇒ PNO2 = 2 × 37.5 = 75 torr
PO2 =
= 18.75 torr⇒ Ptotal = 12.5 + 75 + 18.75
= 106.25 torr.
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