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Question: The reaction, N<sub>2</sub>O<sub>5</sub> (g)\(\rightleftarrows\)2 NO<sub>2</sub> (g) +\(\frac { 1 } ...

The reaction, N2O5 (g)\rightleftarrows2 NO2 (g) +12\frac { 1 } { 2 } O2 (g), is started with initial pressure of N2O5 (g) equal to 600 torr. What fraction of N2O5 (g) decomposed when total pressure of the system is 960 torr ?

A

0.05

B

0.1

C

0.2

D

0.4

Answer

0.4

Explanation

Solution

N2O5 (g) \rightleftarrows 2 NO2 (g) + 12\frac { 1 } { 2 } O2 (g)

Initially : 600 torr — —

At equilibrium: (600 – p) torr 2p torr torr

Total press is torr

\ 600 + = 960

or = 360

or p = 240 torr

\ fraction of N2O5 (g) decomposed = 240600\frac { 240 } { 600 } = 0.4