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siddhant_sahu (0)

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For the reaction, N2O5  2NO2 + 0.5O2 calculate the mole fraction of N2O5 decomposed at a constant volume and temperature, if its initial pressure is 600 mm of Hg and the pressure at any time is 960 mm of Hg. Assume ideal gas behaviour.
    
nikhiljee2007 (88)

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N2O5 -----> 2NO2 + 0.5O2
600............. 0........... 0
600-x .........2x......... x/2
600..+..3x/2....=..960
x=240mm.
at const temp , vol moles are directly proportional to press.

X (N2O5) : (600-x)/960
..................== 360/960
..................==0.375

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iitkgp_bipin (6144)

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N2O5  2NO2 + 0.5O2

Since the volume and temperature are constant , the pressures of the individual components are directly proportional to their respective number of moles.

Let the decrease in pressure of N2O5 be x. Then the increase in the pressure of NO2 will be 2x and that for O2 will be x/2.

Hence at equilibrium :
Pressure(N2O5) = 600-x
Pressure(NO2) = 2x
Pressure(O2) = x/2

Total pressure = 600-x+2x+x/2 = 960

x = 240

Hence mole fraction of N2O5 decomposed = x/600 = 240/600 = 0.4

Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur

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