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Question

At 250C and 1 atmospheric pressure, the partial pressures in equilibrium mixture of gaseous N2O4 and NO2 are 0.7 and 0.3 atm respectively. Calculate the partial pressure of the N2O4 gas when it is in equilibrium 25oC and the total pressure of 10 atm

A
9.0 atm
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B
1.0 atm
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C
0.9 atm
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D
0.1 atm
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Solution

The correct option is A 9.0 atm
N2O42NO2Pressure at equilibrium0.70.3
KP=(PNO2)2PN2O4=0.3×0.30.7=0.1286atm
Now assume decompostion at 10 atm pressure

N2O42NO2Initial mole10Mole at equilibrium(1x)2x


PN2O4=1x1+x×10 Eq. 1PNO2=2x1+x×10 Eq. 2

KP=(PNO2)2PN2O4
0.1286=(2x1+x×10)21x1+x×10
putting values, and solving we get;
x=0.0566

Putting values of x in equation 1 and 2 above,

PNO2=1.07 atm1
PN2O4=8.93 atm9 atm

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