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Question

# When NO2 is cooled to room temperature, some of it reacts to form a dimer N2O4 through the reaction, 2NO2(g)→N2O4(g) In the reaction, 15.2 g of NO2 is placed in a 10.0 L flask at high temperature and the flask is cooled to 25∘C. The total pressure is measured to be 0.500 atm. Thus, the partial pressure of NO2 is found to be:

A
0.31 atm
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B
0.19 atm
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C
0.10 atm
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D
0.40 atm
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Solution

## The correct option is B 0.19 atmInitial moles of NO2=15.246=0.33 Molecular weight of NO2=14+2×16=46 2NO2(g)⇌N2O4(g) After the dimer formation, number of N atom remains unchanged. Let, N-atom in NO2=nNO2 N-atom in N2O4=2nN2O4 ...(i) Also by Dalton's law PNO2+PN2O4=0.5 atm RTVnNO2+RTVnN2O4=0.5 nNO2+nN2O4=0.5VRT nNO2+nN2O4=0.5×100.0821×298 nNO2+nN2O4=0.204 ...(ii) Solving Equations (i) and (ii), we get nNO2=0.078 mol nN2O4=0.126 mol Total moles =0.078+0.126=0.204 Mole fraction of NO2=nNO2nNO2+nN2O4=0.0780.204=0.38 PNO2=χNO2×Ptotal=0.38×0.5=0.191 atm

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