1. 2NO2(g)--->N2O4(g) Suppose 15.7g of NO2 is placed in a 15.0L container at a h

Question

1. 2NO2(g)--->N2O4(g)

Suppose 15.7g of NO2 is placed in a 15.0L container at a high temprature and the container was cooled to 22.0 degrees C. the total pressure is measured to be 0.460atm. What are the partial pressures and mol fractions of each gas at 22 degrees C.

2. Suppose a barometer were designed using water (density +0.997gmL-1) rather than mercury as its fluid. What would be the height, in meters, of the column of water balancing 755 torr pressure?

3. Using the van der Waals equation of state, determine the molor volume of CO2 at 250 degrees C and 90atm.

Explanation / Answer

1. 2NO2 ---> N2O4

Given,

15.7 g of NO2 is reacting

moles of NO2 = g/molar mass of NO2 = 15.7 g/46.0055 g/mol = 0.3413 moles

We know, 2 moles of NO2 react to form 1 mole of N2O4

So, moles of N2O4 = 0.3413/2 = 0.1707 moles

Total moles = 0.3413 + 0.1707 = 0.512 moles

moles fraction of NO2 = 0.3413/0.512 = 0.667

moles fraction of N2O4 = 0.1707/0.512 = 0.333

Total pressure = 0.460 atm

So, partial pressure of NO2 = moles fraction of NO2 x total pressure = 0.667 x 0.460 =0.307 atm

Partial pressure for N2O4 = 0.333 x 0.460 atm = 0.153 atm

2. If water is used instead of mercury in barometer. The height would be 13.6 times higher then mercury.

We would use the relation,

h-H2O = h-Hg x (-Hg / -H2O)

where,

h-water = height of water in barometer

h-Hg = 755 mm Hg

p-H2O = density of water = 0.997 g/ml

p-Hg = density of Hg = 13.6 g/ml

Feed values,

p-H2O = 755 x 13.6/1 = 10268 mm = 10.268 m

Thus, the height in meters of the column would be 10.3 m

3. To calculate molar volume v of CO2 we will use the formula,

v = V/n = RT/P

where,

V = volume in L

n = moles of CO2

P = 90 atm

T = 250 C = 250 + 273 = 523 K

R = gas constant = 0.08206 L.atm/mol.K

Feed values,

v = V/n = 0.08206 x 523 / 90 = 0.48 L/mol

thus, the molar volume would be 0.48 L/mol

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