Question

Challenge Gas Stoichiometry The emission of nitrogen dioxide via fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture as seen in the following reaction 2 CON ), (g) + 4 NO (g)+), (g) à 4 N, (g) + 2 CO, (g)+4 H,O (1) In a 0.950 L reaction vessel, three different sealed tubes containing the three reactants are opened allowed to react. Tube I contained 425 mL of urea at 38.9 C at a pressure of 399 tort. Tube 2 contained 555 ml of nitrogen monoxide at 37.8 C and a pressure of 600.1 torr. Tube 3 contained 410,0 mL of oxygen gas 31.2 C and a pressure 405 torr. If once the reaction completed, the reaction flask was cooled to 25 C, what is the partial pressure of each gas and the total pressure in the flask? Assume water does not contribute any partial pressure at this temperature (a lie), and that the tubes and water do not take up a significant volume of the reaction flask.

Answer 1

According to Dalton's law of partial pressures

Ptotal= P1+P2+P3 ..............................(1) or

Ptotal= R [n1T1/ V1 + n2T2/ V2 + n3T3/ V3 ]

From this we can calculate Ptotal from eqn (1)

Ptotal= 399+600.1+405

= 1404.1 torr

Also ,we know the partial pressure of the individual gas is found out by

P1= Ptotal x Xi where Xi is the mole fraction

Let us denote urea=1 ,NO=2, O2=3

So

X1 =n1 / n1+n2 +n3 [ from the above eqn n1=2,n2=4,n3=1 ]

= 2/7= 0.285 moles

Similarly, X2= 4/7=0.571moles

X3= 1/7 = 0.142 moles

Partial pressure of individual components

1.urea

P1=1404.1 X0.285 =1114.5 torr

2. NO

P2=1404.1X 0.571=801.7 torr

3. O2

P3= 1404.1 X 0.142= 199.3torr