During lecture, the following chemical equilibrium between nitrogen dioxide and
ID: 985859 • Letter: D
Question
During lecture, the following chemical equilibrium between nitrogen dioxide and dinitrogen tetroxide was presented and discussed.
2 NO2(g) N2O4(g) Kp = 7.5 at 300.0
(a) (4 pts) If the total pressure in the container, which has a volume of 100.0 mL, is 0.85 atm, what are the partial pressures (in atm) of NO2(g) and N2O4(g) in this container at 300.0 K?
(b) (2 pts) Using your answer to (b), what are the equilibrium concentrations (in M) of NO2(g) and N2O4(g) in this container at 300.0 K?
c)At low temperatures, close examination of the closed reaction vessel containing essentially all dinitrogen tetroxide shows small amounts of a blue-green liquid that comes from the reaction of dinitrogen tetroxide with trace amounts of water leading to dinitrogen trioxide (blue) and nitric acid (colorless). Write the balanced chemical equation for this reaction
d) With greater amounts of water, dinitrogen tetroxide hydrolyzes into nitrous acid and nitric acid. Write the balanced chemical equation for this process.
Explanation / Answer
Kp = [PN2O4] / [PNO2]2 = 7.5 (1)
Total pressure PN2O4+ PNO2= 0.85
PN2O4= 0.85- PNO2
From (1) (0.85-PNO2] [PNO2]2= 7.5 (2)
This equation can be solved in the form of quadratic equation or by assuming some value of PNO2 and matching LHS and RHS
Let PNO2=0.1
Eq.2 becomes (0.85-0.1)/0.1*0.1= 75
So RHS < less than calculated value By trial and error using solver of excel
PNO2= 0.2761 atm PN2O4= 0.85-0.2761=0.5739 atm
Moles of NO2 at equilibrium =PV/RT= 0.2761*(100/1000)/{(0.08206*300)}=0.001122 moles
For N2O4= 0.5739*(100/1000)/{ 0.08206*300)= 0.002322 moles
Concentrations : NO2= 0.001122 moles/ (100/1000)= 0.01122M and N2O4= 0.02322M
c) The reaction is 2N2O4 + H2O----> N2O3 + 2HNO3
d) balanced reaction is N2O4+ H2O ---> HNO2+ HNO3
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