When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is
ID: 951557 • Letter: W
Question
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibrium, the concentration of N2O4is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction below at 400 K?
N2O4(g) 2 NO2(g)
When 9.2 of frozen is added to a 0.50 reaction vessel and the vessel is heated to 400 and allowed to come to equilibrium, the concentration of is determined to be 0.057 . Given this information, what is the value of for the reaction below at 400 ?
Data for the experiment are provided in this table.
In which bulb would you expect the composition of gases to be closest to equilibrium?
In which bulb would you expect the composition of gases to be closest to equilibrium?
bulb 5
b
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibrium, the concentration of N2O4is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction below at 400 K?
N2O4(g) 2 NO2(g)
When 9.2 of frozen is added to a 0.50 reaction vessel and the vessel is heated to 400 and allowed to come to equilibrium, the concentration of is determined to be 0.057 . Given this information, what is the value of for the reaction below at 400 ?
0.13 2.5 0.36 1.4 0.23Explanation / Answer
The value of Kc for the above problem is 1.435.
Since 9.2 gms of N2O4 is 0.1 moles. Calculate the equilibrium concentration
Let x moles of N2O4 form 2x moles of NO2
Based on this equilibrium moles of N2O4 is 0.1-x
Equilibrium moles of NO2 is 2x.
The equilibrium concentration of N2O4 for 0.5L vessel is given as 0.057. Based on that x can be calculated and x turns out to be 0.0715.
Now Kc = [NO2]2 / [N2O4]
Kc= 0.2862/ 0.057 = 1.435
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