A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm of NO2(g) at 25...
A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm of NO2(g) at 25 ∘C , and the following equilibrium is achieved:
N2O4(g)⇌2NO2(g)
After equilibrium is reached, the partial pressure of NO2 is 0.519 atm .
Calculate the value of Kc for the reaction.
Homework Answers
N2O4(g)⇌2NO2(g)
initially
PN2O4 = 1.500 atm
PNO2 = 1.00 atm
at equilibrium
PNO2 = 0.519 atm
1.00 - 0.519 = 0.481 atm changed
PN2O4 = 1.500 - 2(0.481) = 2.462 atm
Kp = (PNO2)2 / (PN2O4)
Kp = (0.519)2 / ( 2.462)
Kp = 0.109
we have a relation
Kp = Kc (RT)
n
n = 2 - 1 = 1
0.109 = Kc (0.0821 x 298)1
0.109 = Kc 24.46
Kc = 4.46 x 10-3
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