Question

# A flask is charged with 1.800 atm of N2O4(g)and 1.00 atm NO2(g) at 25 ∘C, and...

A flask is charged with 1.800 atm of N2O4(g)and 1.00 atm NO2(g) at 25 ∘C, and the following equilibrium is achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.517 atm .

1)What is the equilibrium partial pressure of N2O4?

2)Calculate the value of Kp for the reaction.

3)Calculate Kc for the reaction.

1)

pN2O4 pNO2

initial 1.8 1.0

change +1x -2x

equilibrium 1.8+1x 1.0-2x

To find x, we will use the equilibrium concentration provided in the question

pNO2 = 0.517

1.0-2x = 0.517

x = 0.2415 atm

[N2O4] = 1.8 + x = 1.8 + 0.2415 =2.0415 atm

2)

Equilibrium constant expression is

Kp = pNO2^2/pN2O4

= 0.517^2 / 2.0415

= 0.131

3)

T= 25.0 oC

= (25.0+273) K

= 298 K

delta n = number of gaseous molecule in product - number of gaseous molecule in reactant

delta n = 1

Kp= Kc (RT)^deltan

0.131 = Kc *(0.0821*298.0)^(1)

Kc = 5.35*10^-3

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