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?

Express your answer with the appropriate units.

2)Calculate the value of Kp for the reaction.

3)Calculate Kc for the reaction.

Homework Answers

Answer #1

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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