A flask is charged with 1.560 atm of N 2 O 4 ( g ) and 0.86 atm NO 2 ( g ) at 25

Question

A flask is charged with 1.560 atm of N2O4(g) and 0.86 atm NO2(g) at 25°C. The equilibrium reaction is given in the equation below.

After equilibrium is reached, the partial pressure of NO2 is 0.512 atm.

(a) What is the equilibrium partial pressure of N2O4?
  atm
(b) Calculate the value of Kp for the reaction.
  (c) Is there sufficient information to calculate Kcfor the reaction?

Yes, because the temperature is specified.

No, because the value of Kc can be determined experimentally only.    

Yes, because the partial pressures of all the reactants and products are specified.

If Kc can be calculated, find the value of Kc. Otherwise, enter 0.

Explanation / Answer

a)
N2O4 <-->2NO2
1.56                0.86   (initial)
1.56-x          0.86+2x (at equilibrium)
given that p (NO2)= 0.512 atm
so,
0.86+2x = 0.512
x = -0.174 atm
p(N2O4) = 1.56 - x
                    = 1.56 - (-0.174)
                     = 1.734 atm

b)
Kp= p(NO2)^2/ P(N2O4)
     = 0.512^2 / 1.734
     = 0.1512

C)
Yes, because the temperature is specified.

Kc = Kp *   (R*T)^(delta n)
      = 0.1512*(8.314*298)^(2-1)
      =374.6

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