A flask is charged with 1.660 atm of N2O4(g) and 1.01 atm NO2(g) at 25°C. The eq

Question

A flask is charged with 1.660 atm of N2O4(g) and 1.01 atm NO2(g) at 25°C. The equilibrium reaction is given in the equation below. N2O4(g) 2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.512 atm. (a) What is the equilibrium partial pressure of N2O4? (b) Calculate the value of Kp for the reaction. (c) Is there sufficient information to calculate Kc for the reaction? No, because the value of Kc can be determined experimentally only. Yes, because the temperature is specified. 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

Answer – We are given, reaction-

N2O4(g) ----> 2 NO2(g)

P NO2 = 1.01 atm, P N2O4(g) = 1.660 atm

At equilibrium, P NO2 = 0.512 atm

So there is reverse reaction

     N2O4(g) ----> 2 NO2(g)

I    1.660                 1.01           

C    -x                    -2x             

E 1.660-x           1.01-2x        

We know, at equilibrium, P NO2 = 0.512 atm

So, 1.01-2x = 0.512 atm

-2x = 0.512-101

      = -0.498 atm

So, x = 0.249 atm

The equilibrium partial pressure of P N2O4(g) = 1.660-x

                                                                        = 1.660-0.249

                                                                          = 1.41 atm                    

Kp = P (NO2(g))2 / P(N2O4(g))

Kp= (0.512 atm)2 / 1.41 atm

Kp = 0.186 atm

We know the relationship between Kp and Kc

Kp = Kc * (RT)n

We know n = sum of moles of product – sum of moles of reactant

                      = 2-1

                      = 1

0.186 = Kc * (0.0821L.atm.mol-1.K-1*298 K )

0.186 = Kc *24.46

Som, Kc = 0.186 /24.46

              = 7.59*10-3 atm

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