An equilibrium mixture contains N2O4, (P=0.26 atm) and NO2 (P=1.2atm) at 350 K.

Question

An equilibrium mixture contains N2O4, (P=0.26 atm) and NO2 (P=1.2atm) at 350 K. The volume of the container is doubled at constanttemperature.
a) Calculate the equilibrium pressure of N2O4 when the systemreaches a new equilibrium. Express your answer using twosignificant figures. P=? b) Calculate the equilibrium pressure of NO2 when thesystem reaches a new equilibrium. Express your answer using twosignificant figures. P=?
I am confused as to how to go about solving thisproblem.
a) Calculate the equilibrium pressure of N2O4 when the systemreaches a new equilibrium. Express your answer using twosignificant figures. P=? b) Calculate the equilibrium pressure of NO2 when thesystem reaches a new equilibrium. Express your answer using twosignificant figures. P=?
I am confused as to how to go about solving thisproblem. a) Calculate the equilibrium pressure of N2O4 when the systemreaches a new equilibrium. Express your answer using twosignificant figures. P=? b) Calculate the equilibrium pressure of NO2 when thesystem reaches a new equilibrium. Express your answer using twosignificant figures. P=?
I am confused as to how to go about solving thisproblem.

Explanation / Answer

We Know that :       The reaction is :         N2O4  (g)<--------> 2 NO2 (g)             Kp = PNO22 /  PN2O4                     = ( 1.2 )2 / 0.26                     = 5.538          N2O4  (g)<--------> 2 NO2 (g)     I :   0.26                             1.2     C:   -x                                2x      E :   0.26 -x / 2                  1.2 + 2x / 2                                                                      Kp = (   1.2 + 2x / 2 )2 /   ( 0.26 -x / 2 )               5.538 = (   1.2 + 2x / 2 )2 /   ( 0.26 -x / 2 )              solving the above equation we get :               x = 0.08870          Theequilibrium partial pressure of N2O4 =   0.26 -x / 2                                                                            = 0.086 atm          Theequilibrium partial pressure of NO2 =   1.2 + 2x / 2                                                                         = 0.69 atm                     = ( 1.2 )2 / 0.26                     = 5.538          N2O4  (g)<--------> 2 NO2 (g)     I :   0.26                             1.2     C:   -x                                2x      E :   0.26 -x / 2                  1.2 + 2x / 2                                                                      Kp = (   1.2 + 2x / 2 )2 /   ( 0.26 -x / 2 )               5.538 = (   1.2 + 2x / 2 )2 /   ( 0.26 -x / 2 )              solving the above equation we get :               x = 0.08870          Theequilibrium partial pressure of N2O4 =   0.26 -x / 2                                                                            = 0.086 atm          Theequilibrium partial pressure of NO2 =   1.2 + 2x / 2                                                                         = 0.69 atm

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