Consider the following reaction: 2NO2(g) N2O4(g) At 298 K for this reaction, H°

Question

Consider the following reaction:

2NO2(g) N2O4(g)

At 298 K for this reaction, H° = -58.0 kJ and S° = -176.6 J/K.

(a) What is G° for the forward reaction at 298 K? Is the reaction spontaneous at 298 K? Assuming the enthalpy and entropy change do not vary with temperature, at what special temperature, T*, does G° = 0.0 J?

(b) Express the equilibrium constant K for this reaction in terms of equilibrium pressures of the reactant and product gases. What is K at 298 K?

(c) Suppose the initial pressures are: pNO2 =1.00 atm and pN2O4 =1.00 atm at 298 K. Predict the direction in which the reaction will shift as equilibrium is approached. Show your work and give a brief sentence to explain your reasoning.

Please explain the answer and name the equations used to find the answer, thanks!

Explanation / Answer

a)
G° = H° - T*S°
     = -58000 J - 298*(-176.6 J/K)
      = -5373.2 J
Since G° is negative, reaction is spontaneous.
use:
G° = H° - T*S°
0= -58000 J - T*(-176.6 J/K)
T= 328.4 K

b)
K= p(N2O4) / p(NO2)^2
use:
G° = -R*T*ln Kc
-5373.2 = -8.314*298*ln Kc
Kc = 8.75
Now use:
Kp = Kc *(R*T)^(delta n)
   =8.75 * (0.0821*298)^(1-2)
    = 0.36

c)
                           2NO2 <--> N2O4
Qp = p(N2O4) /p(NO2)^2
     = 1/1
      = 1
Since Qp > Kp
There is more of products than it was supposed to be at equilibrium.
Hence reaction willmove in backward direction

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