Consider the reaction 2 NO_2(g) leftarrow rightarrow N_2 O_4(g) The values for d

Question

Consider the reaction 2 NO_2(g) leftarrow rightarrow N_2 O_4(g) The values for delta H degree and delta S degree are -58.03 kj/mol and -176.6 J/K-mol, respectively. Calculate delta G degree at 298 K. Assuming delta H degree and delta S degree are independent of temperature, at what temperatures is this reaction spontaneous? Calculate delta G from heats of formation for the following reaction. 2 CH_3 OH(g) + H_2(g) rightarrow C_2H_6(g) + 2 H_2 O_(g) Consider again the reaction 2 NO_2(g) leftarrow rightarrow N_2 O_4(g) Predict the direction of the reaction will proceed if P_NO2 = 0.21 atm and P_N2O4 = 0.50 atm. Temperature is a constant 25 degree C.

Explanation / Answer

a)
       2NO2(g) <---> N2O4 (g)
      
   Given H = -58.03 KJ/mol ; S = -176.6 J/K-mol; T = 298 K ;

   G = ?
  
   G = H -TS = (-58.03 x 10^3 J/mol) - (298 K x -176.6 J/K-mol) = -58030 J/mol + 52626.8 J/mol = -5403.2 J/mol
  
  
   To become the reaction spontaneous the G should become negative;
  
   H = TS at equilibrium
  
   T = H/S = (-58.03 x 10^3 J/mol)/ -176.6 J/K-mol = 329 K
  
   So, the reaction will be spontaneous at temperature below 329 K
  
b)        2CH3OH (g) + H2(g) ---> C2H6 (g) + 2H2O (g)

G reaction = G (products) - G (reactants)
           = {(- 32 KJ/mol)+ 2x (–228.59 kJ/mol)} -{(2x –167.9 KJ/mol) + 0}
          
           = -153.38 KJ/mol
          
          
c) Given reaction ;
       2NO2(g) <---> N2O4 (g)
      
       PNO2 = 0.21 atm and PN2O4 = 0.5 atm

       Qc = [PN2O4]/[PNO2]^2 = 0.5 / (0.21) = 11.34
      
We can find the equilibrium constant from the data given in a)
at 298 K G = -5403.2 J/mol

G = -RT ln Keq

-5403.2 J/mol = -8.314 J·K¹mol¹ × 298 K ln Keq

ln Keq = 2.18

Keq = 8.85

If Q>Keq, the reaction goes from right to left.

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