For the following reaction CuO(s) + H_2(g) rightarrow Cu(s) + H_2O(g) at 298 K,

Question

For the following reaction CuO(s) + H_2(g) rightarrow Cu(s) + H_2O(g) at 298 K, H_r degree = -87.0 kj and DeltaS_r = 47.0 J K^-1. Calculate DeltaG_r degree at 400 K. Calculate DeltaG_r at 298 K for the reaction given DeltaG_r degree = 6.2 kJ at 298 K. If DeltaGr degree = 27.1 kJ at 25 degree C for the reaction Consider the reaction N_2(g) + O_2(g) rightarrow 2NO(g) If the standard molar free energy of formation of NO(g) at 298 K is 86.69 kJ/mol, calculate the value of the equilibrium constant for this reaction at 298 K.

Explanation / Answer

24) G = H - TS

            = - 87000 J - ( 400 K x 47 J/K)

           = -105800 J

          = - 106 kJ

25) K = PC2H5OH = 0.0263

G =  Go + RT In Keq

      = 6200 J + (8.314) (298) In (0.0263)

     = - 2813 J

    = - 2.8 kJ

26)

G = - RT In K

K = e^(-G/RT)

   = e^(-27100/ 8.314 x 298)

= 1.78 x 10-5

27) G = 2NO -( N2 + O2)  

             = 2 x 86.69 kJ/mol - 0- 0 = 173.38 kJ/mol = 173380 J/mol

G = - RT In K

K = e^(-G/RT)

   = e^(-173380/ 8.314 x 298)

= 4.09 x 10-31

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