Question

Using the data in Appendix C in the textbook and given the pressures listed, calculate ΔG for each of the following reactions at 298 K.

Part A) N2(g)+3H2(g)→2NH3(g)

Part B) N2(g)+3H2(g)→2NH3(g)
PN2 = 2.5 atm , PH2 = 7.2 atm , PNH3 = 1.7 atm

Part C) 2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)

Part D) 2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)
PN2H4=PNO2=1.0×10−2atm, PN2 = 2.0 atm , PH2O = 1 atm

Part E) N2H4(g)→N2(g)+2H2(g)

Part F) N2H4(g)→N2(g)+2H2(g)
PN2H4 = 0.6 atm , PN2 = 7.2 atm , PH2 = 9.1 atm

Answer 1

In general, reacall that

dG = dGproducts - dGreactants

so

Part A) N2(g)+3H2(g)→2NH3(g)

dG = 2*(−16.4) - (1*0 + 3*0)

dG = -32.8 kJ/mol

dG° =  -32.8 kJ/mol

Part B) N2(g)+3H2(g)→2NH3(g)

For nonstandard:

dG = dG° + RT*ln(Q)

Q = (P-NH3)^2 / (P-N2)(P-H2)^3

R = 8.314, T = 298K

dG = dG° + RT*ln(Q)

dG = -32.8 *10^3 + 8.314*298*ln(1.7^2)/((2.5)(7.2^3)) = -32800 + 2.8177

dG = -32797.1823 J/mol

dG = -32.78 kJ/mol

Part C) 2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)

dGRxn = (3*0 + 4*-228.61) - (2*159.28488 + 2*51.3)

dG° = -1335.60 kJ/mol

Part D) 2N2H4(g)+2NO2(g)→3N2(g)+4H2O(g)

Q = (N2)^3 * (H2O)^4 / (N2H4)^2 (NO)^2

Q = (2^3)(1^4) / (10^-2)^4 = 800000000

substituet

dG = dG° + RT*ln(Q)

dG = -1335600 + 8.314*298*ln( 800000000)

dG =-1284809.47 J/mol

dG = -1284.809 kJ/mol

E)

Part E) N2H4(g)→N2(g)+2H2(g)

dG = 1*0 + 2*9 - (N2H4) = -159.28488 kJ/mol

Part F) N2H4(g)→N2(g)+2H2(g)

Q = (N2)(H2)^2 / (N2H4)

Q =7.2*(9.1^2)/(9,6) = 62.1075

dG = -159284 + 8.314*298*ln(62.1075)

dG = -149054.435 J/mol

dG = -149.054 kJ/mol