Question

Please explain!

Consider the following reaction N2(g)3H2(g)2NH3(9) Report all answers to three significant figures! a) Determine the standard free energy of the the reaction at 298K AG 33 b) Determine the equilibrium constant at 298K? K-1450x c) Determine the reaction quotient when the partial pressures of the gases are the following N26.70 atm; H2 11.00 atm; NH3 2.30 atm d) Determine the value of ?G when the reaction mixture is the one described in Part C. AG

Answer 1

for the reaction, N2(g)+ 3H2(g) ---------->2NH3,

enthalpy change , deltaH= sum of standard enthalpy of formation of products- sum of standard enthalpy of formation of reactants

and also entropy change = sum of standard entropy of products- sum of standard entropy of reactants

the standard enthalpy of formation for N2 and H2 = 0 Kj/mole and NH3= -46.2 Kj/mole

standard entropy data : ( J/mole.K) : N2= 191.6, H2= 130.6 and NH3= 193.3

standard enthalpy change, deltaH= 2*(-46.2)- (1*0+3*0)= -92.4 Kj

standard entropy change, deltaS= 2*193.3- (1*191.6+3*130.6)=-196.8 J/K

2,1 and 3 are coefficients of NH3, N2 and H2 in the reaction

deltaG0   standard Gibbs free energy change= deltaH-T*deltaS=-92.4*Kj*1000J/KJ-298*(-196.8)=-33753.6 J= -33.754 Kj

but deltaGo is related to Equilibrium constant K as

deltaGo= -RT lnK

lnK= -deltaG/RT= 33753.6/(298*8.314)

k= 825430

if P denotes partial pressure,

Q= reaction coefficient = (PNH3)²/ {PN2*(PH2)³}=(2.3*2.3)/ (6.7*11*11*11)= 0.000593

deltaG= deltaGo+ RT lnQ

=-33754+298*8.314*ln(0.000593)=-52163 J/mole=-52.163 Kj