Question

For the reaction N_2(g) + 3 H_2(g) rightarrow 2 NH_3(g) Delta G degree = -23.6 kJ and Delta S degree = -198.7 J/K at 345 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 345 K. The standard enthalpy change for the reaction of 2.30 moles of N_2(g) at this temperature would be kJ. For the reaction 2 H_2O_2(l) rightarrow 2 H_2O(l) + O_2(g) Delta G degree = -236.9 kJ and Delta H degree = -196.00 kJ at 325 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 325 K. The entropy change for the reaction of 1.57 moles of H_2O_2(l) at this temperature would be J/L. For the reaction Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) Delta H degree = 98.8 kJ and Delta S degree = 142.5 J/K The standard free energy change for the reaction of 2.35 moles of Fe_2O_3(s) at 281 K, 1 atm would be kJ. This reaction is (reactant, product) favored under standard conditions at 281 K. Assume that Delta H degree and Delta S degree are independent of temperature.

Answer 1

Q1.

since dG is negative, this is favoured towad the right

so this reaction is product favoued

for

dG = dH - T*dS

dH = dG + T*dS = -23.6 + 345*-(0.1987)

dH = -92.15 kJ/mol

for 1 mol of N2 --> -92.15

for 2.3 mol --> -92.15*2.3 = -211.945 kJ will be formed

Q2

This is once again -dG, so this is favoured toward products

Find Entopy given:

dG = dH - T*dS

dS = (dG-dH)/(-T)

dS = (-236.9 - -196)/(-325) = 0.1258 kJ/molK

this is per 2 mol of H2O2

for

1.57 moles --> 1.57/2*0.1258 = 0.098753 kJ/K = 0.098753*10^3 J/K = 98.75 J/K

Q3.

finally

dG = dH - T*dS = 98.8 -281*(0.1425) = 58.757 kJ

this must favour reactants, since it is positive with respect to dG value

So

dG when 1 mol reat --> 58.757 kJ

dG when 2.35 moles react --> 58.757*2.35 = 138.07 kJ