Consider the following reaction...HELP!!?

a) Is the reaction product-favoured at 25°C?

Use the equations: ΔG°(rxn) = ΔH°(rxn) - TΔS(rxn)°

ΔH(rxn)° = Sum[ΔHf°(products)] - Sum[ΔHf°(reactants)] =

ΔH(rxn)° = [(1 mole)(-602.9 kJ/mol)] - [(1 mol)(0 kJ/mol) + (4 mol)(-110.525 kJ/mol)]

ΔH(rxn)° = -160.8 kJ

ΔS(rxn)° = Sum[S°(products)] - Sum[(S°(reactants)]

ΔS(rxn)° = [(1 mol)(410.6 J/K·mol)] - [(1 mol)(29.87 J/K·mol) + (4 mol)(197.674 J/K·mol)]

ΔS(rxn)° = (410.6 J/K) - [(29.87 J/K) + (790.696 J/K)] = (410.6 J/K) - (820.566 J/K)

ΔS(rxn)° = -409.966 J/K = -0.409966 kJ/K

ΔG°(rxn) = -160.8 kJ - (298 K)(-0.409966 kJ/K = -160.8 kJ - (-122.169868 kJ) = -38.6 kJ

b) Is the reaction enthalpy or entropy driven? Explain.

If |ΔH(rxn)°| >> |TΔS(rxn)°|: enthalpy-driven reaction

If ΔH(rxn)° << TΔS(rxn)° : entropy-driven reaction.

It appears that the reaction is enthalpy driven at the given temperature.

c) What is the value of K at 25°C?

K = 10^-(ΔG°)/(2.303)(8.314 x 10^-3 kJ/K·mol)(298 K) = 5.98 x 10^6

d) At what temperature, in °C, will this reaction become spontaneous?

When K = 1, ΔG° = 0 and the reaction is at equilibrium. When the reaction is spontaneous K > 1

1 = 10^(-38.6 kJ/mol)/(2.303)(8.314 x 10^-3 kJ/K·mol)(T)

You need to solve for T in this equation. The easiest way is to use logs.

Hope this is helpful to you. JIL HIR

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