What is the standard gibbs free energy of formation, ∆gºf, of nh3(g) at 298 k?

The formation of NH3 can be represented as:

N2(g) + 3H2(g) → 2NH3(g)

the standard free energy change is given as :

ΔG⁰ = ΔH° - TΔS⁰ (1)

where

ΔH° and ΔS⁰ are the reaction enthalpy and entropy changes respectively. Based on the thermochemical data values we can write:

ΔH°(rxn)  = ∑nΔH⁰f(product) - ∑nΔH⁰f(reactant)

              = [2ΔH⁰f(NH3)] - [3ΔH⁰f(H2)+[1ΔH⁰f(N2)]]

              = 2(-46) - [3(0) + 1(0)] = -92 kj

ΔS°(rxn)  = ∑nS⁰f(product) - ∑nS⁰f(reactant)

              = [2S⁰f(NH3)] - [3S⁰f(H2)+[1S⁰f(N2)]]

              = 2(193) - [3(131) + 1(192)] = -199 J/K = -0.199 kJ/K

Substituting for ΔH° and ΔS⁰ in equation (1) at T = 298 K:

∆Gº = ΔH° - TΔS⁰ = -92 - (298)(-0.199) = -32.7kJ

This is the free energy change corresponding to 2 moles of NH3 formed

Therefore:

∆Gºf(NH3) = -32.7kJ/2 moles = -16.4 kJ/mole

Gibbs free energy is the maximum possible work given by chemical reactions at constant pressure and temperature. Gibbs free energy can be used to determine the spontaneity of a reaction

If the Gibbs free energy value is <0(negative) then the chemical reaction occurs spontaneously. If the change in free energy is zero, then the chemical reaction is at equilibrium ,if it is >0, the process is not spontaneous

Free energy of reaction (G) is the sum of its enthalpy (H) plus the product of the temperature and the entropy (S) of the system

Can be formulated: (at any temperature)

or at (25 Celsius / 298 K, 1 atm= standart)

Under standard conditions:

The value of ∆H° can be calculated from the change in enthalpy of standard formation:

∆H° (reaction) = ∑H° (product) - ∑ H° (reagent)

The value of ΔS° can be calculated from standard entropy data

∆S° (reaction) = ∑S° (product) - ∑ S°(reagent)

There is data that needs to be completed from the questions above:

∆H°f for: (kJ / mole )

N₂ (g): 0

H₂ (g): 0

NH₃ (g): -46

∆S for: (J /mole.K)

N₂ (g): 192

H₂ (g): 131

NH₃ (g): 193

From reaction:

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

Value ∆H°

∆H° (reaction) = ∑H° (product) - ∑ H° (reagent)

∆H° (reaction) = 2.NH₃ - (3.H₂ + 1.N₂)

∆H° (reaction) = 2.-46 - (3.0 + 1.0)

∆H° (reaction) = -92 kJ

Value ∆S°

∆S° (reaction) = ∑S° (product) - ∑ S° (reagent)

∆S° (reaction) = 2.NH₃ - (3.H₂ + 1.N₂)

∆S° (reaction) = 2,193 - (3,131 + 1,192)

∆S° (reaction) = -199 J / mole.K = -0.199 kJ / mole.K

Then

∆G° = ∆H° - T∆S°

∆G° = -92 kJ / mole - (298. -0.199 kJ / mol.K)

∆G° = -32,698 kJ (for 2 moles of NH₃ according to the reaction equation)

for 1 mole of NH₃

∆Go = -32,698 kJ / mol : 2 = 16,349 kJ/mole

Delta H solution

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an exothermic reaction

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 as endothermic or exothermic

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an exothermic dissolving process

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Keywords: the standard gibbs free energy of formation,NH₃

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