Question

What is the ∆G for the following reaction under standard conditions (T = 298 K) for the formation of NH4NO3(s)? 2NH3(g) + 2O2(g) NH4NO3(s) + H2O(l)

Given:
NH4NO3(s): ∆Hf = -365.56 kJ ∆Sf = 151.08 J/K.


NH3(g): ∆Hf = -46.11 kJ ∆Sf = 192.45 J/K.
H2O(l): ∆Hf = -285.830 kJ ∆Sf = 69.91 J/K.
O2(g): ∆Hf = 0.00 kJ ∆Sf = 205 J/K.

Answer 1

The ΔG for the formation of NH4NO3(s) under standard conditions is -77.79 kJ.

Step-by-step explanation:

The standard Gibbs free energy change (ΔG) for a reaction can be calculated using the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy and ΔS is the change in entropy. To find the ΔG for the given reaction, we need to calculate the sum of the ΔH and -TΔS values for the reactants and products.

First, we calculate the sum of the ΔH values for the reactants (2NH3(g) and 2O2(g)) and subtract the sum of the ΔH values for the products (NH4NO3(s) and H2O(l)). The result is the change in enthalpy (ΔH) for the reaction.

Next, we calculate the sum of the -TΔS values for the reactants and products. The temperature (T) is given as 298 K. The result is the change in entropy (-TΔS) for the reaction.

Finally, we use the equation ΔG = ΔH - TΔS to calculate the ΔG value.

Using the given data, we have:

ΔH = (2 mol NH3 ×-46.11 kJ/mol) + (2 mol O2 ×0.00 kJ/mol) - (1 mol NH4NO3 ×-365.56 kJ/mol) - (1 mol H2O × -285.830 kJ/mol)

ΔS = (2 mol NH3 ×192.45 J/(mol K)) + (2 mol O2 × 205 J/(mol K)) - (1 mol NH4NO3 ×151.08 J/(mol K)) - (1 mol H2O × 69.91 J/(mol K))

Plugging in the values and converting the units, we have:

ΔH = 92.22 kJ

ΔS = 568.26 J/K

Finally, we can calculate the ΔG:

ΔG = ΔH - TΔS

ΔG = 92.22 kJ - (298 K× 0.56826 kJ)

ΔG = 92.22 kJ - 170.01448 kJ

ΔG = -77.79448 kJ

Answer 2

∆G = ∆H - T∆S

NH3: ∆G = -46.11x10^3 - (298)(192.45) = -103460.1 J
O2: ∆G = 0 - (298)(205) = -61090 J
NH4NO3: ∆G = -365.56x10^3 - (298)(151.08) = -410581.84 J
H2O: ∆G = -285.830x10^3 - (298)69.91) = -306663.18 J

∆Grex = ∆Gproducts - ∆Greactants
∆Grex = (-410581.84 + -306663.18) - (-103460.1/2 + -61090/2)
∆Grex =-634969.97 J/mol = -634.97 kJ/mol