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 the question
To calculate the standard Gibbs free energy change (∆G) for a reaction, you can use the equation:
∆G = ∆H - T∆S
where ∆H is the enthalpy change, T is the temperature in Kelvin, and ∆S is the entropy change.
In this case, you have the enthalpy changes (∆H) and entropy changes (∆S) for each species involved in the reaction. The reaction equation shows the stoichiometric coefficient for each species, which indicates the relative number of moles of each substance.
The enthalpy change for the reaction can be calculated by using the enthalpy changes of the reactants and products, along with their stoichiometric coefficients.
∆H_reaction = (2 * ∆Hf(NH4NO3)) - (2 * ∆Hf(NH3)) - (2 * ∆Hf(O2)) - ∆Hf(H2O)
where ∆Hf represents the enthalpy of formation.
Substituting the given values:
∆H_reaction = (2 * -365.56 kJ) - (2 * -46.11 kJ) - (2 * 0.00 kJ) - (-285.830 kJ)
Next, you can calculate the entropy change for the reaction using the analogous approach, taking into account stoichiometric coefficients:
∆S_reaction = (2 * ∆Sf(NH4NO3)) - (2 * ∆Sf(NH3)) - (2 * ∆Sf(O2)) - ∆Sf(H2O)
Substituting the given values:
∆S_reaction = (2 * 151.08 J/K) - (2 * 192.45 J/K) - (2 * 205 J/K) - 69.91 J/K
Finally, you can calculate ∆G using the equation mentioned at the beginning:
∆G = ∆H_reaction - T∆S_reaction
Substituting the given values (T = 298 K):
∆G = ∆H_reaction - (298 K * ∆S_reaction)
Calculate the value of ∆G using the calculated values for ∆H_reaction and ∆S_reaction.