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.
To find the ∆G (change in Gibbs free energy) for the reaction, you can use the equation:
∆G = ∆H - T∆S
where ∆H is the change in enthalpy and ∆S is the change in entropy.
First, we need to calculate the overall ∆H and ∆S for the reaction. Since we have the ∆Hf (standard enthalpy of formation) and ∆Sf (standard entropy) values for each compound involved, we can use them to calculate the overall ∆H and ∆S.
For the reaction: 2NH3(g) + 2O2(g) → NH4NO3(s) + H2O(l)
The change in enthalpy (∆H) can be calculated by summing the products and subtracting the sum of the reactants:
∆H = (2 * ∆Hf of NH4NO3) + (∆Hf of H2O) - (2 * ∆Hf of NH3) - (2 * ∆Hf of O2)
∆H = (2 * -365.56 kJ) + (-285.830 kJ) - (2 * -46.11 kJ) - (2 * 0.00 kJ)
∆H = -731.12 kJ - 285.830 kJ + 92.22 kJ + 0 kJ
∆H = -924.73 kJ
The change in entropy (∆S) can be calculated in a similar manner:
∆S = (2 * ∆Sf of NH4NO3) + (∆Sf of H2O) - (2 * ∆Sf of NH3) - (2 * ∆Sf of O2)
∆S = (2 * 151.08 J/K) + (69.91 J/K) - (2 * 192.45 J/K) - (2 * 205 J/K)
∆S = 302.16 J/K + 69.91 J/K - 384.90 J/K - 410 J/K
∆S = -423.83 J/K
Now, we can substitute the values of ∆H and ∆S into the ∆G equation:
∆G = ∆H - T∆S
∆G = -924.73 kJ - (298 K * -0.42383 kJ/K)
∆G = -924.73 kJ + 126.37 kJ
∆G = -798.36 kJ
Therefore, the ∆G for the reaction under standard conditions for the formation of NH4NO3(s) is -798.36 kJ.