Solid ammonium nitrate can decompose to dinitrogen oxide gas and liquid water. What is delta G at 298K?
Write the equation.
Balance it. Note: dGo f = free energy of formation at 25C. Then
dGofrxn = (n*dGof products) - (n*dGof reactants)
Values for dGof can be found in an appendix of most texts.
A MIXTURE OF 1.00G OF h2 AND 1.00G hE IS PLACED IN A 1.00L FLASK AT 27.0 WHAT ARE THE PARTIAL PRESSURES OF EACH GAS
To determine the value of ΔG (change in Gibbs free energy) at 298K for the given reaction, we need to use thermodynamic data for the reactants and products involved. The equation for the decomposition of solid ammonium nitrate (NH4NO3) can be written as:
NH4NO3(s) → N2O(g) + H2O(l)
To calculate the ΔG, we'll need to know the standard Gibbs free energy change (ΔG°) for the reaction. We can use the equation:
ΔG = ΔG° + RT ln(Q)
Where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol K)), T is the temperature in Kelvin (298K in this case), and ln(Q) is the natural logarithm of the reaction quotient.
The reaction quotient (Q) is the ratio of the concentration (or partial pressure) of the products raised to their stoichiometric coefficients, divided by the concentration (or partial pressure) of the reactants raised to their stoichiometric coefficients.
In this case, since the reaction involves a solid and liquid, we can treat their concentrations as constant and equal to 1. Therefore, the Q value simplifies to the partial pressure of N2O gas.
We can look up the standard Gibbs free energy change (ΔG°) for the reaction in a thermodynamic data source. For ammonium nitrate decomposition, ΔG° is approximately +245 kJ/mol.
Now we can substitute the known values into the equation:
ΔG = ΔG° + RT ln(Q)
ΔG = (+245,000 J/mol) + (8.314 J/(mol K) * 298 K) * ln(Q)
Simplifying further:
ΔG = 245,000 J/mol + (2,475 J/mol) * ln(Q)
Please note that we cannot directly determine the value of ΔG without the knowledge of the reaction quotient (Q), which depends on the partial pressure of N2O gas. The value of ΔG will change as the reaction proceeds, and when equilibrium is reached, ΔG will be zero.
If you provide the partial pressure of N2O gas, we can calculate the value of ΔG at that specific point.