Answer:
The driving force of a reaction is determined by the difference between the actual reaction's free energy change (ΔG) and the standard free energy change (∆⊖). If the actual free energy change (ΔG) is less than the standard free energy change (∆⊖), the reaction is thermodynamically favorable and will proceed in the forward direction. Conversely, if ΔG is greater than ∆⊖, the reaction will not be spontaneous and will not proceed in the forward direction.
In this case, we know the standard free reaction enthalpy (∆⊖) is 93.26 kJ/mol. However, to determine the driving force of the reaction at the specified conditions (300 bar and 800 K), we need to consider the effect of pressure and temperature on the free energy change.
The actual free energy change (ΔG) can be calculated using the following equation:
ΔG = ∆⊖ + RT * ln(Q)
Where:
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin (800 K in this case)
Q = Reaction quotient
Since all components have the same partial pressure at 300 bar, the reaction quotient (Q) can be expressed as follows:
Q = (P(NH3))^2 / (P(H2))^3 * (P(N2))
Where P(NH3), P(H2), and P(N2) are the partial pressures of ammonia, hydrogen, and nitrogen, respectively, all at 300 bar.
Since we have specified that all components have the same partial pressure, we can simplify the reaction quotient as follows:
Q = (P)^2 / (P)^3 * (P) = P
Where P is the common partial pressure of all components at 300 bar.
Now, let's plug in the values and calculate ΔG:
ΔG = 93.26 kJ/mol + (8.314 J/(mol·K)) * (800 K) * ln(P)
To determine the driving force, we need to compare ΔG to ∆⊖. If ΔG is less than ∆⊖, the reaction is spontaneous and has a driving force for the forward direction.
Remember that the gas pressure (P) must be in units consistent with the gas constant (R), which is J/(mol·K). Therefore, the pressure should be converted to Pascals (Pa) before calculation.
Please note that to provide an exact value for the driving force, the specific pressure value (P) in Pascals needs to be known. Without this value, we cannot determine the exact driving force in this context.