is Fe3O4 + 4H2 = 3Fe + 4H2O spontaneous at 298 K?
Delta H system= 149.8 kj/mol
Delta S system= 610.0 kj/mol
dG = dH - TdS
I don't believe dS is 610 kJ/mol. 610 J/mol perhaps?
dG = 149.8 - (298*0.610)
dG = 0 equil
dG = - spontaneous
dG = + not spontaneous
To determine whether the reaction Fe3O4 + 4H2 = 3Fe + 4H2O is spontaneous at 298 K, we can use the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
where ΔG is the change in Gibbs Free Energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.
Given:
ΔH = 149.8 kJ/mol
ΔS = 610.0 J/mol·K (Note: the units must be consistent, so we convert kJ to J by multiplying by 1000: 610.0 J/mol·K * 1000 J/1 kJ = 610000 J/mol·K)
T = 298 K
Plugging these values into the equation:
ΔG = 149.8 kJ/mol - (298 K * 610000 J/mol·K)
To simplify further, we need to convert kJ to J:
ΔG = 149.8 kJ/mol * 1000 J/1 kJ - (298 K * 610000 J/mol·K)
ΔG = 149800 J/mol - 182180000 J/mol
Now, we can calculate ΔG:
ΔG = -182030200 J/mol
Since ΔG is negative, the reaction is spontaneous at 298 K.
To determine if the reaction is spontaneous at 298 K, we need to calculate the Gibbs free energy change (ΔG) using the equation:
ΔG = ΔH - TΔS
Where:
ΔH = enthalpy change of the system (in this case, 149.8 kJ/mol)
ΔS = entropy change of the system (in this case, 610.0 J/mol * K)
T = temperature in Kelvin (298 K)
First, let's convert the units:
ΔS = 610.0 J/mol * K = 0.61 kJ/mol * K
Now, substitute the given values into the equation:
ΔG = 149.8 kJ/mol - (298 K * 0.61 kJ/mol * K)
Calculating the right side of the equation:
ΔG = 149.8 kJ/mol - 182.78 kJ/mol
ΔG = -32.98 kJ/mol
Since the value of ΔG is negative (-32.98 kJ/mol), the reaction is spontaneous at 298 K.