Using G (measure of free energy) to determine wether the reaction 4Fe(s)+3O2(g)-> 2Fe2O3(s) is spontaneous at 25C, given that H=.393kcal and S=-0.13 kcal/(mol*K).
Is the reaction more or less spontaneous at 1000C assuming that H and S remain constant with temperature?
G at 25C is -354.26(and im not sure what type of units to put it in)
G at 1000C is -227.51.
I do not know how to tell if the reaction is spontaneous at 25C or how to tell if it is more or less spontaneous at 1000C.
Thanks for your help!!!
The sign of DG determines if a reaction is spontaneous or not.
* DG < 0: the reaction is spontaneous
* DG > 0: the reaction is not spontaneous
* DG = 0: the reaction is at equilibrium .
Be certain you worked temperatures in K, not C.
I worked in K.... so at either temp the reaction is not spontaneous? but it is more spontaneous at 1000C?
IT is negative, so spontaneous at both temps. It is more negative at 25C.
To determine if a reaction is spontaneous at a given temperature, you can use the Gibbs free energy (ΔG) equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. If ΔG is zero, the reaction is at equilibrium.
For the given reaction: 4Fe(s) + 3O2(g) -> 2Fe2O3(s)
Given:
ΔH = 0.393 kcal/mol = 0.393 x 4.184 kJ/mol = 1.642 kJ/mol (Converted to SI units)
ΔS = -0.13 kcal/(mol*K) = -0.13 x 4.184 kJ/(mol*K) = -0.542 kJ/(mol*K) (Converted to SI units)
At 25°C (298 K):
ΔG_25°C = ΔH - TΔS
ΔG_25°C = 1.642 kJ/mol - (298 K) x (-0.542 kJ/(mol*K))
ΔG_25°C = 1.642 kJ/mol + 161.316 kJ/mol
ΔG_25°C = 162.958 kJ/mol
Given G_25°C = -354.26, we can rearrange the equation to find ΔG_25°C:
ΔG_25°C = G_25°C - G_standard
162.958 kJ/mol = -354.26 kJ/mol - G_standard
G_standard = -354.26 kJ/mol - 162.958 kJ/mol
G_standard = -517.218 kJ/mol
So, at 25°C, the reaction is non-spontaneous since ΔG_25°C = 162.958 kJ/mol > 0.
To determine if the reaction is more or less spontaneous at 1000°C, we need to calculate ΔG at that temperature using the new temperature and the same ΔH and ΔS values:
T = 1000°C = 1000 + 273.15 = 1273.15 K
ΔG_1000°C = ΔH - TΔS
ΔG_1000°C = 1.642 kJ/mol - (1273.15 K) x (-0.542 kJ/(mol*K))
ΔG_1000°C = 1.642 kJ/mol + 690.687 kJ/mol
ΔG_1000°C = 692.329 kJ/mol
Given G_1000°C = -227.51, we can rearrange the equation to find ΔG_1000°C:
ΔG_1000°C = G_1000°C - G_standard
692.329 kJ/mol = -227.51 kJ/mol - G_standard
G_standard = -227.51 kJ/mol - 692.329 kJ/mol
G_standard = -919.839 kJ/mol
At 1000°C, the reaction is still non-spontaneous since ΔG_1000°C = 692.329 kJ/mol > 0.
Therefore, the reaction is non-spontaneous at both 25°C and 1000°C.