Gibbs free energy
What is the equilibrium constant at 25 C?
deltaG= deltaH - (T(K)+deltaS)
deltaS= -121.1
deltaH= -77.36
so at 25 C
deltaG = -77.36 - (298*(-121.1))
deltaG ~ 3.6*(10^4)
so my question is: what *is* the equilibrium constant, or perhaps more accurately, how does one determine it?
You've thrown up some numbers but you don't say what they are or anything about the reaction. More importantly, the equation you are using is not correct.
DG = DH - T*DS (not T+DS)
sorry.
The reaction is:
2 NO(g) + Cl2 ---><--- 2NOCl
deltaH is the heat of reaction
deltaS is the change in entropy
T is temperature in kelvin
deltaG is a direct measure of spontaneity
DeltaH (kJ/mol) 2NO= 90.29/mol
DeltaH (kJ/mol) Cl2= 0
DeltaH (kJ/mol) 2NOCl= 51.71/mol
DeltaS (J/mol K) 2NO= 210.65/mol
DeltaS (J/mol K) Cl2= 223.0/mol
DeltaS (J/mol K) 2NOCl= 261.6/mol
DeltaG (kJ/mol) 2NO= 86.60/mol
DeltaG (kJ/mol) Cl2= 0
DeltaG (kJ/mol) 2NOCl= 66.07/mol
my delta values from above were from calculating final minus initial values (change) but I just realized as I was typing this that I did not account so deltaS values having only joules in the numerator while the deltaG and deltaH values have kilojoules in the numerator. So my delta G answer is wrong.
Forgetting that DH is in kJ/mol and DS is in J/mol is a common mistake. You only have to make that mistake once so let this be your one time.
If you want to calculate the K for the reaction, I would go about it another way (unless you were given the values you posted in the problem).
Look up delta Go values, then
DGorxn = (n*DGoproducts) - (n*DGoreactants)
DG = DGo + RTlnK
At equilibrium, DG = 0 and this becomes
DGo = -RTlnK
Plug in DGo from above and solve for K.
The equilibrium constant (K) is a mathematical representation of the ratio of products to reactants at equilibrium for a chemical reaction. It helps us understand the extent to which a reaction occurs in either the forward or reverse direction. The equilibrium constant can be determined from the Gibbs free energy change (ΔG) using the equation:
ΔG = -RT ln K
Where:
ΔG is the change in Gibbs free energy
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin (25°C = 298 K)
K is the equilibrium constant
By rearranging the equation, we can solve for K:
K = e^(-ΔG/(RT))
Let's use the values you provided to calculate the equilibrium constant at 25°C (298 K):
ΔG = 3.6 * 10^4 J/mol
R = 8.314 J/mol·K
T = 298 K
Plugging these values into the equation:
K = e^(-3.6 * 10^4/(8.314 * 298))
K ≈ 1.33 * 10^(-24)
So the equilibrium constant at 25°C is approximately 1.33 * 10^(-24). This means that the reaction heavily favors the reactants, as the concentration of the products is extremely low at equilibrium.