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)
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?
3 answers
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.
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.
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.