a) dG = -RTlnK
b) dG at equilibrium is zero
c) I suppose c means immediately after those conditions are used and before it re-establishes equilibrium. Substitute the values, solve for Qsp and recalculate dG.
I2(g) + Cl2(g) 2 ICl(g) Kp = 81.9 at 25°C
Calculate ΔGrxn for the reaction at 25°C under each of the following conditions:
(a) standard conditions
(b) at equilibrium
(c) PICl = 2.48 atm; PI2 = 0.321 atm; PCl2 = 0.219 atm
b) dG at equilibrium is zero
c) I suppose c means immediately after those conditions are used and before it re-establishes equilibrium. Substitute the values, solve for Qsp and recalculate dG.
dG = dGo + RTlnQ
B= 0
C= 0.42
Using the methods suggested by user DrBob222.
(a) Standard conditions? Ugh, so boring. Like a lab coat at a fashion show. Anyway, since we're at standard conditions, ΔGrxn is just ΔG°rxn. No fancy calculations needed. Just plug in the values and let the fun begin!
(b) Ah, equilibrium. It's like trying to balance on a unicycle while juggling flaming swords. The ΔGrxn at equilibrium is zero! Yes, you heard me right. Zero! It's like the reaction is taking a break, saying "Hey, I'm all balanced now, no need to go anywhere."
(c) Time to get specific! PCl = 2.48 atm, PI2 = 0.321 atm, and PCl2 = 0.219 atm. It's like a chemistry circus with these pressures juggling around. But fear not, my chemical acrobat, we can still calculate ΔGrxn using the equation ΔGrxn = ΔG°rxn + RTln(Q). Just plug in the values and let the calculations dance like a bunch of clowns!
I hope I brought a smile to your chemistry-filled face. Just remember, reactions may be serious, but life's too short not to add a little humor into the equation!
ΔGrxn = ΔG°rxn + RT ln(Q)
Where:
- ΔGrxn is the change in Gibbs free energy of the reaction
- ΔG°rxn is the standard Gibbs free energy change of the reaction
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- Q is the reaction quotient, which depends on the concentrations/pressures of the reactants and products.
(a) Standard conditions:
Under standard conditions, the reaction quotient Q = 1 (equilibrium constant Kp). Therefore, ln(Q) = ln(1) = 0.
Thus, the equation simplifies to ΔGrxn = ΔG°rxn + (RT * 0) = ΔG°rxn.
So, under standard conditions, ΔGrxn is equal to the standard Gibbs free energy change of the reaction.
(b) At equilibrium:
At equilibrium, the reaction quotient Q = Kp.
So, ΔGrxn = ΔG°rxn + RT ln(Kp).
(c) Given the partial pressures of the gases, we can calculate Q and then use it to find ΔGrxn:
PICl = 2.48 atm, PI2 = 0.321 atm, PCl2 = 0.219 atm
Q = (PICl^2) / (PI2 * PCl2)
Substitute the values into the equation:
Q = (2.48^2) / (0.321 * 0.219)
Calculate Q and then use it to find ΔGrxn:
ΔGrxn = ΔG°rxn + RT ln(Q)
Note: To calculate ΔG°rxn, you need the standard Gibbs free energies of formation for each species involved in the reaction.