Consider the following reaction:

I2(g) + Cl2(g) 2ICl(g) o

Kp = 81.9 at 25 C.
Calculate ∆Grxn for the reaction at 25 C under each of the following conditions:
a Standard conditions.
b At equilibrium.
c PICl = 2.55 atm; PI2 = 0.325 atm; PCl2 = 0.221 atm.

a) Standard conditions:

Since we are considering standard conditions, we can use the equation ΔG° = -RTlnKp, where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and Kp is the equilibrium constant. Plug in the values:

ΔG° = - (8.314 J/mol·K)(298 K)ln(81.9)
ΔG° = - (8.314 J/mol·K)(298 K)(4.406)

b) At equilibrium:
At equilibrium, the reaction has reached a state in which the forward and reverse reactions occur at equal rates. In this case, ΔG = 0.

c) Using the equation ΔG = ΔG° + RTln(Q), where Q is the reaction quotient, we can calculate ΔG for the given conditions.

Q = (PCl2)^2 / (PI2 x PCl2) = (0.221 atm)^2 / (0.325 atm x 0.221 atm)
Q = 0.2658

ΔG = ΔG° + (8.314 J/mol·K)(298 K)ln(0.2658)
ΔG = - (8.314 J/mol·K)(298 K)(1.327) + (8.314 J/mol·K)(298 K)ln(0.2658)
ΔG = - ((8.314 J/mol·K)(298 K)(1.327) - (8.314 J/mol·K)(298 K)ln(0.2658))

Remember, I'm Clown Bot! I find it fascinating that even chemical reactions can have standard conditions, just like celebrities have their own "standard" red carpet appearances. It's like chemistry is rolling out the red carpet for these reactions!

To calculate ΔGrxn for the reaction under each of the given conditions, we can use the equation:

ΔGrxn = ΔGrxn° + RT ln(Q)

where ΔGrxn is the standard Gibbs free energy change, ΔGrxn° is the standard Gibbs free energy change at standard conditions, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and Q is the reaction quotient.

a) Standard conditions:
At standard conditions, the reaction quotient Q is equal to the equilibrium constant Kp. Therefore, ΔGrxn = ΔGrxn°.

b) At equilibrium:
At equilibrium, the reaction quotient Q is equal to the equilibrium constant Kp. Therefore,

ΔGrxn = ΔGrxn° + RT ln(Kp)

c) Given the partial pressures, we can calculate Q and ΔGrxn using the equation:

Q = (PICl)^2 / (PI2 x PCl2)

Let's substitute the given values into the equation and calculate Q:

Q = (2.55 atm)^2 / (0.325 atm × 0.221 atm)
Q = 405.15

Now, let's use the equation ΔGrxn = ΔGrxn° + RT ln(Q) to calculate ΔGrxn:

a) Standard conditions:
Since the reaction is at standard conditions, ΔGrxn = ΔGrxn°.
(No calculation needed)

b) At equilibrium:
ΔGrxn = ΔGrxn° + RT ln(Kp)
Substitute the given value of Kp and calculate:
ΔGrxn = ΔGrxn° + (8.314 J/(mol·K) × 298 K) × ln(81.9)

c) PICl = 2.55 atm; PI2 = 0.325 atm; PCl2 = 0.221 atm:
ΔGrxn = ΔGrxn° + RT ln(Q)
Substitute the calculated value of Q and calculate:
ΔGrxn = ΔGrxn° + (8.314 J/(mol·K) × 298 K) × ln(405.15)

Note: Make sure to convert the temperature to Kelvin (25°C = 298 K) and use the appropriate units for the gas constant (R).

By following these steps, you can calculate ΔGrxn under each of the given conditions.

a.

dG = -RTlnK

b.
dG = 0 at equlibrium

c.
Are these equilibrium pressures, initial pressures, or some other point?

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