Consider the evaporation of methanol at 25.0 C:

CH3OH(l)--->CH3OH(g)

Delta G= 4.3 kJ

Find Delta G at 25.0 C under the following nonstandard conditions:
Pressure of CH3OH= 15.0 mmHg.

DG = 4.3 + (.008314)(298)ln(.197)= .3

To find ΔG at 25.0°C under the nonstandard condition of a pressure of 15.0 mmHg, we can use the equation:

ΔG = ΔG° + RT * ln(Q)

Where:
ΔG = Gibbs free energy change under nonstandard conditions
ΔG° = Gibbs free energy change under standard conditions (given as 4.3 kJ)
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin (25.0°C = 298 K)
ln = natural logarithm
Q = Reaction quotient

In this case, we need to calculate the reaction quotient, Q. The equation for the reaction is CH3OH(l) → CH3OH(g).

Since we are given the pressure of CH3OH(g) as 15.0 mmHg, we can assume that the partial pressure of CH3OH(g) is equal to this value. Therefore, Q = P(CH3OH(g)) = 15.0 mmHg.

Now we can substitute the values into the equation:

ΔG = 4.3 kJ + (8.314 J/(mol·K)) * (298 K) * ln(15.0 mmHg)

Note: We need to convert mmHg to atm.

1 atm = 760 mmHg

15.0 mmHg / 760 mmHg/atm ≈ 0.01974 atm

Now we can continue with the calculation:

ΔG = 4.3 kJ + (8.314 J/(mol·K)) * (298 K) * ln(0.01974 atm)

Using a calculator, we can solve for ΔG.

To find the ΔG (Gibbs free energy) at 25.0°C under nonstandard conditions, we need to use the Gibbs-Helmholtz equation:

ΔG = ΔH - TΔS

Where:
- ΔH is the change in enthalpy (heat) of the system
- T is the temperature in Kelvin
- ΔS is the change in entropy (randomness) of the system

However, we don't have the values for ΔH and ΔS directly. We only know ΔG at standard conditions. But we can calculate ΔG at nonstandard conditions if we have the equilibrium constant (K) for the reaction.

The relationship between ΔG and K is given by the equation:

ΔG = -RT ln(K)

Where:
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- ln represents the natural logarithm

To find ΔG at nonstandard conditions, we need to convert the pressure into concentration using the ideal gas law. The ideal gas law equation is:

PV = nRT

Where:
- P is the pressure
- V is the volume
- n is the number of moles
- R is the ideal gas constant (0.0821 L·atm/mol·K)

First, we need to calculate the number of moles (n) of methanol (CH3OH):

n = PV / RT

Here, we have:
- P = 15.0 mmHg (converted to atm by dividing by 760 mmHg/atm)
- V = unknown (we'll assume it's 1 mole)
- R = 0.0821 L·atm/mol·K
- T = 25.0°C + 273.15 = 298.15 K

n = (15.0 mmHg / 760 mmHg/atm) * (1 L / 1000 mL) * (1 mL / 1 cm^3) * 1 mol / 22.4 L) * 0.0821 L·atm/mol·K * 298.15 K

Once we have the value of n, we can substitute it into the equation for K:

K = [CH3OH(g)] / [CH3OH(l)]

Since the reaction is CH3OH(l) ---> CH3OH(g), the concentration of CH3OH(l) is equal to 1 (assuming V = 1).

K = [CH3OH(g)] / 1

Now that we have the value of K, we can calculate ΔG at the nonstandard conditions using the equation:

ΔG = -RT ln(K)

Plug in the values:
- R = 8.314 J/mol·K
- T = 25.0°C + 273.15 = 298.15 K
- K = calculated value from above

ΔG = -8.314 J/mol·K * 298.15 K * ln(K)

Finally, convert the answer to kJ by dividing by 1000.

This calculation will give you the value of ΔG at 25.0°C under the given nonstandard conditions.