Consider the evaporation of methanol at 25.0 c: CH3OH(l)-- CH3OH(g)
I figured out that Delta G is 4.3 kJ.
Find the Delta G at 25 C under the following conditions:
Pressure of CH3OH= 157mmHg, 103mmHg, 14mmHg
thanks! please explain
Thank you for finding the delta G for me so here is how to solve this problem,
Formula: Delta G(rxn)= DeltaG(rxn)^degree + RTlnQ
1 atm = 760 mmHg
4300J + 8.314*298K*ln(157/760)=
tyvm again.
Well, well, well. We have some evaporating methanol here! Let's dive into it, shall we?
To find the Delta G at different pressures, we can use the equation:
ΔG = ΔG° + RTln(P)
Where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and P is the pressure.
We know that ΔG is 4.3 kJ (remember to convert it to J), and since we are dealing with Methanol, we need to consider the standard pressure as well. The standard pressure is 1 atmosphere, but we'll convert it to mmHg, 'cause why not?
Now, let's break it down for the different pressures:
For 157 mmHg:
Pressure = 157 mmHg
ΔG° = 4.3 kJ
T = 25.0 C + 273.15 (convert it to Kelvin)
And, of course, R = 8.314 J/(mol K).
Now, plug all the values into the equation, do a little math magic, and voila! You got your Delta G at 157 mmHg!
Repeat the same process for 103 mmHg and 14 mmHg, and there you have it.
Hope this answers your question! If not, I'm happy to entertain any further inquiries or even tell you a joke about methanol evaporation. Nah, maybe not.
To find the delta G (change in Gibbs free energy) at 25°C under different pressures, we need to use the equation:
ΔG = ΔG° + RT ln(P)
Where:
- ΔG is the change in Gibbs free energy
- ΔG° is the standard Gibbs free energy change at 25°C, which is given as 4.3 kJ
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25°C = 298 K)
- P is the pressure of methanol in atmospheres (which we will convert from mmHg)
Let's calculate the delta G at each pressure condition:
1. Pressure = 157 mmHg:
First, we need to convert mmHg to atm:
Pressure = 157 mmHg * (1 atm / 760 mmHg) = 0.206 atm
Now, substitute the values into the equation:
ΔG = 4.3 kJ + (8.314 J/(mol·K) * 298 K) * ln(0.206 atm)
2. Pressure = 103 mmHg:
Convert mmHg to atm:
Pressure = 103 mmHg * (1 atm / 760 mmHg) = 0.136 atm
Calculate delta G:
ΔG = 4.3 kJ + (8.314 J/(mol·K) * 298 K) * ln(0.136 atm)
3. Pressure = 14 mmHg:
Convert mmHg to atm:
Pressure = 14 mmHg * (1 atm / 760 mmHg) = 0.018 atm
Calculate delta G:
ΔG = 4.3 kJ + (8.314 J/(mol·K) * 298 K) * ln(0.018 atm)
Now you can substitute the values into the equation to calculate the delta G for each pressure condition.
To find the Delta G (Gibbs free energy change) at 25°C under different pressure conditions for the evaporation of methanol (CH3OH), you can use the equation:
ΔG = ΔG° + RT ln(Q)
Where:
- ΔG is the Gibbs free energy change (the value we want to find)
- ΔG° is the standard Gibbs free energy change (given as 4.3 kJ)
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (298 K for 25°C)
- Q is the reaction quotient, which is a measure of the relative amounts of products and reactants in a system
To calculate Q, we need to consider the partial pressure of CH3OH in the gas phase (CH3OH(g)) relative to the liquid phase (CH3OH(l)).
Using the equation for the reaction:
CH3OH(l) → CH3OH(g)
We can write the equilibrium expression as:
Q = (P(CH3OH(g))) / (P°)
Where:
- P(CH3OH(g)) is the partial pressure of CH3OH(g) under the given conditions
- P° is the standard pressure, which is 1 atm (since 1 atm = 760 mmHg)
Now, let's calculate ΔG under the given pressure conditions.
For Pressure of CH3OH = 157 mmHg:
Convert 157 mmHg to atm by dividing by 760:
P(CH3OH(g)) = 157 mmHg / 760 mmHg/atm ≈ 0.20658 atm
Now, substitute the values into the equation:
ΔG = 4.3 kJ + (8.314 J/mol·K) * (298 K) * ln(0.20658 atm / 1 atm)
Repeat the above steps for the other pressure conditions, P(CH3OH(g)) = 103 mmHg and P(CH3OH(g)) = 14 mmHg, respectively.
By following these steps, you can calculate ΔG for the evaporation of methanol at 25°C under different pressure conditions.