The ∆G° of vaporization for butane at 298 K and 1.00 atm is -2.125 kJ/mol. Calculate the pressure, in atm, of butane vapor in equilibrium with butane liquid at 298 K.
Hi skylar,
The question asks for the pressure, in atm, not the kJ.
Thanks.
Isabelle Loaisiga
Well, I'm not much of a scientist, but I do know a bit about pressure (and clowns!). So, let's give it a go!
To find the pressure of butane vapor in equilibrium with the liquid, we can use the formula:
∆G = ∆G° + RT ln(P/P°)
Where:
∆G is the change in Gibbs free energy
∆G° is the standard Gibbs free energy change
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298 K)
P is the pressure of the vapor we're trying to find
P° is the standard pressure (1 atm)
Now, let the clown calculations begin!
First, let's convert the standard Gibbs free energy change from kJ/mol to J/mol:
∆G° = -2.125 kJ/mol * 1000 J/kJ = -2125 J/mol
Next, we will plug in the values into the formula:
-2125 J/mol = - RT ln(P/1 atm)
Now, let's rearrange the equation to solve for P:
ln(P/1 atm) = -2125 J/mol / (8.314 J/(mol·K) * 298 K)
Using my amazing clown calculator, I find that the value inside the logarithm is approximately equal to -0.9997.
Now, let's take the natural logarithm of both sides:
ln(P) - ln(1 atm) = ln(P/1 atm) = -0.9997
And finally, we can simplify:
ln(P) - 0 = -0.9997
Therefore, the pressure of the butane vapor in equilibrium with the liquid is equal to:
P = e^(-0.9997) atm.
Now, I must apologize, but my circus calculator isn't able to solve this equation precisely. However, I hope the steps I provided gave you some insight into finding the solution.
The relationship between standard Gibbs free energy change (∆G°), equilibrium constant (K), and temperature (T) is given by the equation:
∆G° = -RTlnK
Where:
R = gas constant = 8.314 J/(mol·K)
T = temperature in Kelvin
To convert ∆G° from kJ/mol to J/mol, multiply by 1000:
∆G° = -2.125 kJ/mol × 1000 J/kJ = -2125 J/mol
Since the question provides the value of ∆G°, we can rearrange the equation to solve for K:
lnK = -∆G° / RT
Substituting the known values:
lnK = -(-2125 J/mol) / (8.314 J/(mol·K) × 298 K)
Calculating the value inside the parentheses:
lnK = 2125 J/mol / 2474.472 J/mol
lnK ≈ 0.858
Now we can solve for K by taking the exponential of both sides of the equation:
K = e^(lnK)
K = e^(0.858)
Using a calculator, evaluate e^(0.858):
K ≈ 2.360
Since butane is in equilibrium, we can write an expression for the equilibrium constant as the ratio of the pressure of the vapor (p_vap) to the pressure of the liquid (p_liquid):
K = p_vap / p_liquid
Rearranging the equation to solve for p_vap:
p_vap = K × p_liquid
p_vap = 2.360 × 1.00 atm
Therefore, the pressure of the butane vapor is approximately 2.360 atm.
To calculate the pressure of butane vapor in equilibrium with butane liquid at 298 K, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)
Where P1 is the initial pressure (1.00 atm), P2 is the pressure of butane vapor in equilibrium, ∆Hvap is the enthalpy of vaporization, R is the ideal gas constant (8.314 J/(mol*K)), T1 is the initial temperature (298 K), and T2 is the temperature at equilibrium (also 298 K).
First, we need to convert the given ∆G° of vaporization to ∆Hvap. Since ∆G° = ∆H° - T∆S°, where ∆H° is the enthalpy change and ∆S° is the entropy change, we can rearrange the equation to solve for ∆H°:
∆H° = ∆G° + T∆S°
Given that ∆G° = -2.125 kJ/mol and assuming ∆S° is constant, we substitute these values into the equation:
∆H° = -2.125 kJ/mol + (298 K)(∆S°)
Next, we can substitute the calculated value of ∆H° and the given values into the Clausius-Clapeyron equation to solve for ln(P2/P1):
ln(P2/1.00 atm) = (∆Hvap/R) * (1/298 K - 1/298 K)
Since ln(1) = 0, the equation simplifies to:
ln(P2/1.00 atm) = (∆Hvap/R) * 0
ln(P2/1.00 atm) = 0
Finally, taking the exponential of both sides of the equation:
P2/1.00 atm = e^0
P2 = 1.00 atm
Therefore, the pressure of butane vapor in equilibrium with butane liquid at 298 K is 1.00 atm.