The boiling point of ethanol is 78.5C and delta H vap = 40.5kJ/mol. Calculate the vapor pressure (in torr) of ethanol at 0.0C ?
I know that I am close to the answer using this formula with plugged in information:
ln p2/760= 40500/8.314(1/273-1/351.1)
Just not sure what I am doing wrong. Is someone could assist me that would be greatly appreciated! thanks :)
I didn't go through it but it appears to me you didn't convert T2 properly.
273.15 + 78.5 = 351.65. Try that. If it doesn't work let me know and I'll go through the math. BUT if would help if you posted you work and I looked for the error. It's better that way.
To calculate the vapor pressure of ethanol at 0.0°C, we can use the Clausius-Clapeyron equation. The equation is as follows:
ln(P2/P1) = -(delta H vap / R) * (1/T2 - 1/T1)
Where:
P1 is the vapor pressure at one temperature (known)
P2 is the vapor pressure at another temperature (which we want to find)
delta H vap is the molar enthalpy change for the phase transition (known)
R is the ideal gas constant (8.314 J/(mol·K) or 0.0821 L·atm/(mol·K))
T1 and T2 are the temperatures in Kelvin (known)
First, we need to convert the boiling point of ethanol from Celsius to Kelvin:
T1 = 78.5°C + 273.15 = 351.65 K
Next, we convert the desired temperature, 0.0°C, to Kelvin:
T2 = 0.0°C + 273.15 = 273.15 K
Now, we can substitute the known values into the equation and solve for ln(P2/P1):
ln(P2/P1) = -(40.5 kJ/mol) / (8.314 J/(mol·K)) * (1/273.15 K - 1/351.65 K)
Calculating the expression in parentheses:
1/273.15 K - 1/351.65 K = 0.003654
Substituting the calculated value back into the equation:
ln(P2/P1) = -(40.5 kJ/mol) / (8.314 J/(mol·K)) * 0.003654
Simplifying the equation:
ln(P2/P1) = -1.826 * 0.003654
Calculating the right-hand side:
ln(P2/P1) = -0.006674
Finally, we can solve for P2 by taking the exponential of both sides:
P2/P1 = e^(-0.006674)
Calculating the exponential value:
P2/P1 ≈ 0.993352
Now, we can solve for P2 by multiplying both sides by P1:
P2 = P1 * 0.993352
Substituting the known value for P1:
P2 = 1 atm * 0.993352
Converting the value to torr (1 atm = 760 torr):
P2 ≈ 754.4 torr
Therefore, the vapor pressure of ethanol at 0.0°C is approximately 754.4 torr.