Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 15 °C?

Question

Accepted Answer

All right. Hi, everyone. So this question says that ethanol has a heat of vaporization of 38.56 kg joules per mole and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 23 °C? And here we have four different answer choices labeled A through D proposing different values for the vapor pressure and units of torque. So for this question, right, our task here is to find the vapor pressure of ethanol at a certain temperature using standard conditions and the heat of vaporization. So, because we're dealing with two different sets of conditions, we can use the classy Claon equation in which the natural logarithm of P one divided by P two is equal to negative delta H of vaporization divided by R and mortified by one over T two, subtracted by one over T one. Now, P one T one, P two and T two represent the vapor pressures and temperatures in both sets of conditions. Whereas the delta age of vaporization is the heat of vaporization in jules promote because of the gas constant or is equal to 8.314 jewels per mole. Kelvin. Now lets review the given conditions, right? T one is the normal boiling point which is 78.4 °C. Now to convert this into Kelvin, I'm going to add 273.15 to this value which equals 351.55 Kelvin. So thats T one now because this is the normal boiling point. P one is equal to 760 tort. Because recall that the normal boiling point of a substance is the temperature in which the vapor pressure is equal to 760 torque. So that gives us P one. And now tea too is the second set of conditions which is 23 °C. And again, we can add 273.15 to this number. The equal 296.15 Kelvin. And in this case, P two is the unknown. That is what we're solving for. Now, before we can go ahead and proceed with our calculation, we do have to convert our delta H of vaporization into units of jewels per mole. This is to make sure that units remain consistent. So delta H of vaporization is equal to 38.56 kilojoules per mole. Now, when converting kilo joules into jewels, we must multiply that value by 1000 because jewels are the smaller units. So this means that delta of April organization is equal to 3.856 multiplied by 10 to the fourth joules per mole. So now we can plug in these data into the classic Laon equation, right. So the Ln of actually, before we do this, there is one quick correction that I want to make, right? So scrolling up here in the classy Claon equation, I initially wrote the Ln of P one over P two, but it is actually the element of P two divided by P one. So apologies for the confusion, but with that being said, we can go ahead and plug in the data that we are given right. So thats the Ln of P two divided by 760 tor equals the negative delta H vaporization which is 3.856 multiplied by 10th, the fourth jules per more divided by 8.314 jewels per mole. Kelvin. So now this is going to be multiplied, buy one over 296.15. Kelvin subtracted by one over 351.55 Kelvin. OK. So here on the right side of the equation, there are no variables which means that we can go ahead and evaluate the right side of the equation. So first, because there is a term in parentheses, we're going to find the inverse of both T two and T one and calculate the difference between them. And then we're going to multiply this term, write this difference buy the value of the delta age of vaporization multiplied by negative one. And then we divide by the gas constant 8.314 joules per mole Kon. This means that the Ln of P two divided by 760 tour is equal to negative 2.46796. After simplifying the right side of the equation. So now to eliminate the natural logarithm, we're going to make both sides of the equation an exponent of E. This means that P two divided by 760 tour equals E to the power of negative 2.46796 which equals 0.08476. So lastly to solve for P two were going to multiply both sides of the equation by 760 tour. So 0.08476 multiplied by 760 equals 64 tour after rounding to two significant figures and there you have it. So our answer is 64 tour which matches option B in the multiple choice. So with that being said, thank you so very much for watching. And I hope you found this helpful.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 15 °C?

Verified Solution

Key Concepts

Heat of Vaporization

Boiling Point

Vapor Pressure