Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a normal boiling point of 46.3 °C. Find ΔH_vap for carbon disulfide.

Question

Accepted Answer

Welcome back everyone to another video compound X has a vapor pressure of 355 tor at 25 °C and a normal boiling point of 63.2 °C by the entropy of vaporization. For compound X were given for answer choices. A 19 B 3.3 C 36 D 27. All of them are given in kilo juice per mole. Those are standard units for the entropy of vaporization for the purposes of this problem. Let's recall the Cla Cla Peron equation which states that Ln the natural logarithm of the ratio of P two and P one should be equal to the negative delta H map. The entropy of vaporization divided by R which is the universal gas constant and multiplied by one divided by T two minus one, divided by T one. Those are the absolute temperatures add pressures P one and P two, right. So what we're solving for is the entropy of the polarization, meaning we can rearrange this formula for delta H VA and we end up with negative R multiplied by Ln of P two divided by P one. And we're dividing everything bye, the difference in the reciprocal values of those temperatures. So one divided by T two minus one, divided by T one. So let's start substituting the values and now our R value is 8.314 joules per Kelvin per mole. We multiply by the natural logarithm of the ratio of our pressures. Now, essentially what we can do here is first of all, take P two as one atmosphere because the normal boiling point is measured at one atmosphere, which is also equivalent to 760 millimeters mercury, right? We're going to use those units because we're given 355 tor. So it would be more accurate to say that instead of millimeters mercury, let's just add Tor as our units, they are basically equivalent. So let's fix this 760 tour. Now, our pressure one is given to us is 325 tour. We are done with our numerator. What about our denominator? We have one divided by T two. Now, the question is, what is our T two or basically what is our temperature at which we have 760 tour? Because we have assigned P two as one atmosphere. So the normal boiling point which is measured at one atmosphere corresponds to a temperature of 63.2 °C, meaning we're going to convert that into Kelvin. So what we have to do here is take 63.2 °C and add 273.15 Kelvin. And we're going to subtract the temperature that corresponds to 355 Torre, which is 25 °C. So T one is 25 °C. We're going to take 25 °C and add 273.15 Kelvin. When we perform the calculation, we end up with 19 kilojoules per mole, right? We essentially have to divide our answer by 1000 to get the answer in kilojoules per mole instead of jules. And now we can essentially state that the correct answer to this problem corresponds to option a 19 kilojoules per mole. That's the entropy of A IZATION. Thank you for watching.

Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a normal boiling point of 46.3 °C. Find ΔH_vap for carbon disulfide.

Verified Solution

Key Concepts

Vapor Pressure

Enthalpy of Vaporization (ΔH<sub>vap</sub>)

Clausius-Clapeyron Equation

Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a normal boiling point of 46.3 °C. Find ΔHvap for carbon disulfide.

Verified Solution

Key Concepts

Vapor Pressure

Enthalpy of Vaporization (ΔH<sub>vap</sub>)

Clausius-Clapeyron Equation

Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a normal boiling point of 46.3 °C. Find ΔH_vap for carbon disulfide.