Enthelpy change of a vaporization rxn

In summary, we are given the standard enthalpy of vaporization for a certain liquid and must calculate q, w, delta H, and delta U when 0.75 mol of the liquid is vaporized at 260 K and 765 Torr. The solution also states that q is equal to delta H, but it is unclear when we should make this assumption. The reason why work is nonzero in this case is because the volume of the system increases and therefore does work on its surroundings. Enthalpy takes into account the change in volume during processes like heating and vaporization. To solve for delta H, we must adjust the enthalpy of vaporization for the number of moles in the system.
  • #1
yungwun22
20
0

Homework Statement



A certain liquid has a standard enthalpy of vaporization of 32 KJ/mol. Calculate q, w, delta H and delta U when 0.75 mol is vaporized at 260 K and 765 Torr.



Homework Equations





The Attempt at a Solution



I was able to get delta H, but the solution states that q is equal to delta H. If delta H is equal to q at constant pressure, I don't know how I'm supposed to know when to make that assumption. Also, why does work not equal zero if the pressure is constant, is this additional work?
 
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  • #2
The amount of work is nonzero because the volume increases and the system therefore does work on its surroundings. This is the big difference between energy and enthalpy; enthalpy takes into account that a system's volume tends to vary during processes such as heating and vaporization.
 
  • #3
I have a similar question, just wondering how you solved ∆H? I can do the rest.

Thanks
 
  • #4
Hi hinglis, welcome to PF. ∆H is the enthalpy of vaporization, adjusted to the number of moles in the system.
 

Related to Enthelpy change of a vaporization rxn

1. What is the definition of enthalpy change of a vaporization reaction?

The enthalpy change of a vaporization reaction is the amount of energy that is absorbed or released when a substance changes from a liquid to a gas state at a constant pressure and temperature.

2. How is the enthalpy change of a vaporization reaction calculated?

The enthalpy change of a vaporization reaction can be calculated by taking the difference between the enthalpy of the gaseous state and the enthalpy of the liquid state at the same temperature and pressure.

3. What factors affect the enthalpy change of a vaporization reaction?

The enthalpy change of a vaporization reaction is affected by the pressure and temperature at which the reaction takes place, as well as the chemical properties of the substance being vaporized.

4. Why is the enthalpy change of a vaporization reaction important?

The enthalpy change of a vaporization reaction is important because it provides information about the amount of energy required to change a substance from a liquid to a gas state. This information is useful in various industries, such as in the design of refrigeration systems or in the production of fuels.

5. Can the enthalpy change of a vaporization reaction be negative?

Yes, the enthalpy change of a vaporization reaction can be negative. This indicates that the reaction releases energy, meaning that the substance is more stable in the gaseous state compared to the liquid state. This is often the case for substances with strong intermolecular forces, such as water.

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