Enthalpy is a thermodynamic property of a system. It is the sum of the internal energy added to the product of the pressure and volume of the system. It reflects the capacity to do non-mechanical work and the capacity to release heat. Enthalpy is denoted as **H**; specific enthalpy denoted as **h**. Common units used to express enthalpy are the joule, calorie, or BTU (British Thermal Unit). Enthalpy in a throttling process is constant.

It is change in enthalpy that is calculated rather than enthalpy, in part because total enthalpy of a system cannot be measured. However, it's possible to measure the difference in enthalpy between one state and another. Enthalpy change may be calculated under conditions of constant pressure.

### Enthalpy Formulas

H = E + PV

where H is enthalpy, E is internal energy of the system, P is pressure, and V is volume

*d* H = T *d* S + P *d* V

### What Is the Importance of Enthalpy?

- Measuring the change in enthalpy allows us to determine whether a reaction was endothermic (absorbed heat, positive change in enthalpy) or exothermic (released heat, negative change in enthalpy).
- It is used to calculate the heat of reaction of a chemical process.
- Change in enthalpy is used to measure heat flow in calorimetry.
- It is measured to evaluate a throttling process or Joule-Thomson expansion.
- Enthalpy is used to calculate minimum power for a compressor.

- There are many other applications of enthalpy in thermal engineering.
- Enthalpy change occurs during a change in the state of matter.

### Example Change in Enthalpy Calculation

You can use the heat of fusion of ice and heat of vaporization of water to calculate the enthalpy change when ice melts into a liquid and the liquid turns to a vapor.

The heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram of ice melts). The heat of vaporization of liquid water at 100°C is 2257 J/g.

**Part a:** Calculate the change in enthalpy, ΔH, for these two processes.

H_{2}O(s) → H_{2}O(l); ΔH = ?

H_{2}O(l) → H_{2}O(g); ΔH = ?**Part b:** Using the values you calculated, find the number of grams of ice you can melt using 0.800 kJ of heat.

**Solution****a.)** The heats of fusion and vaporization are in joules, so the first thing to do is convert to kilojoules. Using the periodic table, we know that 1 mole of water (H_{2}O) is 18.02 g. Therefore:

fusion ΔH = 18.02 g x 333 J / 1 g

fusion ΔH = 6.00 x 10^{3} J

fusion ΔH = 6.00 kJ

vaporization ΔH = 18.02 g x 2257 J / 1 g

vaporization ΔH = 4.07 x 10^{4} J

vaporization ΔH = 40.7 kJ

So, the completed thermochemical reactions are:

H_{2}O(s) → H_{2}O(l); ΔH = +6.00 kJ

H_{2}O(l) → H_{2}O(g); ΔH = +40.7 kJ**b.)** Now we know that:

1 mol H_{2}O(s) = 18.02 g H_{2}O(s) ~ 6.00 kJ

Using this conversion factor:

0.800 kJ x 18.02 g ice / 6.00 kJ = 2.40 g ice melted

**Answer****a.)**

H_{2}O(s) → H_{2}O(l); ΔH = +6.00 kJ

H_{2}O(l) → H_{2}O(g); ΔH = +40.7 kJ**b.)** 2.40 g ice melted