Enthalpy Definition

Definition and Examples of Enthalpy

Enthalpy in an internal combustion engine is calculated as internal energy plus pressure multiplied by volume.
Enthalpy in an internal combustion engine is calculated as internal energy plus pressure multiplied by volume. Dorling Kindersley, Getty Images

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.

Also Known As: heat content, total heat

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.

H2O(s) → H2O(l); ΔH = ?
H2O(l) → H2O(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 (H2O) is 18.02 g. Therefore:

    fusion ΔH = 18.02 g x 333 J / 1 g
    fusion ΔH = 6.00 x 103 J
    fusion ΔH = 6.00 kJ

    vaporization ΔH = 18.02 g x 2257 J / 1 g
    vaporization ΔH = 4.07 x 104 J
    vaporization ΔH = 40.7 kJ

    So, the completed thermochemical reactions are:

    H2O(s) → H2O(l); ΔH = +6.00 kJ
    H2O(l) → H2O(g); ΔH = +40.7 kJ

    b.) Now we know that:

    1 mol H2O(s) = 18.02 g H2O(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.)
    H2O(s) → H2O(l); ΔH = +6.00 kJ
    H2O(l) → H2O(g); ΔH = +40.7 kJ
    b.) 2.40 g ice melted

More Worked Enthalpy Example Problems