How to Calculate S for the Combustion of Propane

How to Calculate S for the Combustion of Propane thumbnail
Propane generates a large increase in entropy when combusted.

Whenever two or more molecules react to form new molecules, energy is released and stored to form or break bonds between atoms. In many cases, the energy released and stored is not equal, which causes a change in the total energy present in a system. This is known as a change in entropy, and is represented in thermodynamics by S. When S is negative in a reaction, the total energy increases, which makes temperatures rise. The combustion of propane releases a large amount of energy, and by calculating S it can be determined exactly how much.

Things You'll Need

  • Pencil
  • Paper
  • Calculator
Show More

Instructions

    • 1

      Write out the equation for the combustion of propane. Propane combusts with 5 oxygen molecules to form 3 carbon dioxide molecules and 4 water molecules. The equation is then 1(C3H8) + 5(O2) = 3(CO2) + 4(H2O).

    • 2

      Write out the heat of formation below C3H8 and O2. The heat of formation is the energy that is released or stored when the bonds of a molecule are broken or formed. The heat of formation of propane is '104.7 kilojoule/mole (kJ/mol), while oxygen is 0 kJ/mol. Kilojoules per mole is a measure of the energy involved in each chemical reaction.

    • 3

      Write out the heat of formation below CO2 and H2O. The heat of formation of carbon dioxide is '393.52 kJ/mol and of water is -285.83 kJ/mol.

    • 4

      Carry down the multipliers from the top equation to the enthalpy equation. Five molecules of oxygen are consumed, so the energy is released five times. The final equation is 1('104.7) + 5(0) = 3('393.52) + 4(-285.83). The difference between the two sides is S. If the left side is larger, S is positive, while if the right side is larger, S is negative.

    • 5

      Total each side of the equation, then find the difference between the two. 2,323.88 - 104.7 = 2,219.18 kJ/mol = S. As a result of the combustion of propane, entropy in the system is increased by 2,219.18kJ/mol.

Related Searches:

References

  • Photo Credit hot-air balloon propane burner - vertical image image by monamakela.com from Fotolia.com

Comments

You May Also Like

Related Ads

Featured