Lecture 2

Lecture 2 - University of Southern California Version 1.2...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
University of Southern California ©Hai Wang Version 1.2 1 AME599 Combustion Chemistry and Physics Lecture 2 2. CHEMICAL, PHYSICAL, AND THERMOCHEMICAL PROPERTIES OF HYDROCARBONS We learned from Lecture 1 that there are three important thermochemical properties for combustion analysis. These are specific heat (or sensible enthalpy), enthalpy of formation, and entropy. In this lecture, we will learn the methods with which these properties may be estimated. To do this we will have to understand a bit of organic chemistry, and specifically the structure of organic molecules. 2.1. Chemical Bonds in Organic Molecules Fossil fuels are basically hydrocarbons. While some of them are oxygenated (e.g., methanol and ethanol), others are pure hydrocarbons composed of only carbon and hydrogen atoms. The structure and chemical make-up of hydrocarbons are closely related to the electronic structures of carbon. 2.1.1 Alkanes A carbon atom has 6 electrons distributed into several atomic orbitals. These are termed as 1s, 2s, 2p x , 2p y , 2 p z etc. The Paulis exclusion principle says that each orbital can have a maximum of 2 electrons in opposite spin. Therefore we expect that 2 electrons will fill the 1 s orbital, 2 electrons for the 2 s orbital, and 1 electron each for the 2 p x and 2 p y orbitals (i.e., 1s 2 2 s 2 ,2 p x 1 ,2 p y 1 ). For a C-H bond to form, the electron in the 2 p x or 2 p y orbital will have to be shared, and with the unpaired electron in an H atom the two electrons form a molecular orbital or a covalent bond that again satisfies Paulis’ exclusion principle. The trouble is that if the electronic structure of a carbon atom is indeed 1s 2 2 s 2 ,2 p x 1 ,2 p y 1 , we would expect that the stable form of a C 1 hydrocarbon is CH 2 since there are only two unpaired electrons in carbon. Of course, we know that the most stable form of C 1 hydrocarbon is methane, CH 4 . It turns out that the four electrons in the 2 s , 2 p x and 2 p y orbitals like to undergo hybridization to promote bond formation, since in general the formation of a chemical bond releases energy and gives more stability to the molecule resulting from bond formation. The sp 3 hybridization is the result of promotion of an electron from the 2 s orbital to the 2 p z orbital and in doing so it produces four identical orbitals organized in a tetrahedral fashion (see, Figure 2.1). Of course, each of these orbitals has an unpaired electron, ready for chemical or covalent bonding. Paired with four unpaired electrons in four H atoms, a methane molecule is formed (see, Figure 2.2). The covalent bonds formed in this process are known as the σ bonds. Because the four sp 3 orbitals are identical, the four C-H bonds have the same bond length (1.09 Å) and all H-C-H angles are equal (109º).
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
University of Southern California ©Hai Wang Version 1.2 2 Figure 2.3 . Structures of typical alkane molecules. Top row (from left): methane, ethane, propane; bottom row: normal butance, iso-butane.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/23/2008 for the course AME 599 taught by Professor Wang during the Spring '08 term at USC.

Page1 / 34

Lecture 2 - University of Southern California Version 1.2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online