{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

themochemistry notes

themochemistry notes - THERMOCHEMISTRY Chemistry CHEM 213W...

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

View Full Document Right Arrow Icon
THERMOCHEMISTRY Chemistry CHEM 213W David Ronis McGill University 1. Enthalpy Calculations: Chemical Reactions and Hess’ Law The enthalpy change for a process, H , is equal to the heat absorbed by the system if that process is done under constant pressure conditions (and assuming that only P-V work is possi- ble). Since the enthalpy of a system, H = E + PV, is a state function, we can systematize enthalpy calculations by considering a path whereby the compounds first turn into their con- stituent elements in their standard states (by convention at 25 o C and 1 atm pressure) and then recombine to form the products. The enthalpy change in the latter step is just the enthalpy of for- mation of the products and the former is the enthalpy of destruction (i.e., the negative of the enthalpy of formation) of the reactants. Hence, H = Σ H 0 f ( products ) - ∆ H 0 f ( reactants ) . (1) Since we are interested in calculating a difference, the absolute enthalpy of the elements in their standard states is unimportant [it cancels out of Eq. (1)], and we adopt the convention that the enthalpy of formation of an element in its standard state is zero. Consider the following example (reduction of iron oxide): Fe 2 O 3 ( s ) + 3 H 2 ( g ) 25 o C , 1 atm 2 Fe ( s ) + 3 H 2 O ( l ). A table of thermochemical data gives: Enthalpies of Formation at 1 atm and 25 C Compound H 0 f (kJ/mole) Fe 2 O 3 (s) -824.2 H 2 (g) 0.0 Fe(s) 0.0 H 2 O (l) -285.830 By using these in Eq. (1), we find that H = [3( - 285. 830) - ( - 824. 2)]kJ/mole = - 33. 29kJ/mole. Winter Term 2001-2002
Image of page 1

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

View Full Document Right Arrow Icon
Thermochemistry -2- Chemistry CHEM 213W Note that the calculated enthalpy change depends on how the reaction was written. For example, if we wrote 1 2 Fe 2 O 3 ( s ) + 3 2 H 2 ( g ) 25 o C , 1 atm Fe ( s ) + 3 2 H 2 O ( l ), then H = - 16. 65 kJ/mole. 2. Measuring H 0 f There are a number of ways in which to measure the enthalpy of formation of a compound; here are two. The most obvious is to simply carry out the formation reaction from the con- stituent elements in their standard states in a constant pressure calorimeter (recall that H = Q p ). For example, consider the combustion of graphite to form carbon dioxide C ( graphite ) + O 2 ( g ) 25 o C , 1 atm CO 2 ( g ). The heat released in this reaction is -∆ H 0 f ( CO 2 ), since the standard enthalpy of formation of the reactants is zero. For this method to work, two conditions must be met: 1) the reaction goes to completion and 2) only one product is formed. Thus, the reaction C ( graphite ) + 2 H 2 ( g ) 25 o C , 1 atm CH 4 ( g ) is not suitable for this method since it doesn’t readily go to completion and we get a complicated mixture of hydrocarbons.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern