{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter6LEcturepart2

# Chapter6LEcturepart2 - Chapter 6 Enthalpies in Reactions...

This preview shows pages 1–5. Sign up to view the full content.

6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies in Reactions Any reaction has an associated enthalpy change: It either heats the surroundings or cools the surroundings. Combustion heats the surroundings. The sign of H is negative, because heating surroundings means the system loses internal enthalpy, and all our numbers refer to the system, that is, the molecules. C(s) + O 2 (g) CO 2 (g) H = -393.5 kJ • If we have a good set of reactions, we can combine them to learn the enthalpy changes in other reactions. The US government spent much money on this to support the economy. • Mostly, they did combustion reactions, because they are fairly easy to do. Burn everything! • If you turn the reaction around, the sign of the enthalpy change switches: CO 2 (g) C(s) + O 2 (g) H = +393.5 kJ • You have to put energy in, either by doing work or by heating the CO 2 to break the bonds and make the atoms come apart. That makes sense.

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

View Full Document
6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies in Reactions • Let’s do another example of the use of Hess’s Law. • Given: C(s) + 2H 2 (g)  CH 4 (g) H f = -74.9 kJ 3H 2 (g) + N 2 (g) 2NH 3 (g) H = -91.8 kJ H 2 (g) + 2C(s) + N 2 (g) 2HCN(g) H = +270.3 kJ What would be the enthalpy change if you made HCN out of methane and ammonia? Reverse some and divide some by two: CH 4 (g) C(s) + 2H 2 (g) H = +74.9 kJ NH 3 (g) 1.5H 2 (g) + 0.5N 2 (g) H = +45.9 kJ ½H 2 (g) + C(s) + ½N 2 (g) HCN(g) H = +135.1 Add: CH 4 (g) + NH 3 (g) 3H 2 (g) + HCN(g) H = +256 This is endothermic. If the reaction is to stay at standard T, it has to absorb energy from the surroundings. 256 kJ as written.
6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies in Reactions • We can now deal with any reaction, but some questions remain: • The first is: Why are we using H? I sneaked the answer into the last statement. We must define the initial state of the system exactly and the final state of the system exactly. We choose a convenient T, 25  C, and a convenient p, 1 atm (or 1 bar). This is convenient for chemist working in open containers on a lab bench, and it is really what biologists and physicians and pharmacists need. We assume we start at these standard conditions; and the enthalpy exchange with the surroundings is to bring things back to standard conditions at the end. Since H is a state function, it does not matter what happens in between. We just compare the end points. You are used to doing that for the atoms. When you write chemical equations for reactants and products, you have no idea what may happen in between.

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

View Full Document
6 Copyright: 2010 Prof. Magde Chapter 6: Enthalpies in Reactions
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

Chapter6LEcturepart2 - Chapter 6 Enthalpies in Reactions...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online