Bond Enthalpy and Activation Energy

Transcription

1 Bond Enthalpy and Activation Energy

2 Energy of a Chemical Reaction ΔH = ΔH (bonds broken) - ΔH (bonds formed) Add up all the energies of the broken bonds Add up all the energies of the bonds that are reformed Subtract the difference and that is the net energy of the reaction.

3 Bond Enthalpies

4 Exothermic vs. Endothermic If the final ΔH is negative, then it is an exothermic reaction. Energy released is ΔH because it is leaving the system. There is less energy than before. If the final ΔH is positive, then it is an endothermic reaction. Energy released is +ΔH because it is joining the system. There is more energy than before.

5 Example One 2H 2 + O 2 2H 2 O H-H = 436 KJ/mol O=O = 498 KJ/mol H-O = 463 KJ/mol ΔH = ΔH (bonds broken) - ΔH (bonds formed) ΔH = [2(436) + 498] [4(463)] = = -482 KJ/mol

6 2H 2 + O 2 2H 2 O Example One ΔH = [2(436) + 498] [4(463)] = = -482 KJ/mol This is an exothermic reaction!

7 Getting Chemical Reactions Going Activation energy is the term for the energy needed to break the bonds so new bonds can form Activation Energy Energy Released Net

8 Activation Energy The net energy difference is the ΔH. But, first you must add activation energy to initially break the chemical bonds.

9 Catalysts Catalysts simply lower the activation energy needed to do the chemical reaction going.

10 Catalyst Catalysts provide a surface for chemical reactions to occur without themselves being used up in the reaction. Enzymes are critical catalysts for reactions in animals

11 Getting Chemical Reactions Going Energy is usually needed to break the bonds and get the reaction going. This can be a spark, flame, heat, or even a feather.

12 Getting Chemical Reactions Going Activation energy is the energy needed to break the bonds so new bonds can form Activation Energy Energy Released Net

13 Example Question What is the enthalpy value of the equation below: C 2 H 2 + O 2 CO 2 + H 2 O

14 Example Question What is the enthalpy value of the equation below: 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O 2 C C +839*2 8 C=O -804*8 4 C-H +413*4 4 H-O -463*4 5 O=O + 498* kj kj ΔH = kj But, despite the /mol it is for the entire equation or kj/mol of C 2 H 2.

15 Example Question What is the enthalpy value of the equation below: What about 10 g of acetylene? 2C 2 H 2 + 5O 2 4CO 2 + 2H 2 O 2 C C +839*2 8 C=O -804*8 4 C-H +413*4 4 H-O -463*4 5 O=O + 498* kj kj ΔH = kj But, despite the /mol it is for the entire equation or kj/mol of C 2 H 2.

16 Bond Enthalpy Example #2 What is the ΔH for the decomposition of NI 3? NI 3 N 2 + I 2 N-I bond is 169 kj/mol

17 Entrance Question #1

18 Entrance Question #1 H-H = +436 kj/mol I-I = +151 kj/mol +587 kj/mol to break bonds

19 Entrance Question # kj/mol to break bonds ΔH = -12 kj/mol 2 H-I = -299 * 2 = -598 kj/mol

20 Entrance Question #1 Which mole does kj/mol refer to? kj/mol refers to the balanced equation as a whole. If you are focused on one particular substance, you need to divide it by that coefficient kj/mol to break bonds 2 H-I = -299 * 2 = -598 kj/mol ΔH = -12 kj/mol

21 Entrance Question #1 What is the ΔH per mole of HI produced? Since -12kJ/mol refers to 2 moles of HI, divide by 2 to get -6 kj/mol of HI molecules +587 kj/mol to break bonds 2 H-I = -299 * 2 = -598 kj/mol ΔH = -12 kj/mol

22 Endothermic and Exothermic Reactions Energy Signs Examples Exothermic Reactions Energy is Released to the Surroundings Gets Hot Flame Light produced Combustion of Methane Endothermic Reactions Energy is Absorbed from the Surroundings Gets Cold Dark color Cold Pack Photosynthesis

23 Factors That Influence Reaction Rates Affect on Reaction Rate Catalyst Increases the Reaction Rate Concentration of Reactants Surface Area Temperature Stirring Increase in Concentration increases the reaction rate because of an increase in particle collisions. More surface area increases the reaction rate. (Smaller particles give a greater overall surface area when all the particles surfaces are added together.) Increase in Temperature increases the reaction rate because of an increase in particle collisions. Increases the Reaction Rate by increasing the number of collisions.

24 Concentration Higher concentrations mean more collisions of molecules and a faster reaction rate.

25 Surface Area Powdered forms are much more reactive than solid because there is a larger surface area for the reaction to occur.

26 Temperature We cook food to make the chemical reactions occur more quickly that denature proteins and make it more edible/digestible.

27 Stirring Stirring moves the particles around and speeds up the rate of collisions.

28 Factors That Influence Reaction Rates Affect on Reaction Rate Catalyst Concentration of Reactants Surface Area Temperature Stirring

29 Many Chemical Reactions Can Be Reversed If energy is added to a system, chemical reactions can be reversed. These would by definition be endothermic reactions. In Entropy, we ll discuss spontaneity and the likelihood of reversal.

30 EXAMPLE The complete combustion of propane can be represented by the following equation: C 3 H 8 + 5O 2 3CO 2 + 4H 2 O It s helpful to draw the molecules, so you can see which bonds are present to be broken and reformed. The Bond Enthalpies Are: C-H = 413 C-C = 347 O=O = 498 C=O = 805 H-O = 464

31 Example The complete combustion of propane can be represented by the following equation: C 3 H 8 + 5O 2 3CO 2 + 4H 2 O The Bond Enthalpies Are: C-H = 413 C-C = 347 O=O = 498 C=O = 805 H-O = 464 ΔH = ΔH (bonds broken) - ΔH (bonds formed) [8(413) + 2(347) + 5(498)] [6(805) + 8(464)] = KJ/mol Burning Propane Is Highly Exothermic

32 Entrance Question: Bond Enthalpy How much energy can be produced by burning 5 grams of kerosene (C 14 H 30 )? C 14 H 30 + O 2 CO 2 + H 2 O 2C 14 H O 2 28CO H 2 O

33 Entrance Question: Bond Enthalpy How much energy can be produced by burning 5 grams of kerosene (C 14 H 30 )? 2C 14 H O 2 28CO H 2 O? 26 C-C 56 C=O 60 C-H 60 O-H 43 0=0 [(26*348)+(60*413)+(43*495)]-[(56*804)+(60*463) = = kj kj/mol of kerosene 5 g / 198 g/mol = mol mol * kj/mol = kj