Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is cnsidered late, and it is wrth 9 pints extra credit tward exam. -Exam may be mved t the Tuesday after next Thursday. If this is an issue please cntact the prfessr. We will knw fr certain n next Tuesday's class (the scheduled review sessin) -The test is up t and including chapter 5 heat f a reactin [ delta H = q / (mles f ne f the chemicals in the reactin)] when tw slutins each cntaining ne reactant are mixed. The prblem must specify mles f which chemical t put in the denminatr. Reminder: When a system releases heat it is an exthermic reactin with a (-) q. When a substance heats up (absrbs heat) it is an endthermic reactin with a (+)q.hess's Law Delta H sum = Delta H + Delta H3 Simple Break Dwn f Hess's Law: 1. a --> b = x b---> a = ( -x). a---> b = x a----->b= (x) 3. a--->b = x and c---->d= f therefre, a+c---> b +d= x + f
Example Prblems fr Lecture 15 Summary f Hess s Law A reactin and its reverse have equal magnitude, ppsite sign ΔH values If A B has ΔH = 100, then B A has ΔH = 100 If yu multiply a reactin by a factr, then yu multiply the ΔH by the same factr If A B has ΔH = 100, then A B has ΔH = 00 When yu add tw reactins, yu add the ΔH values If A B has ΔH = 100, and C D has ΔH = 50, then A + C B + D has ΔH = 100 + 50 = 150 Yu can use Hess s Law t calculate ΔH fr a reactin f interest if yu can sum ther reactins t equal the reactin f interest. 1. Calculate ΔH fr the reactin C (s) + H (g) C H (g) given the fllwing chemical equatins and their respective enthalpy changes. 5 CH ( g) + O ( g) CO ( g) + HO ( l) Δ H = 199.6 C( s) + O ( g) CO ( g) Δ H = 393.5 1 H ( g) + O ( g) HO ( l) Δ H = 85.8 T d this prblem yu need t first determine hw yu change all the cmpnents. 1. Yu need C(s) n the reactants side. Yu have this in equatin #, hwever yu nly have ne therefre yu need t multiply the whle equatin thrugh by making a. C(s) + O(g) CO(g) <>H = ()(-393.5 ). Then yu need t have H(g) n the prduct side, fr this yu need t include the reverse f equatin 3. Making: a. HO(l) H(g) + ½ O(g) <>H =(-1)(-85.8) 3. Then yu need t get CH(g) n the prduct side, yu d this by multiplying equatin 1 by (-1), as we just did in part. This makes equatin 1 lk like: a. CH(g) + 5/ O(g) CO(g) + HO(l) <> H= (-1)(-199.6) In summary: 5 CO ( g) + HO ( l) CH ( g) + O ( g) Δ H =+ 199.6 C( s) + O ( g) CO ( g) Δ H = 787.0 1 H ( g) + O ( g) HO ( l) Δ H = 85.8 CO(g) + HO(l)+ H(g) + ½ O(g) + C(s) + O(g) CH(g) + 5/ O(g)+ HO(l)+ CO(g) The clrs cancel each ther ut leaving nly the black parts in the equatin. Then t find the <>H sum yu simply add tgether all the numbers <>Hsum= ((-1)(-199.6)) + ((-1)(-85.8))+ (()(-393.5 ))
. Calculate ΔH fr the reactin NO (g) + O (g) NO (g) given the fllwing infrmatin NO( g) + O ( g) NO ( g) + O ( g) Δ H = 198.9 3 3 O 3 ( g) O ( g) Δ H = 14.3 O ( g) O ( g) Δ H =+ 495.0 Yu d the fllwing t all the equatins t transfrm them int what yu need. NO(g )+ O 3(g ) --->NO(g )+ O (g ) 198.9KJ (x1) ( just leave it alne) O3(g)---->3/O(g) - 14.3kj(x-1) ( reverse it ) O(g)------> O 495.0 kj (x -1/) ( divide by, r multiply by ½, and reverse it) Nte: There is n O 3 in the desired equatin, why d we need it then? Because the equatin #1 has O 3 in it as a prduct, therefre we need t use equatin in such a way as t eliminate the O 3 in the reactin. The numbers wrk ut as fllws : <>H = -198.9 +((-1/)( 495.0)) + ((-1)(-14.3 )) = 304.1 Answer: 304.1
Yu can use a special categry f reactins, called frmatin reactins, as the reactins that are summed t equal an verall reactin. Enthalpy f Frmatin An enthalpy f frmatin, Δ H f, is defined as the enthalpy change fr the reactin in which a cmpund is made frm its cnstituent elements with all substances in their standard states. Standard enthalpies f frmatin are measured under standard cnditins (5 C and 1.00 atm pressure) and they are reprted per mle f the substance. Practice writing frmatin reactins fr several f the cmpunds in the table abve. Entrpy change = <>H Cmpund: a substance with r mre elements in it Cnstitute- parts Standard Space= what it lks like at rm temp. (5 C and 1.00 atm pressure) Acetylene CH(g) C(s) + H (g) CH(g) In these there is nly ne prduct and the cefficient f that prduct can nly be ne!!!!! Sdium bicarbnate NaHCO3(s) Na(s) + ½ H(g) + C(s) + 3/ O(g) NaHCO 3 (s)
THIS IS LIKELY TO BE A TEST QUESTION!!!! (r smething very similar t it may be) 3. Use standard enthalpies f frmatin t calculate the enthalpy f reactin fr the cmbustin f ethanl, C H 5 OH (l). Standard enthalpies f frmatin Δ H f [ HO( l) ] = 85.83 ml Δ H f [ CO ( g) ] = 393.5 ml Δ H f [ CH5OH( l) ] = 77.7 ml 1. Write a balanced equatin a. CH5OH(l) + 3O(g) CO(g) + 3HO(l). use the numbers f the standard enthalpies f frmatin as needed a. Ethanl <>H = {<>Hf [ CO(g)] + 3 <>Hf [HO(l)] -<>Hf [CH5OH(l)] } b. plug and chug i. ( -393.5/ml) + 3( -85.83 /ml) 77.7 /ml In summary (shrtcut): where n and m are stichimetric cefficients rxn f f prducts reactants Δ H = nδh mδh