110c-lecture4

110c-lecture4 - Problem Set #1 Due after class on Wed April...

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Problem Set #1 Due after class on Wed April 7 th . Answer key posted afterwards on website. Problem Set #2 Due next Wed April 14 th . Office Hours Cancelled for Wed April 14 th . Held on Tue April 13 th , 10:30-11:30 am. Midterm 1 Wed April 21 st . Material covered through end of Fri April 16 th . Chapters 19, 20, 21, and 22.

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Enthalpy and Thermochemistry (pp 786-800) Enthalpy Change for Chemical Reactions ( r H). Reactants Products r H = H prod - H react Exothermic reaction (q p = r H < 0) heat released as a product (favorable rxn). Endothermic reaction (q p = r H > 0) heat must be supplied as a reactant to drive rxn. r H(reverse) = - r H(forward): r H A B = - r H B A A—A + B—B 2A—B + heat r H<±0 Enthalpy of Combustion ( c H): CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat c Hº = -890 kJ/mol Enthalpy of Formation ( f Hº): heat + 2C(s) + H 2 (g) C 2 H 2 (g) f Hº(C 2 H 2 ) = +226 kJ/mol Bond Dissociation Enthalpy ( H(A—B)): heat + H—Cl(g) H(g) + Cl(g) Hº(H-Cl) = +431 kJ/mol Phase Transition Enthalpy ( fus H): H 2 O(s) H 2 O(l) fus H = +6.01 kJ H 2 O(l) H 2 O(g) vap H = +40.7 kJ
Hess’ Law and Combustion Calorimetry n-C 4 H 10 (g) + 13/2 O 2 (g) 4CO 2 (g) + 5H 2 O(l) c Hº = -2877 kJ/mol 4CO 2 (g) + 5H 2 O(l) i-C 4 H 10 (g) + 13/2 O 2 (g) - c Hº = +2869 kJ/mol n – C 4 H 10 (g) i – C 4 H 10 (g) + heat H(tot) = -2877 + 2869 = -8 kJ/mol q V = U c H= U + (PV) ≅∆ U + n gas RT T=298K T=298K C 4 H 10 (g) + 13/2 O 2 (g) 4CO 2 (g) + 5H 2 O(l) T=T1 T=T2 C 4 H 10 (g) + 13/2 O 2 (g) 4CO 2 (g) + 5H 2 O(l) Bomb Calorimeter dT products C H C H dT react C H T P T c P c ) ( ) ( ) ( 298 2 10 1 298 4 298 + + =

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Molar Heats of Formation ( f Hº) Heats of Formation (pp 791-797). Balanced chemical reactions have the same # of elements on both sides of arrow. All that changes is formation of chemical bonds or the physical state of a substance. C(s) + 1/2O 2 (g) C—O(g) f Hº = -110 kJ/mol DEFN: The standard heat of formation ( f Hº) —the enthalpy for the formation of one mole of a molecule from its constituent elements under standard conditions. ) ( ) ( 2 / 1 ) ( 2 2 2 l O H g O g H + f Hº = -285.8 kJ at T = 298 K C(s) + O 2 (g) CO 2 (g) f Hº = -393.5 kJ at T = 298 K C(s) C(s) graphite f Hº = 0 C(s) C(s) diamond f Hº = +1.897 kJ at T = 298 K See Table 19.2 or CRC handbook for listings of f
Enthalpy of Reaction ( r H) from Heats of Formation ( f Hº) aA + bB yY + zZ Step 1 determines heat to break down reactants into constituent elements (- f H(react)). Step 2 determines heat of formation of products from constituent elements ( f H(prod)). Overall Enthalpy of the net reaction is r H = f H(products) – f H(reactants). If bonds formed in products are stronger than bonds broken in reactants, then r H < 0. r H(step 1) = -a f Hº(A) + -b f Hº(B) r H(step 2) = y f Hº(Y) + z f Hº(Z) r H = r H(step 1) + r H(step 2)

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Thermochemistry Practice Problem Use the f Hº data in Table 19.2 to calculate r Hº for the reaction at 298K: C 2 H 2 (g) + 5/2O 2 (g) 2CO 2 (g) + H 2 O(l) r Hº = - ∑∆ f Hº(reactants) +
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110c-lecture4 - Problem Set #1 Due after class on Wed April...

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