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Chemistry Unit 13

Chemistry Unit 13. Heat of Reaction. A calorie is the amount of heat needed to raise the temperature of 1 gram of H 2 O through 1 o C. 4.18 joules is the amount of heat needed to raise the temperature of 1 gram of H 2 O through 1 o C.

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Chemistry Unit 13

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  1. Chemistry Unit 13 Heat of Reaction

  2. A calorie is the amount of heat needed to raise the temperature of 1 gram of H2O through 1o C. • 4.18 joules is the amount of heat needed to raise the temperature of 1 gram of H2O through 1 oC. • Chemical reactions are either endothermic (heat into) or exothermic (heat out of)Endothermic H2O (s) + 6.03 KJ  H2O (L) • Exothermic 2 H2(g) + O2(g)  2 H2O (g) + 480.9 KJ • Almost all of the energy changes in a chemical reaction are the result of bonds breaking plus bonds forming.

  3. 1a. 2 hour walking 10 hr sitting 8 hr sleeping 2 hr standing 1 hr running 1 hr swimmingb. 6 oz hamburger 3 cup milk 2 eggs 4 bread 1 cup corn 20 French fries 1 choc. Cake 1 cup peas 2 sq. butter 2 oz choc 1 apple

  4. 9. Heat absorbed by H2O H = m c DT10. Heat given off by food0 = HH2O + Hfood11. joules/gram joules from #10/mass from #412. Kj/gram #11/1000

  5. DH means change in enthalpy (heat) of a chemical reaction.IfDHis negative, the reaction is exothermic. The heat term is added to the right side of the chemical reaction.2 H2(g) + O2(g)  2 H2O (g) + 115.6 Kcal2 H2(g) + O2(g)  2 H2O (g) + 480.9 KJ If DH is positive, the reaction is endothermic. The heat term is added to the left side of the chemical reaction.H2O (s) + 1.44 Kcal  H2O (L)H2O (s) + 6.03 KJ  H2O (L)

  6. Demonstration on endothermic and exothermic • Dissolving NH4Cl in water • a. Write the equation • b. 2 grams of compound used in 50 mL H2O, calculate moles • c. Beginning temperature • d. Final temperature • e. DT • f. Number of joules involved with H2O • g. Number of joules with NH4Cl dissolving • h. Joules per mole of compound

  7. Dissolving Pb(NO3)2 in water • a. Write the equation • 6.12 grams of compound used in 50 mL H2O, Calculate moles • c. Beginning temperatured. Final temperaturee. DT • f. Number of joules involved with H2Og. Number of joules for Pb(NO3)2dissolvingh. Joules per mole of compound

  8. Combining the 2 solutions: a. Write the equationb. Moles of solid formedc. Beginning temperatured. Final temperaturee. DTf. Number of joules involved with H2Og. Number of joules involved with precipitate h. Number of joules per mole of precipitate

  9. Methane demonstration CH4(g) + O2(g) CO2(g) + H2O(g)1. Balance the equation2. Seconds to collect 100 mL3. Moles of methane we collected4. Moles of methane burned in 2 minutes5. H = m c D T (for the water) mL of water & tin 6. Joules of combustion for 1 mole of methane% efficiency =

  10. Determination of the Heat of Combustion for Candle Wax • Mass initial - Massfinal = • mL of H2O = g of H2O • DT = T final - T initial • H = m c DT (for the water) • Remember that the candle wax will be the (–) of that heat • 5. - #4 / #1 • 6. #1 / molar mass of candle wax = moles • - #4 / moles = J/mole

  11. Mg + H2O  Mg(OH)2 + H2

  12. 62.5 g of propane ( C3H8) fuel is burned completely. The heat from the combustion is used to heat 12,500 g H2O from 20 oC to 80 oC. What is the heat of combustion in joules/gram and in joules/mol?

  13. Examples:N2 (g) + O2 (g) NO (g)DH = +90.42 KJ/ mole NON2(g) + H2 (g)  NH3 (g)DH = -30.56 KJ/ mole H2

  14. C(s) + O2(g) CO(g) DH = -110.51 KJ/mole COC(s) + H2(g)  C3H8 (g)DH = -103.81 KJ/mole C3H8S(s) + O2(g)  SO3(g)DH = -395.16 KJ/moleSO3N2(g) + O2(g)  NO(g)DH = +90.42 KJ/mole NO

  15. N2(g) + O2(g) NO2(g)DH = +33.91 KJ/mole NO2 N2(g) + H2(g)  N2H4(aq)DH = +34.16 KJ/mole N2H4 Ca+2(aq) + OH-1(aq)  Ca(OH)2 (s)DH = -987.06KJ/mole Ca(OH)2C(s) + H2(g)  C6H6(s)DH = +82.88 KJ/mole C6H6

  16. Hess’s Law of Summation You can add equations for chemical reactions algebraically. The result is a net equation which includes the heat for the net reaction. This principle can be used to calculate the heat of a reaction with unknown heat involved.

  17. To use Hess’s Law of summation, the following manipulations may be used: Equations may be multiplied by multiplying all coefficients and heat term by the same number. Multiple the following equation by 2 C(s} + ½ O2(g) CO(g)DH = -110.51KJ/mole CO Multiple the following equation by 3 ½ H2(g) + ½ I2(s)  HI(g)DH = +25.95KJ/mole HI

  18. Equations may be reversed by writing everything on other side of arrow. Reverse the following equation 1/2 N2(g) + 3/2 H2(g) NH3(g) DH = -46.0 KJ/mole NH3

  19. Equations may be divided by dividing all coefficients by the same number Divide the following equation by 4 4 H2(g)+ 2 O2(g) 4 H2O (L) DH = -1,143.6KJ/mole H2O Reverse the following equation 3 C(s) + 4 H2(g)  C3H8(g)DH = -103.8KJ/mole C3H8

  20. Equations may be added by adding the coefficients of like terms Add the following 2 equations 2 C(s) + 3 H2(g) C2H6(g)DH = -84.6 KJ/mole C2H6 3 C(s) + 4 H2(g)  C3H8(g)DH = -103.8 KJ/mole C3H8

  21. Add the following 2 equations C(s) + 1/2 O2(g) CO(g)DH = -110.5 kJ/mole COC(s) + O2(g)  CO2(g)DH = -393.5 kJ/mole CO2

  22. When adding equations, cancel any like terms on opposite sides of the equations Add the following 2 equations canceling any like terms before adding them NO(g) ½ N2(g) + ½ O2(g) DH = -90.4 KJ/mole NO ½ N2(g) + O2(g)  NO2(g)DH = +33.9 KJ/mole NO2

  23. Add the following equations, cancel any like terms SO2(g) S(s) + O2(g)DH = +297.2 KJ/mole SO2 S(s) + 3/2 O2(g)  SO3(g)DH = -395.2 KJ/mole SO3

  24. Add the following equations, cancel any like terms H2O(L) H2(g) + ½ O2(g)DH = +285.9 KJ/mole H2O SO2(g)  S(s) + O2(g)DH = +297.2 KJ/mole SO2 H2(g) + S(s) + 2 O2(g)  H2SO4(L)DH = +814.0 KJ/mole H2SO4

  25. Find the heat of reaction for:3 C(s) + 5 H2(g) CH4(g) + C2H6(g) Knowns: 2 C(s) + 3 H2}(g) C2H6(g)DH = -84.56 KJ C(s) + 2 H2(g)  CH4(g)DH = -74.93 KJ

  26. Find the heat of the reaction for:CO(g) + ½ O2(g) CO2(g) Knowns: C(s) + ½ O2(g) CO(g)DH = -110.5 KJ C(s) + O2(g)  CO2(g)DH = -393.5 KJ

  27. To determine the heat of reaction from heat of formation:DHorxn = (S npDHf products) - (S nrDHfreactants)DHorxn= heat of reaction in KJS = the sum ofnp andnr = coefficientsDHf = heat of formation Heat of formation is the heat required to make 1 mole of substance from elements in their natural state. Heat of formation for elements in their natural state is zero.

  28. Calculate the heat of reaction for the following: NO2(g) + H2O(g) HNO3(l) + NO(g)

  29. Determine the heat of reaction for the following: (be sure to balance first) SO2(g) + O2(g) SO3(g)

  30. Acid – Base Calorimeter LabNaOH(s) + HCl(aq) NaCl(aq) + H2O(l) + joules0 = DHH2O + DHAl + DHRxn

  31. Review for exam1a. DT = 30 – 20 = 10 OC b. DH = (550 g) (4.18j/goC ) ( 10 oC) = + 22,990 joules for the H2O - 22,990 joules for the pine c. -22,990 joules/1.25 g pine = - 18,392 J/g2. DH = (100 g ) (4.18j/goC) (-2.5oC) = -1045 joules for the H2O + 1045 joules for the Ba(NO3)2 b. 1g/261.3 g/mole = .0038 mole, +1045 J /.0038 mole = 275,000 J/mol3. DH = -501.5 KJ CO + ½ O2 CO2 + 501.5 KJ4. 2 HNO3 + Na2CO3  H2O + CO2 + 2 NaNO3 -207 -1131 -286 -393.5 -467 x2 x2 -414 -934 DHrxn = (-286 + -393.5 + -934) - (-414 + -1131)DHrxn = (-1,613.5) – ( -1,545) = -68.5 KJ

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