Chemistry Unit 13 Heat of Reaction

Almost all of the energy changes in a chemical reaction are the result of bonds breaking plus bonds forming. 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

1a. 2 hour walking. 10 hr sitting. 8 hr sleeping. 2 hr standing 1a. 2 hour walking 10 hr sitting 8 hr sleeping 2 hr standing 1 hr running 1 hr swimming b. 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

9. Heat absorbed by H2O. H = m c DT 10. Heat given off by food 9. Heat absorbed by H2O H = m c DT 10. Heat given off by food 0 = HH2O + Hfood 11. joules/gram joules from #10/mass from #4 12. Kj/gram #11/1000

DH means change in enthalpy (heat) of a chemical reaction DH means change in enthalpy (heat) of a chemical reaction. If DH is 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 Kcal 2 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)

Dissolving NH4Cl in water 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

Dissolving Pb(NO3)2 in water a. Write the equation 6.12 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 for Pb(NO3) 2 dissolving h. Joules per mole of compound

Combining the 2 solutions:. a. Write the equation b Combining the 2 solutions: a. Write the equation b. Moles of solid formed c. Beginning temperature d. Final temperature e. DT f. Number of joules involved with H2O g. Number of joules involved with precipitate h. Number of joules per mole of precipitate

Methane demonstration. CH4(g) + O2(g)  CO2(g) + H2O(g) 1 Methane demonstration CH4(g) + O2(g)  CO2(g) + H2O(g) 1. Balance the equation 2. Seconds to collect 100 mL 3. Moles of methane we collected 4. Moles of methane burned in 2 minutes 5. H = m c D T (for the water) mL of water & tin 6. Joules of combustion for 1 mole of methane % efficiency =

Determination of the Heat of Combustion for Candle Wax Mass initial - Mass final = 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

Mg + H2O  Mg(OH)2 + H2

62. 5 g of propane ( C3H8) fuel is burned completely 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?

Examples: N2 (g) + O2 (g)  NO (g). DH = +90 Examples: N2 (g) + O2 (g)  NO (g) DH = +90.42 KJ/ mole NO N2 (g) + H2 (g)  NH3 (g) DH = -30.56 KJ/ mole H2

C(s) + O2(g)  CO(g) DH = -110. 51 KJ/mole CO C(s) + H2(g)  C3H8 (g) C(s) + O2(g)  CO(g) DH = -110.51 KJ/mole CO C(s) + H2(g)  C3H8 (g) DH = -103.81 KJ/mole C3H8 S(s) + O2(g)  SO3(g) DH = -395.16 KJ/mole SO3 N2(g) + O2(g)  NO(g) DH = +90.42 KJ/mole NO

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 = -987.06 KJ/mole Ca(OH)2 C(s) + H2(g)  C6H6(s) DH = +82.88 KJ/mole C6H6

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.

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 = -110.51 KJ/mole CO Multiple the following equation by 3 ½ H2(g) + ½ I2(s)  HI(g) DH = +25.95 KJ/mole HI

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

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 = -1,143.6 KJ/mole H2O Reverse the following equation 3 C(s) + 4 H2(g)  C3H8(g) DH = -103.8 KJ/mole C3H8

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

Add the following 2 equations C(s) + 1/2 O2(g)  CO(g). DH = -110 Add the following 2 equations C(s) + 1/2 O2(g)  CO(g) DH = -110.5 kJ/mole CO C(s) + O2(g)  CO2(g) DH = -393.5 kJ/mole CO2

½ N2(g) + O2(g)  NO2(g) DH = +33.9 KJ/mole NO2 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

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

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

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

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

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. To determine the heat of reaction from heat of formation: DHorxn = (S np DHf products) - (S nr DHf reactants) DHorxn = heat of reaction in KJ S = the sum of np and nr = coefficients DHf = heat of formation

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

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

Acid – Base Calorimeter Lab NaOH(s) + HCl(aq)  NaCl(aq) + H2O(l) + joules 0 = DHH2O + DHAl + DHRxn

Review for exam 1a. DT = 30 – 20 = 10 OC b. DH = (550 g) (4 Review for exam 1a. 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/g 2. 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/mol 3. DH = -501.5 KJ CO + ½ O2  CO2 + 501.5 KJ 4. 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