Thermochemistry combination problems part 2 1) Heat of reaction (Use the following 3 reactions.)

2NaHCO3 (s) + 129 kJ Na2CO3 (s) + H2O (g) + CO2 (g)

CaO (s) + H2O (l) Ca(OH)2 (s) + 65.2 kJ

Fe2O3 (s) + 3CO (g) 2Fe (s) 3CO2 (g) H = 26.3 kJ

a) Calculate the energy absorbed when 68.7 grams of sodium bicarbonate is decomposed. 68.7 g NaHCO3 x 1 mol NaHCO3 x 129 kJ = 52.7 kJ 84.01 g NaHCO3 2 mol NaHCO3

b) Calculate the mass of calcium hydroxide formed when 742.9 kJ is released. 742.9 kJ x 1 mol Ca(OH)2 x 74.10 g Ca(OH)2 = 844 g Ca(OH)2 65.2 kJ 1 mol Ca(OH)2

c) Calculate the energy released when 88.4 liters of carbon monoxide is reacted with excess iron (III) oxide.

88.4 L CO x 1 mole CO x 26.3 kJ = 34.6 kJ released 22.4 L CO 3 mol CO 2) Heat of reaction, temperature changes, and phase changes

a) A sample of 92.22 g of sodium bicarbonate is decomposed. If the heat needed to decompose that sodium bicarbonate comes from the cooling of 498 g of silver, calculate the temperature change in the sample of silver.

Decomposing the NaHCO3 92.22 g NaHCO3 x 1 mol NaHCO3 x 129 kJ = 70.8 kJ 84.01 g NaHCO3 2 mol NaHCO3 Cooling the silver q= 70.8 kJ or 70800 J m= 498 g Ag C= 0.235 J/gAgK T = q = 70800 J = 605 K C x m 0.235 J/gAgK x 498 g Ag

b) Calculate the mass of ice that can be melted from the heat released by the reaction of 168.5 g of iron (III) oxide reacted with excess carbon monoxide.

Heat released from the reaction 168.5 g Fe2O3 x 1 mol Fe2O3 x 26.3 kJ = 27.7 kJ 159.68 g Fe2O3 1 mol Fe2O3 Melting the ice 27.7 kJ x 1 mol H2O x 18.02 g H2O = 83.2 g H2O ice melted 6.01 kJ 1 mol H2O

c) Calculate the final temperature of 14.36 g of water with a starting temperature of 4.57 C if 82.7 g of calcium hydroxide is formed from calcium oxide and water.

Heat released from the reaction 82.7 g Ca(OH)2 x 1 mol Ca(OH)2 x 65.2 kJ = 72.8 kJ 74.10 g Ca(OH)2 1 mol Ca(OH)2 Affecting the water Starting temp = 4.57 C Final temp = ? Mass = 14.36 g H2O Heat water to 100C from 4.57 C

q= CmT q= ? m= 14.36 g H2O C= 4.18 J/gC T= 95.43C

14.36 g H2O x 4.18 J/gC x 95.43 C = 5730 J

Remaining energy: 72800 J- 5730 J = 67100 J

Boiling water:

14.36 g H2O x 1 mol H2O x 40.7 kJ = 32.4 kJ or 32400 J

18.02 g H2O 1 mol H2O

Remaining energy: 34700 J

Heating steam:

q= 34700 J C= 1.7 J/gC m= 14.36 g H2O T=?

T= q = 34700 J = 1400 C

Cm 1.7 J/gC x 14.36 g H2O

Final temperature 1400C + 100C = 1500 C

3) Heat of solution

a) Calculate the heat released when 98.54 g of NaOH is completely dissolved. 98.54 g NaOH x 1 mol NaOH x -445.1 kJ = 1097 kJ released 40.00 g NaOH 1 mol NaOH

b) Calculate the mass of calcium chloride that is dissolved when 42.97 kcal of energy is released by the dissolving of the calcium chloride.

-42.97 kcal x 4.184 kJ x 1 mol CaCl2 x 110.91 g CaCl2 = 241 g CaCl2 dissolved 1 kcal -82.8 KJ 1 mol CaCl2 4) Heat of solution, temperature changes, and phase changes

a) Calculate the mass of ammonium nitrate that dissolves when the 89.34 g of water changes its temperature from 79.34 C to 21.76 C.

Cooling of the water q=CmT q= ? C= 4.18 J/gC m= 89.34 g H2O T= 57.58C 4.18 J/gC x 89.34g H2O x 57.58 C = 21500 J Dissolving the ammonium nitrate 21500 J x 1 KJ x 1 mol NH4NO3 x 80.06 g NH4NO3 = 67.0 g NH4NO3 dissolved 1000 J 25.7 KJ 1 mol NH4NO3

b) Calculate the mass of phosphorus (at it melting point temperature) that can be melted using the heat given off by the dissolving of 76 g of sodium hydroxide.

Dissolving the NaOH 76 g NaOH x 1 mol NaOH x -445.1 kJ = -850 kJ 40.00 g NaOH 1 mol NaOH Melting the phosphorus 850 kJ released x 1 mol P x 30.97 g P = 42000g P 0.63 kJ 1 mole P

c) Calculate the final temperature of 543 g of water starting at 87.54 C using the heat released when 1298 g of calcium chloride dissolves.

Dissolving the calcium chloride (releases energy) 1298 g CaCl2 x 1 mol CaCl2 x -82.8 kJ = 968 kJ released 110.98 g CaCl2 1 mol CaCl2 Affecting the water

Heating the water to 100C from 87.54 C

q=? m= 543 g H2O T= 12.46 C C= 4.18 J/gC

543 g H2O x 12.46C x 4.18 J/gC = 28300 J or 28.3 kJ

Remaining energy: 969 kJ 28.3 kJ = 941 kJ

Boiling the water 543 g H2O x 1 mol H2O x 40.7 kJ = 1226 kJ needed to boil all the water

18.02 g H2O 1 mol H20 Since only part of the water is boiled and part remains as a liquid, the average temperature will equal the boiling point (i.e. 100 C) 5) Heat of formation

a) Using the standard heats of formation, write a balanced equation and calculate the heat of reaction when hydrogen peroxide reacts with nitrogen monoxide to form steam and nitrogen dioxide.

H2O2 + NO H2O + NO2 Find the Heat of Formation for each Element Hf for H2O2 = -187.8 kJ/mol Hf for NO = 90.37 kJ/mol Hf for H2O = -241.8 kJ/mol Hf for NO2 = 33.85 kJ/mol Find the Heat of the Reaction Hrxn = Hf - Hf products reactants Hrxn = 1 mol H2O x -241.8 kJ + 1 mol NO2 x 33.85 kJ 1 mol H2O 1 mol NO2

- (1 mol H2O2 x 187.8 kJ + 1 mol NO x 90.37 kJ ) = -110.5 kJ

1 mol H2O2 1 mol NO b) Show a balanced equation and calculate the heat of reaction when sodium reacts with

calcium carbonate in a single displacement reaction. 2 Na + CaCO3 Na2CO3 + Ca Find the Heat of Formation for each Element Hf for Na2CO3 = -1131.1 kJ/mol Hf for Ca = 0 kJ/mol Hf for Na = 0 kJ/mol Hf for CaCO3 = -1207.0 kJ/mol Find the Heat of the Reaction Hrxn = Hf - Hf products reactants Hrxn = 1 mol Na2CO3 x -1131.1 kJ + 1 mol Ca x 0 kJ 1 mol H2O 1 mol Ca

- (1 mol Na x 0 kJ + 1 mol CaCO3 x -1207.0 kJ ) = 75.9 kJ

1 mol Na 1 mol CaCO3

c) Hydrogen sulfide reacts with chlorine to form monoclinic sulfur and hydrogen chloride. Calculate the energy released when 657 g of hydrogen chloride is produced with this reaction.

Balanced Equation 8 H2S + 8 Cl2 S8 + 16 HCl Find the Heat of Reaction Hrxn = Hf - Hf products reactants

Hrxn = 1 mol S8 x 0.30 kJ + 16 mol HCl x 92.31 kJ 1 mol S8 1 mol HCl

(8 mol H2S x 20.1 kJ + 8 mol Cl2 x 0 kJ ) = -1315.9 kJ 1 mol H2S 1 mol Cl2 Find the Energy Released by 657 g HCl 657 g HCl x 1 mol HCl x -1315.9 kJ = 1480 kJ released 36.45 g HCl 16 mol HCl 6) Heat of formation, temperature changes, and phase changes

a) Calculate the temperature change of water when 729.87 g of water is heated by the decomposition of 0.4378 g of hydrogen peroxide decomposes to form liquid water and oxygen gas.

Find the Balanced Equation 2H2O2 2H2O (l) + O2 Find the Heat of Formation for each Element Hf for H2O2 = -187.8 kJ/mol Hf for H2O = -285.8 kJ/mol Hf for O2 = 0 kJ/mol Find the Heat of the Reaction Hrxn = Hf - Hf products reactants Hrxn = (2 mol H2O x 285.8 kJ + 1 mol O2 x 0 kJ ) - (2 mol H2O2 x 187.8 kJ ) = -196.0 kJ 1 mol H2O 1 mol O2 1 mol H2O2 Find the Energy Released 0.4378 g H2O2 x 1 mol H2O2 x 196.0 kJ = -1.261 kJ 34.02 g H2O2 2 mol H2O2

Find the Heat of the Water T= q/Cm = 1261 J = 0.413 C 4.18J/gC x 72.87 g H2O

b) Calculate the mass of cesium at it melting point that can be melted when the heat from the reaction of 96.2 L of ozone at STP with an excess of sulfur dioxide produces sulfur trioxide.

Find the Balanced Equation O3 + 3 SO2 3 SO3 Find the Heat of Formation for each material Hf for O3 = 142.0 kJ/mol Hf for SO2 = -296.8 kJ/mol Hf for SO3 = -395.7 kJ/mol Find the Heat of the Reaction Hrxn = Hf - Hf products reactants Hrxn= 3 mol SO3 x 395.7 kJ - (1 mol O3 x 142.0 kJ + 3 mol SO2 x 296.8 kJ ) = -438.7 kJ 1 mol SO3 1 mol O3 1 mol SO2 Calculate the heat released and melt the cesium 96.2 L O3 x 1 mole O3 x 438.7 kJ x 1 mol Cs x 132.91 g Cs = 1.20 x 105 g Cs 22.4 L O3 1 mole O3 2.09 kJ 1 mol Cs

c) Calculate the mass of liquid water that can be boiled and heated from 89.65 C to 178.09 C using the energy released when 88.30 g of calcium oxide is produced when calcium reacts with excess iron (III) oxide. Iron metal is also produced by the reaction.

Find the Balanced Equation 3Ca + Fe2O3 3 CaO + 2Fe Find the Heat of Formation for each Element Hf for CaO = -635.1 kJ/mol Hf for Fe2O3 = -822.1 kJ/mol

Hf for Ca = 0 kJ/mol Hf for Fe = 0 kJ/mol Find the Energy Released: Hrxn = Hf - Hf products reactants Hrxn = 3 mol CaO 3 x -635.1 kJ + 2 mol Fe x 0 kJ 1 mol CaO 1 mol Fe

- (3 mol Ca x 0 kJ + 1 mol Fe2O3 x -822.1 kJ ) = -1083.2 kJ

1 mol Ca 1 mol Fe2O3

88.30 g CaO x 1 mol CaO x 1083.2 kJ = -568.5 kJ 56.08 g CaO 3 mol CaO Affecting the water (solve each energy calculation but leave the variable M for mass)

Heat Water from 89.65 C to 100 C

q= ? m= M C= 4.18 J/gC T= 10.35 C

q= 4.18 J/gC x M x 10.35 C = 43.3 J/g M

Boil the Water:

M g H2O x 1 mol H2O x 40.7 kJ = 2.26 kJ/g M

18.02 g H2O 1 mol H2O

Heat Water from 100.00 C to 178.09 C

q= ? m= M C= 1.7 J/gC T= 78.09 C

q= 1.7 J/gC x M x 78.09 C = 130 J/g M

Total energy = 43.26 J/g M + 2260 J/g M + 130 J/g M = 2400 J/g M Mass calculation using energy from CaO reaction 2.4 kJ/g M = 568.5 kJ released M = 240 g water 7) Calculating with different thermochemical equations

a) Given the following thermochemical equations: i) H2CO3 H2CO + O2 H =112 kJ ii) H2CO + O2 H2O + CO2 H =-50 kJ Calculate the heat of reaction for the following:

H2CO3 H2O + CO2 H =? Adding the chemical equations as stated above will give us the reaction desired. Therefore we just add the heat of reactions together to calculate the heat of reaction of the reaction desired.

H =112 kJ + H =-50 kJ H = 62 kJ

b) Given the following thermochemical equations: i) 4PCl3(g) P4(s) + 6Cl2(g) H = 650 kJ ii) 4PCl5(g) P4(s) + 10Cl2(g) H = 917 kJ Calculate the heat of reaction for the following: PCl3(g) + Cl2(g) PCl5(g) H =? Reverse the second equation and then add the two reactions together: 4PCl3(g) P4(s) + 6Cl2(g) H = 650 kJ P4(s) + 10Cl2(g) 4PCl5(g) H =-917 kJ 4PCl3(g) + 4Cl2(g) 4PCl5 (g) H 270 kJ Divide by 4 to get the equation desired. H = -67.5 KJ

8) Using thermochemical equations, temperature, and phase changes a) Given the following thermochemical equations:

i) 4PCl5(g) P4(s) + 10Cl2(g) H =1222 kJ ii) PCl5(g) PCl3(g) + Cl2(g) H =89 kJ Calculate the mass of sulfur that will be boiled by the heat given off by the production of 754 g of phosphorus trichloride according to the reaction below.

P4(s) + 6Cl2(g) 4PCl3(g) Multiply equation ii) by 4, reverse equation i) and then add the two equations. P4(s) + 10Cl2(g) 4PCl5(g) H =-1222 kJ 4PCl5(g) 4PCl3(g) + 4Cl2(g) H =356 kJ P4(s) + 6Cl2(g) 4PCl3(g) H =-866 Find the Energy Released: 754 g PCl3 x 1 mol PCl3 x 866 kJ = -1190 kJ (but dont use the rounded number) 137.32 g PCl3 4 mol PCl3 Melting the sulfur

1190 kJ kJ released x 1 mol S x 32.06 g S = 4000g S 10 kJ 1 mole S 9) Mixed bag

a) Calculate the energy needed to raise the temperature of 980 grams of mercury liquid by 387.65 C.

q= ? m= 980 g Hg C= 0.140 J/gHgK

q = C m T = 0.140 J/gHgK x 980 g Hg x 387.65 C = 53000 J b) Calculate the temperature change of 738.9 grams of water if 9.65 grams of calcium

chloride is dissolved in the water. Find the Energy Released 9.65 g CaCl2 x 1 mol CaCl2 x -82.8 kJ = 7.20 kJ released 110.97 g CaCl2 1 mol CaCl2 Find the Change in Temperature T=q/cm T= ? _ q= 7200 J C= 4.18 J/gC m= 738.9 g H2O T= 7200 J = 2.3 C 4.18 J/gC x 738.9 g H2O

c) Calculate the final temperature of 80765 grams of water with a starting temperature of 2.98 C if the water is heated first by the freezing of 114.76grams of ruthenium at its freezing point and then the water is further heated by the cooling of the same ruthenium by 47.8 C.

Freeze the ruthenium 114.76 g Ru x 1 mol Ru x 25.52 kJ = 28.98 kJ 101.07 g Ru 1 mol Ru Cooling the ruthenium

q= ? m= 114.76 g Ru C= 0.238 J/gRuK

q = C m T = 0.238 J/gRuK x 114.76 g Ru x 47.8 C = 1.30 kJ Total energy from the Ru 28.98 kJ + 1.30 kJ = 30.28 kJ Find the Change in Temperature of the Water q= 30280 J m= 80765 g H2O C= 4.18 J/gC T = q/C m = 30280 J = 0.0897 C 80765 g H2O x 4.18 J/gC Final Temperature 0.090 C + 2.98 C = 3.07 C

d) Calculate the mass of gold that can be warmed up to its melting point from a starting

temperature of 727.00 K and melted using 5624 kJ of heat. We are doing two processes to the gold, heating and then melting, without knowing the mass or the individual energies. So, we use M for the mass and solve the two processes with the unknown mass. Heating the gold q= ? m= M g Au C= 0.129 J/gRuK T = 1337.58 K - 727.00 K

q = C m T = 0.129 J/gAuK x M g Au x 610.58 C = 78.8 J/gAu x M g Au Melting the gold M g Au x 1 mol Au x 12.36 kJ = 0.06275 kJ/gAu x M g Au = 62.75 J/gAu x M g Au 196.97 g Au 1 mol Au

Totals 5624000 J = M g Au (78.8 J/gAu + 62.75 J/gAu ) M g Au = 9210 g Au

e) Calculate the heat in Joules needed to change a 76.23 g sample of 27.89 C ice to steam at 149.33 C.

Find the Energy of the Ice q=cmT q= ? C= 2.1 J/ gC m= 76.23 g H2O T= 27.89 C q= 2.1 J/gC x 76.23 g x 27.89 C = 4500 J Find the Energy Needed to Melt the Ice 76.23 g H2O x 1 mol H2O x 6.01 kJ = 25.4 kJ 18.02 g H2O 1 mol H2O Energy Needed to get from 0C to 100C q=cmT q= ? C= 4.18 J/gC m= 76.23 g H2O T= 100 C q= 4.18 J/gC x 76.23 g x 100 C = 31900 J Energy Needed to Boil 76.23 g H2O x 1 mol H2O x 40.7 kJ = 172 kJ 18.02 g H2O 1 mol H2O Energy Needed to get from 100C to 149.33C

q=cmT q= ? C= 1.7 J/gC m= 76.23 g H2O T= 49.33 C q= 1.7 J/gC x 76.23 g x 49.33 C = 6400 J Add up all the Energy Needed to get Total 4500 J + 25400 J + 172000 J + 6400 J = 2.40 x 102 kJ needed

f) Calculate the mass of hafnium (at it boiling point) that can be boiled by the heat generated when 245 g of ozone is reacted with an excess of sulfur to produced sulfur dioxide.

Find the Balanced Equation 3 S + 2 O3 3 SO2 Find the Heat of Formation for each material Hf for O3 = 142.0 kJ/mol Hf for SO2 = -296.8 kJ/mol Hf for S = 0 kJ/mol Find the Heat of the Reaction Hrxn = Hf - Hf products reactants Hrxn= 3 mol SO2 x 296.8 kJ - (2 mol O3 x 142.0 kJ + 3 mol S x 0 kJ ) = -1174.4 kJ 1 mol SO2 1 mol O3 1 mol S Find the heat released 245 g O3 x 1 mol O3 x -1174.4 kJ = 2997 kJ (not rounded yet as we still are doing x & /) 48.00 g O3 2 mol O3 Boil the hafnium 2997 kJ x 1 mol Hf x 178.49 g Hf = 809 g Hf 661.07 kJ 1 mol Hf

g) Calculate the change in temperature of 269.0 g of tin using 9.76 kJ.

Find the Change in Temperature T = q/Cm q= 9760 J C= 0.228 J/gK m= 269.0g Sn T= ? T = 9760 J = 159 K 0.228 J/gK x 269.0 g Sn

Cooling the silverHeat released from the reactionMelting the ice

Heat released from the reactionAffecting the waterFinal temperature

Cooling of the waterDissolving the ammonium nitrateDissolving the NaOHMelting the phosphorusDissolving the calcium chloride (releases energy)Affecting the waterHeating the water to 100C from 87.54 CBoiling the water

Find the Heat of Formation for each ElementFind the Heat of the ReactionFind the Heat of Formation for each ElementFind the Heat of the ReactionBalanced EquationFind the Heat of ReactionFind the Energy Released by 657 g HCl

Find the Balanced EquationFind the Heat of Formation for each ElementFind the Heat of the ReactionFind the Energy ReleasedFind the Heat of the WaterFind the Balanced EquationFind the Heat of Formation for each materialFind the Heat of the ReactionCalculate the heat released and melt the cesiumFind the Heat of Formation for each ElementMelting the sulfurFind the Energy ReleasedFind the Change in Temperature

Freeze the rutheniumCooling the rutheniumTotal energy from the RuFind the Change in Temperature of the WaterFinal Temperature

Heating the goldMelting the goldTotalsFind the Energy of the IceFind the Energy Needed to Melt the IceEnergy Needed to BoilAdd up all the Energy Needed to get TotalFind the Balanced Equation

Find the Heat of Formation for each materialFind the Heat of the ReactionFind the heat released

Boil the hafnium