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Chemistry

Chemistry. Thermodynamics The Study of Energy in a Chemical Reaction. What about that heat?. In order to measure the heat released or absorbed from a chemical reaction, we must study the instrument that we are to use to measure it. Water, Aqua, H 2 O is the Stuff

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Chemistry

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  1. Chemistry Thermodynamics The Study of Energy in a Chemical Reaction

  2. What about that heat? • In order to measure the heat released or absorbed from a chemical reaction, we must study the instrument that we are to use to measure it. • Water, Aqua, H2O is the Stuff • If I took Ice at a temperature below the freezing Point, how would a heating curve look?

  3. Measurement Lets plot a graph of the temperature of water versus time. Let start when water is a solid (ice ) at -20 0C and continue to heat until the water changes from ice to steam and 125 0C

  4. Let’s Graph It ΔHv = Hv x mw qs =(mCΔt)steam qw = (mCΔt)water ΔHf = Hf x mice qi = (mCΔt)ice

  5. Symbols • q = heat • m = mass • C = specific heat: the amount of heat a substance can hold per gram, per degree Celsius. • Δt = change in temperature • Heat=mass x specific heat x change temperature. • q = m C Δt

  6. Some Standards • Cice = 2.06 j/g0C Hf = 334 j/g • CWater = 4.18 j/g0C Hv = 2,260 j/g • Csteam = 2.02 j/g0C • Qtotal = qice + ΔHf + qwater + ΔHv + qsteam • Problem: How much heat is required to change the temperature of 250.0 g of water at 25.0 0C to 65.0 0C? • Solution: • We only need the change in temperature of water and nothing else.

  7. Solution Continued • Problem: How much heat is required to change the temperature of 250.0 g of water at 25.0 0C to 65.0 0C? • qw = m CwΔt • = ( 250.0 g )( 4.18 j/g0C ) ( 40.00C ) • = 41,800 j or 41.8 kj

  8. Use the Curve qs =(mCΔt)steam ΔHv = Hv x mw Locate where you finish ☻ qw = (mCΔt)water ΔHf = Hf x mice Locate where you start ☻ qi = (mCΔt)ice How much heat is required to change the temperature of 345 g of ice from -12.50C to water at 45.60C?

  9. Solve How much heat is required to change the temperature of 345 g of ice from -12.50C to water at 45.60C? QT = qi + ΔHf + qw = (mi CiΔti) + (Hf mi) + (mw CwΔtw) = (345 g x 2.06j/gC x 12.5C) +( 334 j/g x 345 g) + ( 345 g x 4.18j/gC x 45.6 C) = 190 kj

  10. Lets Get Steamed • How much heat is required to take 150.0 g of ice at -14.40C to steam at 123.50C? • Qtotal = qice+ ΔHf+qwater+ΔHv+ qsteam • = (mi CiΔti) + (Hf mi) + (mw CwΔtw) + (Hvmv) + (ms CsΔts) • =(150.0 g x 2.06 j/gC x 14.4C) + (334 j/g x 150.0 g) + (150.0g x 4.18 j/gC x 100.0C) + (2,260 j/g x 150.0g) + ( 150.0 g x 2.02 j/gC x 23.5 C) • = 463 kj

  11. It is not a choice but the Law • According to the law of conservation energy and matter are not destroyed but conserved. They are converted. In heat, a form of energy the law holds true. If two systems are mixed the heat transfer is conserved. • Therefore: Heat gained by a system is equal to the heat lost by another system in a closed environment.

  12. A Full Metal Jacket • If a 124 g piece of a metal at 2560C is placed in 500.0 g of water at 25.00C and the final temperature after the system reaches equilibrium is 65.70C, what is the specific heat of the metal? • SOLUTION: • List variables • mm = mass of metal = 124 g • Δtm = change in temperature of metal • = ti – tf = 2560C – 65.70C = 190.30C • mw = 500.0 g • Cw = 4.18 j/g0C • Δtw = tf – ti = 65.70C- 25.00C = 40.70C

  13. Put the Peddle to the Metal • Formula: • Heat gained by water = Heat lost by metal • mw x Cw x Δtw = mm x Cm x Δtm • Solve equation for Cm • Cm = mw x Cw x Δtw= 500.0 g x 4.18j/gC x 40.70C • mm x Δtm 124 g x 190.30C • = 3.60 j/g0C

  14. Are we there yet? • Problem: • If a 55.0 g ice cube at -12.10C is put into an insulated container that has 345 g of water at 54.60C, what is the final temperature when the system reaches equilibrium? • Solution: Heat gained by the ice = Heat lost by the water. • Go to the curve for the set up.

  15. Use the Curve qs =(mCΔt)steam ΔHv = Hv x mw Locate where Water starts ☻ ☻ qw = (mCΔt)water ΔHf = Hf x mice Locate where ice starts Locate where you finish ☻ qi = (mCΔt)ice

  16. The Solution • If a 55.0 g ice cube at -12.10C is put into an insulated container that has 345 g of water at 54.60C, what is the final temperature when the system reaches equilibrium? • Heat Gain = Heat Lost • (mi x Ci x Δti) + (Hf x mi) + (miw x Ciw x Δtiw) = (mw x Cw x Δtw) • Δti = [0.00C – (-12.10C)] = 12.10C • Δtiw = tf - 0.00C = tf • Δtw = ti - tf • Therefore: • (mi x Ci x Δti) + (Hf x mi) + (miw x Ciw x tf) = [mw x Cw x (ti - tf)] • Solve for tf

  17. Solution • (mi x Ci x Δti) + (Hf x mi) + (miw x Cw x tf) = [mw x Cw x (twi - tf)] • (mi x Ci x Δti) +(Hf x mi)+(miw x Cw x tf) = mw x Cw x twi - mw x Cw x tf • (miw x Cw x tf) + mw x Cw x Δtf = (mw x Cw x twi)-[(mi x Ci x Δti)+(Hf x mi)] • tf(miw x Cw + mw x Cw ) = (mw x Cw x twi)-[(mi x Ci x Δti)+(Hf x mi)] • tf = (mw x Cw x twi)-[(mi x Ci x Δti)+(Hf x mi)] • Cw (mi + mw )

  18. Problems • # 9 and 10 • mcCwΔtc = mhCwΔth • mc(tf - tic) = mh (tih - tf) • mctf – mctic = mhtih - mhtf • mctf + mhtf = mhtih + mctic • tf (mc + mh) = mhtih + mctic • tf = mhtih + mctic • mc + mh

  19. Lab Procedure on Heat • 1. Measure out as close to 2.00 g of NaOH as possible on ¼ piece of notebook paper. • 2. Add this to 50.0 ml of water in a glass beaker as measured in a graduated cylinder. This is one of the reactants for Reaction # 3. Set aside until you get of the 3rd reaction. • 3. Measure out as close to 2.00 g of NaOH as possible on ¼ piece of notebook paper. • 4. Pour 100.0 ml of water in the reaction vessel • ( styrafoam cup) and measure the initial temperature. • Add the NaOH and measure the highest Temp Reached. This is Reaction # 1

  20. Continue • 5. Measure out as close to 2.00 g of NaOH as possible on ¼ piece of notebook paper. • 6. Measure 100.0 ml of HCl and determine the initial temperature after it is poured into the reaction vessel. • 7. Add the crystals of NaOH and record the highest temperature reached. This is Reaction # 2 • 8. Obtain 50.0 ml of the HCl and determine the initial temperature. Measure the initial of the solution prepared in step one. Add the two temperatures and divide by 2 to get the initial Temperature of Reaction 3. • 9. Pour the two solutions together in the reaction vessel and measure the highest temperature reached.

  21. Data 1.99 0.0498 21.3 28.4 7.1 2.97 59.6 2.05 21.5 33.4 11.9 4.97 96.9 0.0513 2.03 0.0508 23.3 28.1 2.01 39.6 4.8

  22. Heat of Reaction Experiment • Experimental values from chart. 1.NaOH(s) ---- > Na+1 + OH-1; ΔH1 = 59.6 kj/mol 2.NaOH(s) + H+1 + Cl-1 ---> Na+1+ Cl-1 + H2O;ΔH2 = 96.9 ‘ 3.Na+1+ OH-1 + H+1+ Cl-1 ---> Na+1+ Cl-1+ H2O;ΔH3=39.6

  23. Calculated value of the experiment Calculated Values from adding Reaction 1 to Reaction 3 and eliminating the ions that appear on both sides of the arrow. 1. NaOH(s) ---- > Na+1 + OH-1; ΔH1 = 59.6 kj 3. Na+1+OH-1 + H+1+ Cl-1 --> Na+1+Cl-1+ H2O;ΔH3=39.6 kj • ---------------------------------------------- 2. NaOH(s) + H+1 + Cl-1 --> Na+1+ Cl-1 + H2O;ΔH2 =99.2 kj

  24. Look at the Data • On the Data sheet the ΔH value is 96.9 kj/mol • PERCENT ERROR • %Error = H - L x 100% • H • = 99.2 – 96.9 x 100% • 99.2 • = 2.32 %

  25. Interpretation • The Law of Additivity: The energy of reaction 2 can be calculated by adding the energy of reaction 1 and reaction 3. 1. NaOH(s) ---- > Na+1 + OH-1; ΔH1 = 34 kj 3. Na+1+OH-1 + H+1+ Cl-1 ---> Na+1+ Cl-1+ H2O;ΔH3=60 kj • ---------------------------------------------- 2. NaOH(s) + H+1 + Cl-1 --- > Na+1+ Cl-1 + H2O;ΔH2 =94 kj

  26. The Law of Additivity • Calculate the heat of reaction for the following using Hess’s Law of Additivity • Given the combustion equation, calculate the energy from this reaction. • C2H4 + O2 ----- > CO2 + H2O • Balance Equation for 1.00 mol of the reactant compound C2H4 • C2H4 + 3O2 ----- > 2 CO2 + 2 H2O • Write formation equations • C2H4 ---- > 2 C + 2H2 ; ΔHf = -52.3 kj/mol • 2C + 2O2 ----- > 2CO2 ΔHf = 2(-393.5 kj/mol) • 2 H2 + O2 ------ > 2H2O ΔHf = 2(-285.8 kj/mol) • C2H4 + 3O2 ----- > 2 CO2 + 2 H2O; C2H4 = -1410 kj/mol

  27. Problem 35, page 319 • 1. Write the problem and determine the reactants and products. • Fe2O3 + 3CO ---- > 2Fe + 3CO2; ΔH = -23 Kj/mol • 3Fe2O3 + CO ---- > 2Fe3O4 + CO2 ; ΔH = -39 Kj/mol • Fe3O4 + CO ---- > 3FeO + CO2 ; ΔH = +18 Kj/mol • ------------------------------------------------------ • FeO + CO ---- > Fe + CO2 • The first equation the Fe is on the correct side of the arrow. • The third equation the FeO needs to be on the left side. • The second equation the Fe3O4 needs to be flipped so it can be canceled.

  28. Let’s Get It Right • Equation one is OK because Fe is a product. • Fe2O3 + 3CO ---- > 2Fe + 3CO2; ΔH = -23 Kj/mol • Flip equation two so the Iron III oxide can be eliminated. • 2Fe3O4 + CO2 ---- > 3Fe2O3 + CO; ΔH = +39 Kj/mol • Flip equation to put the FeO on the left. • 3FeO + CO2 ---- > Fe3O4 + CO; ΔH = -18 Kj/mol

  29. Next Step • Divide or multiply each equation by the number that gives it the same mole value as in the final equation. • Reaction 1: Divide by 2 • 1/2Fe2O3 + 3/2CO ---- > Fe + 3/2CO2 ; ΔH = 1/2(-23 Kj/mol) • Reaction 2: divide by 6 ( least common multiple for Iron oxides.) • 1/3Fe3O4 + 1/6CO2 ---- > 1/2Fe2O3 + 1/6CO; ΔH = 1/6(+39 Kj/mol) • Reaction 3: divide by 3 to make one mole of FeO. • FeO + 1/3CO2 ---- > 1/3Fe3O4 + 1/3CO; ΔH = 1/3(-18 Kj/mol)

  30. Finally • 1/2Fe2O3 + 3/2CO ---- > Fe + 3/2CO2 ;ΔH = 1/2(-23 Kj/mol) • 1/3Fe3O4 + 1/6CO2 ---- > 1/2Fe2O3 +1/6CO;ΔH = 1/6(+39 Kj/mol • FeO + 1/3CO2 ---- > 1/3Fe3O4 + 1/3 CO; ΔH = 1/3(-18 Kj/mol) • ------------------------------------------------------- • FeO + ½ CO2 + 3/2 CO -- Fe + ½ CO + 3/2 CO2 11 • ------------------------------------------------------- • FeO + CO ---- > Fe + CO2; ΔH = -11 Kj/mol

  31. Hess’s Law Formula • To Calculate energy from formation energies use the following formula • ΔHr = Σ ΔHf products - Σ ΔHf reactants • C2H4 + 3O2 ----- > 2 CO2 + 2 H2O • ΔHr = [ 2(CO2) + 2 (H2O )] - ( C2H4 )] • = [ 2 ( -393.5 ) + 2 (-285.8)] – ( +52.6 ) • = -1410 kj/mol

  32. HOMEWORK • 1. Given the Equation: • H2SO4 + Fe ---- > Fe2(SO4)3 + H2 • Calculate the ΔHr for one mole of sulfuric acid using the Hess formula. ΔHf Fe2(SO4)3= +79.4 kj/mol • 2. Given the Equation • C2H2 + O2 ----- > CO2 + H2O • Calculate the ΔHr for one mole of acetylene using the Hess formula. • 3. Problem 61 on page 320 in text. Problem 64 is a bonus problem.

  33. Example Law of Additivity • C6H6 + O2 --- > CO2 + H20 C6H6 + O2 --- > 6CO2 + 3H20 15 2 6 3 C6H6 6C + 3H2 ;ΔH = +32.1 6C + 6O2 6CO2 ;ΔH = 6(-393.5) 3 2 3H2 + O2 3H2O ;ΔH = 3(-285.8) 15 2 ;ΔHR = -1390 kj/mol

  34. Hess’ Equation • C6H6 + 15/2O2 --- > 6CO2 + 3H20 • ΔHR = Σ ΔHf prod - Σ ΔHf react • = [ 6(CO2 ) + 3(H2O)] – ( C6H6) • = [ 6(-393.5)+3(-285.8)] – (-32.1) • = -1390 kj/mol

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