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Ch17 – Thermochemistry Ch17.1 Temp vs Heat

Ch17 – Thermochemistry Ch17.1 Temp vs Heat All chemical reactions and changes in state either release or absorb energy. Energy- ability to do work or supply heat. Heat- a form of energy that always flows from hotter to cooler objects.

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Ch17 – Thermochemistry Ch17.1 Temp vs Heat

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  1. Ch17 – Thermochemistry Ch17.1 Temp vs Heat All chemical reactions and changes in state either release or absorb energy. Energy- ability to do work or supply heat. Heat- a form of energy that always flows from hotter to cooler objects. Law of Conservation of Energy- in a chemical or physical progress, energy is neither created or destroyed it only changes form. Exs: 2CO(g) + O2(g) 2CO2(g) + 221,000 Joules of Energy given off H2O(g) H2O(l) + 40,700 Joules Metric system unit of energy

  2. Heat Capacity - Each substance absorbs or releases heat differently. Specific Heat Capacity (Cp) - amount of heat necessary too raise the temp of 1 gram of a substance 1˚C. Ex: H2O: Requires 4.186 Joules of heat energy to raise the temp of 1g 1˚c. (originally defined as a calorie) 1 Cal = 4.186J 1000 cal = 1 kcal = 1 Calorie(food) Units: Specific Heat Capacity Formula Q = m · Cp · ΔT Change in Temp Heat Energy Mass Specific Heat

  3. Temp Review K = C + 273 “K is always bigger” Ex1) 20˚C = ____ K “Even through C & K differ by 273, Ex2) 100K = _____˚C they use the same increments.” Ex3) How much heat is required to raise the temp of 50.0g of water from 20˚C to 100˚C? Cp for H2O = 4.186 J/g·˚C Ex4) How much heat is required to raise the temp of 50.0g of iron from 20˚C to 100˚C? Cp for iron = 0.46 J/g·˚C

  4. Ex5) What is the specific heat capacity of a piece of copper with a mass of 95.4g, if it absorbs 849J of heat to change its temp from 298K to 321K? Ch17 HW#1 1 – 9

  5. Ch17 HW#1 1) 500oC = ____K 2) 500K = ____ oC 3) 20oC = ____ K 4) 100oC = ____ K 5) Law of Cons of Energy 6) 1 Cal = J

  6. Ch17 HW#1 1) 500oC = 773 K 2) 500K = 227oC 3) 20oC = 293 K 4) 100oC = 373 K 5) Law of Cons of Energy 6) 1 Cal = J

  7. Ch17 HW#1 1) 500oC = 773 K 2) 500K = 227oC 3) 20oC = 293 K 4) 100oC = 373 K 5) Law of Cons of Energy – Energy cannot be created or destroys, it can only change forms. 6) 1 Cal =J

  8. Ch17 HW#1 1) 500oC = 773 K 2) 500K = 227oC 3) 20oC = 293 K 4) 100oC = 373 K 5) Law of Cons of Energy – Energy cannot be created or destroys, it can only change forms. 6) 1 Cal =J

  9. 7) How much heat to raise temp of 20.0g of wood from 20oC to 100oC? Cp= 1.8 J/ g.˚C 8) How much heat to raise 20.0g iron from 20oC to 100oC? Cp= 0.46 J/ g.˚C 9) Which hot faster – iron or wooden spoon?

  10. 7) How much heat to raise temp of 20.0g of wood from 20oC to 100oC? Cp= 1.8 J/ g.˚C Q = m · Cp · ΔT = (20g)(1.8 J/ g.˚C)(80˚C) = 2800J 8) How much heat to raise 20.0g iron from 20oC to 100oC? Cp= 0.46 J/ g.˚C 9) Which hot faster – iron or wooden spoon?

  11. 7) How much heat to raise temp of 20.0g of wood from 20oC to 100oC? Cp= 1.8 J/ g.˚C Q = m · Cp · ΔT = (20g)(1.8J/ g.˚C)(80˚C) = 2800J 8) How much heat to raise 20.0g iron from 20oC to 100oC? Cp= 0.46 J/ g.˚C Q = m · Cp · ΔT = (20g)(0.46J/ g.˚C)(80˚C) = 736J 9) Which hot faster – iron or wooden spoon? Iron

  12. Ch17.2 Calorimetry Ex1) 100.00g of an unknown substance is placed in a test tube, which is placed in a breaker of boiling water at 99.5oC. The sample is removed and placed in a calorimeter with 100.00g of water at 20.0oC. The contents come to equilibrium at 22.4oC. What is the Cp of the unknown substance?

  13. Ex2) Lab group # ____ masses their unknown at 18.5g. They place it in a test tube in boiling water measured at 94oC. Meanwhile, they mass an empty calorimeter at 2.2g, half–fill it with tap water and re-mass it at 186.6g. (Mwater = 184.4g) They measure the temp of the water with the same thermometer to be 23oC. They not-so-carefully place the unknown in the calorimeter, and the temp rises to 24oC. What Cp did they get? If the actual is 0.700 J/g.oC, will they get unknown points? % error: Ch17 HW#2 10 – 13

  14. Ch17 HW#2 10 – 13 10) A 100.00g sample of iron at 0°C absorbs 1350 joules of heat. What is the final temp? Cp for iron is 0.46 J/g ·°C 11) 50.04g unknown metal heated to 98.2°C. Placed in calorimeter with 51.12g water at 20.0°C. Equilibrium at 23.0°C. Find Cp:

  15. Ch17 HW#2 10 – 13 • 10) A 100.00g sample of iron at 0°C absorbs 1350 joules of heat. • What is the final temp? Cp for iron is 0.46 J/g ·°C • Q = m · Cp · ΔT • Q = m · Cp · (Tf – Ti) • 1350J = (20g)(0.46J/g.˚C)(Tf – 0°C) • Tf = 29.3°C 11) 50.04g unknown metal heated to 98.2°C. Placed in calorimeter with 51.12g water at 20.0°C. Equilibrium at 23.0°C. Find Cp:

  16. Ch17 HW#2 10 – 13 • 10) A 100.00g sample of iron at 0°C absorbs 1350 joules of heat. • What is the final temp? Cp for iron is 0.46 J/g ·°C • Q = m · Cp · ΔT • Q = m · Cp · (Tf – Ti) • 1350J = (20g)(0.46J/g.˚C)(Tf – 0°C) • Tf = 29.3°C 11) 50.04g unknown metal heated to 98.2°C. Placed in calorimeter with 51.12g water at 20.0°C. Equilibrium at 23.0°C. Find Cp. Metal Water Qlost = Qgained m · Cp · ΔT = m · Cp · ΔT (50.0)(Cp)(73.2) = (51.13)(4.186)(3) Cp = 0.175

  17. 12) 23.95g unknown heated to 95.1°C. Placed in calorimeter with 50.01g water at 21.1°C. Equilibrium at 22.2°C. Cp? 13) 34.55g unknown heated to 99.5°C. Placed in calorimeter with 49.98g water at 19.3°C. Equilibrium at 23.1°C. Cp?

  18. 12) 23.95g unknown heated to 95.1°C. Placed in calorimeter with 50.01g water at 21.1°C. Equilibrium at 22.2°C. Cp? Unknown Water Qlost = Qgained m · Cp · ΔT = m · Cp · ΔT (23.95)(Cp)(72.9) = (50.00)(4.186)(1.1) Cp = 0.132 13) 34.55g unknown heated to 99.5°C. Placed in calorimeter with 49.98g water at 19.3°C. Equilibrium at 23.1°C. Cp?

  19. 12) 23.95g unknown heated to 95.1°C. Placed in calorimeter with 50.01g water at 21.1°C. Equilibrium at 22.2°C. Cp? Unknown Water Qlost = Qgained m · Cp · ΔT = m · Cp · ΔT (23.95)(Cp)(72.9) = (50.00)(4.186)(1.1) Cp = 0.132 • 13) 34.55g unknown heated to 99.5°C. Placed in calorimeter with 49.98g • water at 19.3°C. Equilibrium at 23.1°C. Cp? • Unknown Water • Qlost = Qgained • m · Cp · ΔT = m · Cp · ΔT • (34.55)(Cp)(76.4) = (49.98)(4.186)(3.8) • Cp = 0.301

  20. Lab 17.1 Write Up 1. Mass of Empty Test Tube: _________ g 2. Mass of test tube and metal: _________ g 3. Mass of metal: _________ g 4. Mass of empty cup: _________ g 5. Mass of cup and distilled H20: _________ g 6. Mass of water in cup: _________ g 7. Temperature of boiling water: _________˚C 8. Initial temperature of metal: _________˚C 9. Initial temp of H20 in cup: _________˚C 10. Temp of H20 with metal: _________˚C 11. Change in temp. of metal: _________˚C How would you calculate heat lost by the metal? Heat lost by metal = (Cpmetal)(Mmetal)(ΔTmetal) = 12. Change in temp. H20: _________˚C How do you find the heat gained by the water? Heat gained by water = (Cpwater)(Mmetal)(ΔTmetal) =

  21. What is the relationship between the heat lost by the metal and the heat gained by the water? That’s right!!! They are equal. Then set them equal and solve for Cp metal. (Cpwater)(Mmetal)(ΔTmetal) = (Cpwater)(Mwater)(ΔT water) Scores out of 12 points: %error < 5% = 20 %error < 10% = 18 %error < 15% = 16 %error < 20% = 14 %error < 25% = 12 %error > 25% = 10 And lose points for lack of precise measuring!

  22. Ch17.2 cont Ex1) 45.05g unknown heated to 92.5°C. Placed in calorimeter with 50.50g water at 20.5°C. Equilibrium at 23.2°C. Cp? Ex2) 122.25g unknown heated to 101.2°C. Placed in calorimeter with 75.00g water at 20.1°C. Equilibrium at 22.2°C. Cp?

  23. Ch17 HW#2B 1. 50.00g unknown heated to 95.0°C. Placed in calorimeter with 50.00g water at 20.0°C. Equilibrium at 21.0°C. Cp? 2. 100.00g unknown heated to 100.0°C. Placed in calorimeter with 50.00g water at 20.0°C. Equilibrium at 25.0°C. Cp?

  24. Lab13.1 – Calorimetry - unknown due at end of period - lab due in 3 days - Ch17 HW#2 due at beginning of period

  25. Ch17.3 – Heat & Changes of State 120 For H2O: 100 Temp (°C) 0 -10 Time

  26. Ch12.3 – Heat & Changes of State 5) Raise temp of gas: Q = m ∙Cp ∙ ∆T 120 For H2O: l g 100 4) Vaporize it: Q = n ∙ Hv Temp (°C) 3) Raise temp of liquid: Q = m ∙Cp ∙ ∆T moles! s l 2) Melt it: Q = n ∙ Hf 0 1) Raise temp of solid: Q = m ∙Cp ∙ ∆T -10 Time 6) Add steps • Step 1: Cp for ice = 2.03 J/g.˚C • Step 2 is called Heat of Fusion (melting/freezing): Hf = 6010 J/mol • Step 3: Cp for water = 4.186 J/g.˚C • Step 4 is called Heat of Vaporization (evap/condensing): Hv = 40,700 J/mol • Step 5: Cp for steam = 1.7 J/g.˚C

  27. Step 1: Raise temp of ice from initial temp to melting point (0°C) Q= m·Cp·ΔT Cp ice = 2.03 J/g·°C Step 2: Melt solid→ Heat of Fusion Q=n·HfYou”ll have to convert grams to moles Hffor H2O= 6010 J/mole During step 2, all heat energy (Q) goes into breaking the bonds of the solid into liquid. Use same formula for solidifying a liquid , only Q is (-) Heat given off rather than absorbed. Step 3; Raise temp of liquid from 0°C to 100°C Q= m·Cp·ΔT Cp water= 4.186 J/g·˚C Step 4: Boil liquid→ Heat of Vaporization Q=n·HvHv for H2O = 40,700 J/mol Step 5: Raise temp of gas from 100˚C to final temp. Q=m·Cp·ΔT Cp=1.7 J/g·°C Step 6: Add up steps.

  28. Ex1) How much heat is required to turn 63.25g of ice at -50.4˚C to a vapor at 124˚ 124 100 Temp (°C) 0 -50.4 Time

  29. Ex1) How much heat is required to turn 63.25g of ice at -50.4˚C to a vapor at 124˚ Ch17 HW#3 124 100 Temp (°C) 0 -50.4 Time • Step 1: Q = m.Cp.ΔT = (63.25g)(2.03J/g.˚C)(50.4˚C) = 6694.4J • Step 2: Q = n.Hf = (3.51mol)(6010J/mol) = 21,100J • Step 3: Q = m.Cp.ΔT = (63.25g)(4.186J/g.˚C)(100˚C) = 26,477J • Step 4: Q = n.Hv = (3.51mol)(40,700J/mol) = 143,000J • Step 5: Q = m.Cp.ΔT = (63.25g)(1.7J/g.˚C)(24˚C) = 2623J • Step 6: Add: = 199,894J

  30. Ch17 HW#3 • 14) 100.00g sample of Al at 0oC absorbs 1350J of heat. • Cp = .90J/g.oC. Final temp? • 15) How much heat to bring 5.0g ice from -10oC to 120oC? • Step 1: Q = m.Cp.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = • Step 2: Q = n.Hf = (.28mol)(6010J/mol) = • Step 3: Q = m.Cp.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = • Step 4: Q = n.Hv = (.28mol)(40,700J/mol) = • Step 5: Q = m.Cp.ΔT = (5.0g)(1.7J/g.˚C)(20˚C) = • Step 6: Add: = • 16) How much heat given off 10.5g steam at 150oC to water at 75oC? • Step 5: Q = m.Cp.ΔT = (10.5g)(1.7J/g.˚C)(-50˚C) = • Step 4: Q = n.Hv = (.56mol)(-40,700J/mol) = • Step 3: Q = m.Cp.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) = • Step 6: Add: =

  31. Ch17 HW#3 • 14) 100.00g sample of Al at 0oC absorbs 1350J of heat. • Cp = .90J/g.oC. Final temp? • Q = m · Cp · ΔT  Q = m · Cp · (Tf – Ti) • 1350J = (100g)(0.90J/g.˚C)(Tf – 0°C) • Tf = 15°C • 15) How much heat to bring 5.0g ice from -10oC to 120oC? • Step 1: Q = m.Cp.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = • Step 2: Q = n.Hf = (.28mol)(6010J/mol) = • Step 3: Q = m.Cp.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = • Step 4: Q = n.Hv = (.28mol)(40,700J/mol) = • Step 5: Q = m.Cp.ΔT = (5.0g)(1.7J/g.˚C)(20˚C) = • Step 6: Add: = • 16) How much heat given off 10.5g steam at 150oC to water at 75oC? • Step 5: Q = m.Cp.ΔT = (10.5g)(1.7J/g.˚C)(-50˚C) = • Step 4: Q = n.Hv = (.56mol)(-40,700J/mol) = • Step 3: Q = m.Cp.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) = • Step 6: Add: =

  32. Ch17 HW#3 • 14) 100.00g sample of Al at 0oC absorbs 1350J of heat. • Cp = .90J/g.oC. Final temp? • Q = m · Cp · ΔT  Q = m · Cp · (Tf – Ti) • 1350J = (100g)(0.90J/g.˚C)(Tf – 0°C) • Tf = 15°C • 15) How much heat to bring 5.0g ice from -10oC to 120oC? • Step 1: Q = m.Cp.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = 105J • Step 2: Q = n.Hf = (.28mol)(6010J/mol) = 1669J • Step 3: Q = m.Cp.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = 2093J • Step 4: Q = n.Hv = (.28mol)(40,700J/mol) = 11,306J • Step 5: Q = m.Cp.ΔT = (5.0g)(1.7J/g.˚C)(20˚C) = 170J • Step 6: Add: = 15,342J • 16) How much heat given off 10.5g steam at 150oC to water at 75oC? • Step 5: Q = m.Cp.ΔT = (10.5g)(1.7J/g.˚C)(-50˚C) = • Step 4: Q = n.Hv = (.56mol)(-40,700J/mol) = • Step 3: Q = m.Cp.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) = • Step 6: Add: =

  33. Ch17 HW#3 • 14) 100.00g sample of Al at 0oC absorbs 1350J of heat. • Cp = .90J/g.oC. Final temp? • 15) How much heat to bring 5.0g ice from -10oC to 120oC? • Step 1: Q = m.Cp.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = 105J • Step 2: Q = n.Hf = (.28mol)(6010J/mol) = 1669J • Step 3: Q = m.Cp.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = 2093J • Step 4: Q = n.Hv = (.28mol)(40,700J/mol) = 11,306J • Step 5: Q = m.Cp.ΔT = (5.0g)(1.7J/g.˚C)(20˚C) = 170J • Step 6: Add: = 15,342J • 16) How much heat given off 10.5g steam at 150oC to water at 75oC? • Step 5: Q = m.Cp.ΔT = (10.5g)(1.7J/g.˚C)(-50˚C) = -893J • Step 4: Q = n.Hv = (.56mol)(-40,700J/mol) = -23,742J • Step 3: Q = m.Cp.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) = -1099J • Step 6: Add: = -25,734J

  34. Ch17.4 Enthalpy Enthalpy – heat content (H) Every chemical reaction, every change of state, and any 2 objects at different temperatures will have a change in enthalpy. (ΔH=Q) Exothermic reaction – a chemical reaction where heat is given off (ΔH=(-)) Ex1) CaO(s) + H2O(l) Ca(OH)2(aq)ΔH = – 82kJ the (-) is just an indicator that heat is given off (feels hot). - how much heat is given off when 1 mole CaO reacts? - how much heat is given off when 2 moles CaO reacts? - how much heat is given off when 49.36g CaO reacts?

  35. Endothermic reaction – chemical reaction where heat is absorbed (ΔH=(+)) Ex2) Ba(OH)(aq) + NH4Cl(s) BaCl2(aq) + NH3(aq) + H2O(l) ΔH = +50kJ (feels cold) Write the equation to include ΔH in the equation, and balance. - how much heat is absorbed when 10.42g of BaCl2 is produced? Ch17 HW#4 17 – 20

  36. Lab17.2 – Heat of Fusion - Lab due tomorrow - Ch17 HW#4 due at beginning of period

  37. Ch17 HW#4 17 – 20 17) How much heat required to raise temp of 56.02g of water from 20.0˚C to 85˚C? 18) a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔH= -890 kJ b) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH= -1368 kJ

  38. Ch17 HW#4 17 – 20 17) How much heat required to raise temp of 56.02g of water from 20.0˚C to 85˚C? Q = m.Cp.ΔT = (56.02g)(4.186J/g.˚C)(65˚C) = 15,242J 18) a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔH= -890 kJ b) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH= -1368 kJ

  39. Ch17 HW#4 17 – 20 17) How much heat required to raise temp of 56.02g of water from 20.0˚C to 85˚C? Q = m.Cp.ΔT = (56.02g)(4.186J/g.˚C)(65˚C) = 15,242J 18) a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔH= -890 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890 kJ b) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH= -1368 kJ C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) +1368 kJ

  40. 19) 2NaHCO2(s) Na2CO3(s) + H2O(l) + CO2(g) ΔH= +129 kJ a) Rewrite with ΔH b) How much heat to produce 125.00g Na2CO3 c) How much heat to produce 44.8L C02 20) NH4NO3 NH4+ + NO3-ΔH= +25.7 kJ a)Rewrite b) How much heat required to dissolve 95.00g NH4N03?

  41. 19) 2NaHCO2(s) Na2CO3(s) + H2O(l) + CO2(g) ΔH= +129 kJ a) Rewrite with ΔH 2NaHCO2(s) +129 kJNa2CO3(s) + H2O(l) + CO2(g) b) How much heat to produce 125.00g Na2CO3 c) How much heat to produce 44.8L C02 20) NH4NO3 NH4+ + NO3-ΔH= +25.7 kJ a)Rewrite b) How much heat required to dissolve 95.00g NH4N03?

  42. 19) 2NaHCO2(s) Na2CO3(s) + H2O(l) + CO2(g) ΔH= +129 kJ a) Rewrite with ΔH 2NaHCO2(s) +129 kJNa2CO3(s) + H2O(l) + CO2(g) b) How much heat to produce 125.00g Na2CO3 c) How much heat to produce 44.8L C02 20) NH4NO3 NH4+ + NO3-ΔH= +25.7 kJ a)Rewrite b) How much heat required to dissolve 95.00g NH4N03?

  43. 19) 2NaHCO2(s) Na2CO3(s) + H2O(l) + CO2(g) ΔH= +129 kJ • a) Rewrite with ΔH • 2NaHCO2(s) +129 kJNa2CO3(s) + H2O(l) + CO2(g) • b) How much heat to produce 125.00g Na2CO3 • c) How much heat to produce 44.8L C02 • 20) NH4NO3 NH4+ + NO3-ΔH= +25.7 kJ • a)Rewrite NH4NO3 +25.7 kJ NH4+ + NO3- • b) How much heat required to dissolve 95.00g NH4N03?

  44. Ch17 Mid Ch Rev 27) How much heat will be absorbed by 320g of water when its temp is raised by 35˚C? 28) How much heat will be given off by 55g of water as it cools from 87˚C to 25˚C?

  45. 29) Calculate the specific heat of glass if 65 grams of it increases its temp by 26˚C when it absorbs 840 J of energy. 30) Calc temp change if 160 g of Hg absorbs 1500J of heat. (Cp is 0.14 J/g·˚C)

  46. 31) An unknown 15.00g metal sample is heated in a hot water bath until it reaches 96.2˚C. It is then immediately placed in a calorimeter, filled with 75.00g of H20 at 19.6˚C. The calorimeter temp levels at 25˚C. Find Cpof unknown. 32) An unknown 25.00g metal sample is heated to 94.3˚C. Placed in calorimeter with 115.00g of H20 at 21.1˚C. Calorimeter temp levels at 22.5˚C. Find Cp of unknown. 33) a) 2K + Br2 2 KBr + 784 kJ ΔH= b) H20 + 286 kJ H2+ ½O2ΔH=

  47. Ch17.5 – Thermo Equations In all chemical processes, heat is absorbed or given off. Old bonds break (requires energy) New bonds form (releases energy) (One of these 2 will be a greater amount.) These signs are Look at ΔH: Heat given off  ΔH = – only indicators of Heat absorbed  ΔH = + which side to put Exothermic reaction – heat given off ΔH. DON’T USE! CaO + H2O  Ca(OH)2 + 82,000J

  48. Ch17.5 – Thermo Equations In all chemical processes, heat is absorbed or given off. Old bonds break (requires energy) New bonds form (releases energy) (One of these 2 will be a greater amount.) These signs are Look at ΔH: Heat given off  ΔH = – only indicators of Heat absorbed  ΔH = + which side to put Exothermic reaction – heat given off ΔH. DON’T USE! CaO + H2O  Ca(OH)2 + 82,000J Exothermic energy diagram Enthalpy (kJ) Time

  49. Ch17.5 – Thermo Equations In all chemical processes, heat is absorbed or given off. Old bonds break (requires energy) New bonds form (releases energy) (One of these 2 will be a greater amount.) These signs are Look at ΔH: Heat given off  ΔH = – only indicators of Heat absorbed  ΔH = + which side to put Exothermic reaction – heat given off ΔH. DON’T USE! CaO + H2O  Ca(OH)2 + 82,000J Exothermic energy diagram Enthalpy (kJ) Time ΔH = Hproducts – Hreactants = (small #) – (big #) ΔH = (–) (Just an indicator that heat is given off)

  50. Endothermic Reaction – heat absorbed (feels cold) Ba(OH)2+ NH4Cl + 50,000J  BaCl2 + NH3 + H2O Endothermic energy diagram Enthalpy (kJ) Time ΔH = Hproducts – Hreactants = (big #) – (small #) ΔH = (+)(Just an indicator that heat is absorbed)

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