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Test REVIEW

Test REVIEW. Unit 10 Thermochemistry. 1. How much heat energy is needed to raise the temperature of a 55 g sample of aluminum from 22.4  C to 94.6  C (a) in J, and (b) in calories? The specific heat of aluminum is 0.900 J/g  C. Q = mc ΔT. Q = (55 g). (0.900 J/g o C). (72.2 o C).

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Test REVIEW

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  1. Test REVIEW Unit 10 Thermochemistry

  2. 1. How much heat energy is needed to raise the temperature of a 55 g sample of aluminum from 22.4C to 94.6C (a) in J, and (b) in calories? The specific heat of aluminum is 0.900 J/gC. Q = mcΔT Q = (55 g) (0.900 J/goC) (72.2 oC) = + 3,600 J Q = 3,573.9 J Q = 3,573.9 J 1 cal x = 854.18 cal 4.184 J = + 850 cal

  3. 2. A piece of copper metal of mass 6.22 kg absorbs 728 kJ of energy (heat) and its temperature increases from 20.5C to 324.3C. Calculate the specific heat of the copper metal in J/gC. Q = mcΔT 728000 J = (6220 g) (x) (324.3 oC – 20.5 oC) 728000 J = (6220 g) (x) (303.8 oC) x = 0.385 J/goC

  4. Enthalpy (H) 3. What are the two driving forces in a chemical reaction? 4. What is the natural trend or what do they naturally move towards? 5. What to their + and – signs mean? Entropy (S) Maximize entropy and minimize enthalpy! Enthalpy (H): + (endothermic); - (exothermic) Entropy (S): + (favorable); - (unfavorable)

  5. (a) 2,000,000 cal 6. 2000 Calories (kcal) are equal to (a) how many calories? (b) how many joules? 7. What does specific heat mean? 8. Does specific heat of a substance depend upon the amount of the substance? (b) 8,368,000 J The heat (energy) needed to raise 1 gram of a substance by 1oC. No!!!

  6. = - 783 kJ/mol of C7H8 9. When 5.00 g of liquid toluene, C7H8 is burned, it causes 500.0 mL of water at 20.0C to rise to 40.3C. What is the heat of combustion (Hc) of liquid toluene in kJ/mol? Q = mcT Q = (500.0 g)(4.184J/goC)(40.3 oC – 20.0 oC) Q = 42467.6 J = 42500 J = 42.5 kJ 5.00 g C7H8 1 mol C7H8 = 0.0543 mol C7H8 92.15 g C7H8 exothermic

  7. mcT = mcT mAlcAl(Thot – Tfinal) = mwcw(Tfinal – Tcold) = 35.4 oC TF TH = 100.0 oC TC = 10.0 oC TF = ? 10. If a student puts a 100.0°C piece of aluminum that weighs 228 g into 125 mL of 10.0°C water, what will the final temperature of the mixture be? mAl = 228 g mw = 125 g cw = 4.184 J/g oC cAl = 0.900 J/g oC = (125 g H2O)(4.184 J/g oC)(TF – 10.0 oC) (228 g Al))(0.900 J/goC)(100.0 oC – TF) = 523TF - 5230 20520 – 205.2Tf = 728.2 TF 25750

  8. C7H8(l) + 9 O2(g) 7 CO2(g) + 4 H2O(l) 11. Calculate the ΔHrxn and ΔHc of liquid toluene (C7H8). (ΔH°f of toluene is 11.950 kJ/mol). Horxn = [7(-393.5) + 4(-285.83] − [11.950 + 9(0)] Horxn = − 3909.8 kJ = − 3909.8 kJ/mol C7H8 12. Know the following: • Define activation energy. • Define a catalyst. • Which has greater entropy? Solid (s); Liquid (l); or Gas (g)

  9. 13. Example: Calculate the ΔH, ΔS, and ΔG of this reaction. State whether the reaction is exothermic or endothermic, whether the entropy is favorable or unfavorable, and whether the reaction is spontaneous or non-spontaneous. (@ standard temp.) 2 KClO3(s) 2 KCl(s) + 3 O2(g) = - 89.34 kJ Horxn = [2(–435.87) + 3(0)] − [2(–391.20)] exothermic = + 494.4 J/K Srxn = [2(82.68) + 3(205.0)] − [2(142.97)] favorable G = H – TS G = (– 89.34 kJ) – (298 K)(0.4944 kJ/K) spontaneous G = -237 KJ

  10. 14. Example: Given ΔH and ΔS, calculate ΔG for this reaction. At what temperatures is this reaction spontaneous? ΔH = +847.6 kJ; ΔS = +41.2 J/K Al2O3(s) + 2 Fe(s) 2 Al(s) + Fe2O3(s) (at standard conditions) G = H – TS G = (847.6 kJ) – (298 K)(0.0412 kJ/K) Not spontaneous G = + 835.3 kJ TS > H = 2.06 x 104 K

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