Additional Calculations: 10. How many joules are required to change the temperature of 80.0 g of water from 23.3 C to 38.8 C?

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1 Additional Calculations: 10. How many joules are required to change the temperature of 80.0 g of water from 23.3 C to 38.8 C? q = m C T 80 g (4.18 J/gC)( C) = 5183 J 11. A piece of metal weighing 5.10 g at a temperature of 52.6 C was placed into ml of water in a calorimeter at 22.1 C, and the final equilibrium temperature was found to be 24.2 C. What is the specific heat of the metal? q metal + q water = 0 5.1g (C) ( C) + 20 g(4.18 J/gC)( C) = (C) = J/ o C 12. If the specific heat of methanol is 2.51 J/gK, how many joules are necessary to raise the temperature of 50.0 g of methanol from 18 C to 33 C? q = m C T 50 g (2.51 J/gC) (33-18) = J = 1880 J 13. When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in g of water, the temperature rose from 23.9 C to 32.2 C. Calculate H (in /mol NaOH) for the solution process: NaOH(s) Na + (aq) + OH - (aq) Assume it s a perfect calorimeter and that the specific heat of the solution is the same as pure water. q = m C T 100 g (4.18 J/gC) ( ) = J J g 3.25 g NaOH 1000 J 1 mol NaOH mol NaOH 14. Consider the reaction of 50.0 ml of 1.09 M CH 3 COOHwith 50.0 ml of M NH 3. a. How many moles of CH 3 COOH and of NH 3 are initially present? b. How many moles of CH 3 COOH and of NH 3 are actually neutralized? 15. Following the procedure described in Steps 4-13 of this experiment, this data was obtained when a glass calorimeter was calibrated. Calculate the heat capacity (calorimeter constant) for this calorimeter. Temperature of the calorimeter and 50.0 ml of cool water 20.6 C Temperature of 50 ml warm water 38.9 C Temperature of the calorimeter and mixture at time = C q hot water + q cold water + q calorimeter = 0 50(4.18)( ) + (50)(4.18)( ) + x = The complete neutralization of moles of a monoprotic acid with excess NaOH evolves 2.62 of heat. What is the molar heat of neutralization of the acid with NaOH.

2 17. A student measured the molar heat of neutralization of a monoprotic acid, HA(1.02 M), with NaOH (0.974 M) and obtained these data. Calculate the molar heat of neutralization of this acid with NaOH. temperature of HA and NaOH before mixing C volume of the acid, HA, used 50.0 ml volume of the base, NaOH, used 50.0 ml temperature of calorimeter and mixture at time = C heat capacity of calorimeter 6.9 J/ C 100 g (4.18 J/g o C) (29.1 o C o C) = J q calorimeter = 6.9 J/ o C (6.4 o C) = J q reaction = J The acid and base react in a one-to-one molar ratio. Equal volumes of both are used thus the less concentrated solution is the limiting reactant and will be used to calculate the moles that react ml sol n 1 L mol NaOH 1 mol H 2 O 1000 ml 1 L sol n 1 mol NaOH mol H 2 O J 1 mol H 2 O 1000 J - mol H 2 O Calorimetry Calculations: 18. The heat evolved in the decomposition of g of ammonium nitrate can be measured in a bomb calorimeter. The reaction that occurs is NH 4 NO 3 (s) N 2 O(g) + 2H 2 O(g) The temperature of the calorimeter, which contains 415 grams of water, increases from C to C. The heat capacity of the calorimeter is 155 J/ C. What quantity of heat is evolved in this reaction in /mol? (6.80) 415 g (4.18 J/g o C) (20.72 o C o C) = 3157 J q calorimeter = 155 J/ o C (1.82 o C) = J q reaction = J J g g NH 4 NO J 1 mol NH 4 NO mol NH 4 NO 3

3 19. A bomb calorimetric experiment was run to determine the head of combustion of ethanol (a common fuel additive derived from corn). The reactions is C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l) The bomb had a heat capacity of 550 J/ C, and the calorimeter contained 650g of water. Burning 4.20 grams of ethanol resulted in a rise in temperature from 18.5 C to 22.3 C. Calculate the heat of combustion of ethanol in /mol. (6.81) 650 g (4.18 J/g o C) (22.3 C o C) = 10,324.6 J q calorimeter = 550 J/ o C (3.8 o C) = 2090 J q reaction = -12,414.6 J -12,414.6 J g 4.20 g -136 C 2 H 5 OH(l) 1000 J 1 mol C 2 H 5 OH(l) mol C 2 H 5 OH(l)

4 20. In a coffee cup calorimeter, 50.0ml of M AgNO 3 and 50.0 ml of M HCl are mixed and react in a double replacement reaction. Write the molecular and net ionic equations. The two solutions are initially at C and the final temperature is C. Calculate the heat that accompanies the reaction in /mol of AgCl formed. Assume that the combined solution has a mass of grams and has a specific heat of 4.18 J/g C and the heat capacity of the calorimeter is 31.2 J/ C. AgNO 3 (aq) + HCl (aq) HNO 3 (aq) + AgCl (s) Ag + (aq) + NO 3 - (aq) + H + (aq) + Cl - (aq) H + (aq) + NO 3 - (aq) + AgCl (s) Ag + (aq) + Cl - (aq) AgCl (s) 100 g (4.18 J/g o C) (23.40 C o C) = J q calorimeter = 31.2 J/ o C (0.9 o C) = 28.1 J q reaction = J The solutions are equimolar and the same volume and react in a one to one molar ratio therefore both solutions completely react ml sol n 1 L 0.10 mol Ag+ 1 mol AgCl mol 1000 ml 1 L sol n 1 mol Ag+ AgCl J mol AgCl 1000 J mol AgCl Hess s Law 21. The enthalpies of the following reactions can be measured. (6.50) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O(l) ΔH = CH 3 OH(g) + 3/2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH = -676 a. Use these values and Hess s Law to determine the enthalpy change for the reaction CH 4 (g) + ½O 2 (g) CH 3 OH(g) ½ O 2 (g) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O(l) ΔH = Reverse CO 2 (g) + 2H 2 O(l) CH 3 OH(g) + 3/2O 2 (g) - ΔH = -676 CH 4 (g) + ½O 2 (g) CH 3 OH(g) ΔH = b. Draw an energy level diagram that shows the relationship between the energy quantities.

5 22. The enthalpies of the following reactions can be measured. (6.51) C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) ΔH = C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l) ΔH = Use these values and Hess s Law to determine the enthalpy change for the reaction C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) Reverse 2CO 2 (g) + 3H 2 O(l) C 2 H 5 OH(l) + 3O 2 (g) - ( ) ΔH = Enthalpy changes for the following reactions can be determined experimentally. (6.52) N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH = NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) ΔH = H 2 (g) + ½O 2 (g) H 2 O(g) ΔH = Use these values and Hess s Law to determine the enthalpy change for the formation of NO(g) directly from the elements (an enthalpy that cannot be measured directly because the reaction is reactant favored) ½N 2 (g) + ½O 2 (g) NO(g) 2 N 2 (g) + 6H 2 (g) 4NH 3 (g) Times 2 2 [ N 2 (g) + 3H 2 (g) 2NH 3 (g) ] 2 ΔH = O 2 (g) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) ΔH = Reverse x 6 6 [ H 2 O(g) H 2 (g) + ½O 2 (g) ] - 6 ΔH = H 2 O(g) 6 H 2 (g) + 3 O 2 (g) 2 N 2 (g) + 2 O 2 (g) 4 NO(g) ΔH = This forms 4 moles of NO instead of only one so need to multiply reaction by ¼ and do same to ΔH So ΔH = Use the following data to calculate the enthalpy of formation for 1.00 mol PCl 3 (l) from its elements P 4 (g) and Cl 2 (g). The enthalpy change for the formation of PCl 5 (s) can be determined experimentally, as can the enthalpy change for the reaction of PCl 3 (g) with more chlorine to give PCl 5 (s). (6.54) P 4 (g) + 10Cl 2 (g) 4PCl 5 (s) ΔH = PCl 3 (l) + Cl 2 (g) PCl 5 (s) ΔH = Goal ¼ P 4 (g) + 3 / 2 Cl 2 (g) PCl 3 (l) 6 Cl 2 (g) P 4 (g) + 10Cl 2 (g) 4PCl 5 (s) ΔH = reverse x4 4 [ PCl 5 (s) PCl 3 (l) + Cl 2 (g) ] - 4 ΔH = PCl 5 (s) 4 PCl 3 (l) + 4 Cl 2 (g) 4 P 4 (g) + 6 Cl 2 (g) 4 PCl 3 (l) Four moles of PCl 3 produced instead of only one so need to multiply reaction and ΔH by ¼ 462.7

6 25. Yeast can produce ethanol by fermentation of glucose (C 6 H 12 O 6 ); this is the basis of the production of most alcoholic beverages. C 6 H 12 O 6 (aq) 2C 2 H 5 OH(l) + 2CO 2 (g) Calculate ΔH, ΔS, and ΔG for the reaction. Is the reaction spontaneous as written? In addition thermodynamic values provided, you will need the following data for C 6 H 12 O 6 (aq) ΔH f = /mol, ΔS = 289 J/molK, and ΔG f = /mol. (19.70) ΔH = ΔH f products - ΔH f reactants [2 ΔH f C 2 H 5 OH(l) + 2 ΔH f CO 2 (g)] - [ ΔH f C 6 H 12 O 6 (aq) ] 2 mol ( /mol) + 2 mol ( /mol) - ( /mol ) = ΔH = S products - S reactants [2 S C 2 H 5 OH(l) + 2 S CO 2 (g)] - [S C 6 H 12 O 6 (aq) ] 2 mol (160.7 J/molK) + 2 mol ( J/molK) - (289 J/molK) = J/K ΔG = ΔH + T ΔS standard temp is 25 C or 298 K ΔG = K ( /K) = A cave in Mexico was recently discovered to have some interesting chemistry. Hydrogen sulfide, H 2 S, (the rotten egg sulfur smell) reacts with oxygen in the cave to give sulfuric acid, which drips from the ceiling in droplets with a ph less than 1. If the reaction occurring is H 2 S(g) + 2O 2 (g) H 2 SO 4 (aq). calculate the ΔH, ΔS, and ΔG at 25 C. Is the reaction product-favored? Is it enthalpy or entropy driven? (19.80) ΔH = ΔH f products - ΔH f reactants [ ΔH f H 2 SO 4 (aq) ] - [ ΔH f H 2 S(g) + 2 ΔH f O 2 (g) ] 1 mol ( /mol) - [ 2 mol ( /mol) + /mol ] = ΔH = S products - S reactants [ S H 2 SO 4 (aq) ] - [ S H 2 S(g) + 2 S O 2 (g) ] 1 mol ( J/molK) - [ 2 mol ( J/molK) + J/molK] = J/K ΔG = ΔH + T ΔS standard temp is 25 C or 298 K ΔG = K ( /K) = 27. Some metal oxides can be decomposed to the metal and oxygen under reasonable conditions. Is the decomposition of silver(i) oxide product favored at 25 C? 2Ag 2 O(s) 4Ag(s) + O 2 (g). If not, can it become so if the temperature is raised? At what temperature is the reaction product favored? (19.82) ΔH = ΔH f products - ΔH f reactants [4 ΔH f Ag(s) + ΔH f O 2 (g)] - [ 2 ΔH f Ag 2 O(aq) ] [ 4 mol ( /mol) + 1 mol ( /mol) ] - 2 ( /mol ) = ΔH = S products - S reactants [4 S Ag(s) + S O 2 (g)] - [ 2 S Ag 2 O(aq) ] 4 mol ( J/molK) + 1 mol ( J/molK) - 2 ( J/molK) = J/K ΔG = ΔH + T ΔS standard temp is 25 C or 298 K ΔG = K ( /K) = 28. The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the

7 following steps: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) 2NO(g) + O 2 (g) 2NO 2 (g) 3NO 2 (g) + H 2 O(g) 2HNO 3 (aq) + NO(g) a. Use the values of ΔHf to calculate the ΔH for each step above. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Calculate the ΔH for the overall reactions. Is the process endothermic or exothermic? 4H 2 O(g) Forward 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Forward x3 Forward x2 2NO(g) + O 2 (g) 2NO 2 (g) 6NO(g) + 3O 2 (g) 6NO 2 (g) 3NO 2 (g) + H 2 O(g) 2HNO 3 (aq) + NO(g) 6NO 2 (g) + 2 H 2 O(g) 4HNO 3 (aq) + 2NO(g) 4NH 3 (g) + 8O 2 (g) 4HNO 3 (aq) + 4H 2 O(g)