Experiiment 24 calorimetry (2)

Name: Ngo Dang Quang Hieu Trieu Vinh Khang Dinh Xuan Phuoc Duong Khac Trieu 22125084 22125112 22125226 22125333 Experiment 25 Report Sheet Calorimetry B Enthalpy (Heat) of Neutralization for Acid-Base Reaction HCl + NaOH Volume of Acid (mL) Temperature of acid (oC) Volume of NaOH (mL) Temperature of NaOH (oC) Trial 50.00 32.00 50.00 32.50 Exact molar concentration of NaOH (mol/L) Maximum temperature from graph (oC) Trial 50.00 32.00 50.00 33.00 0.05 39.0 39.0 Calculations for Enthalpy (Heat) of Neutralization for an Acid – Base Reaction Average initial temperature of acid and NaOH (oC) 32.75 33.5 5.75 5.5 Temperature change, ΔT (oC)T (oC) Volume of final mixture (mL) 100 100 Mass of final mixture (g) (Assume the density of the 100 100 solution is 1.0 g/mL) Specific heat of mixture 4.18 J/g• oC Heat evolved (J) -0.2403 -0.2299 Moles of OH- reacted, the limiting reactant (mol) 0.05 0.05 Moles of H2O formed (mol) 0.05 0.05 -2403 kJ/mole -2299 kJ/mole ΔT (oC)Hn (kJ/mol H2O), equation 25.8 -2351 kJ/mole 10 Average ΔT (oC)Hn (kJ/mol H2O) Calculations for Trial 1: Show calculations for Trial using the correct number of significant figures △ T =T final −T initial =38.5−3 75=5.75 oCV final =V HCl + V NaOH =50+ 50=100 mL m final=V final x Density=100 ( mL ) x Heat evolved=m final x 4.18 Moles of OH ( mLg )=100 g=0.01 kg ( Jg •° C ) x △ T =0.01 x 4.18 x 5.75=−0.24 ( J ) −¿reacted= Extractmolar concentration of NaOH x V NaOH =1 ( molL ) x50 x 10 −3 ( L ) =0.05 mol¿ △ H n=−specific heat ( H O ) x combined massse sacid +base x △ T =−4.18 x 100 x 5.75=−2403 Average △ H n= kJ ( mol ) △ H ntrial 1+ △ H n trial kJ =−2403 mol Trial Trial Temp Temp 36.0 36.0 The5stemperature change in theinfirst Temperature change theminute following 39.5 10s 39 15s 38.5 20s 38 37.5 137 minute 38.0 38.0 38.5 38.0 39.1 39 39.0 38.5 25s 38.939.0 39.0 30s 38.839.0 39.0 Temperature (oC) Temperature (oC) Time 38.739.0 39.0 39.0 39.0 39.0 39.0 36.5 35s 36 40s 35.5 45s 35 50s 38.5 39.0 39.0 55s 10 38.439.0 15 20 39.0 25 60s 38.339.0 39.0 Time (second) 39.0 34.5 15s minutes minutes minutes minutes minutes 30s 38.6 39.0 38.2 30 35 40 45 50 55 minutes 60 Trial Trial120 135 150 165 180 195 210 225 240 255 270 285 300 39.0 39.0 15 30 45 60 175 90 105 45s 39.0 39.0 60s 39.0 39.0 75s 39.0 39.0 90s 39.0 39.0 105s 39.0 39.0 120s 39.0 39.0 135s 39.0 39.0 150s 39.0 39.0 165s 39.0 39.0 180s 39.0 39.0 195s 39.0 39.0 210s 39.0 39.0 225s 39.0 38.5 240s 39.0 38.5 255s 38.5 38.5 270s 38.5 38.5 285s 38.5 38.5 300s 38.5 38.5 Time (second) Trial Trial C Enthalpy (Heat) of Solution for the Dissolution of a Salt NH4Cl Trial Trial Mass of salt (g) 5.000 5.000 Moles of salt (mol)) 0.093 0.093 Mass of calorimeter (g) 207.470 207.440 Mass of calorimeter + water (g) 225.280 226.310 Mass of water (g) 18.810 18.870 32.000 33.000 o Initial temperature of water ( C) Final temperature of mixture from 25.000 24.000 graph (oC) Calculations for Enthalpy (Heat) of Solution for the dissolution of Salt 1.Change in temperature of solution, △T (℃)) Heat change of water (J) -7 -9 550.90 710.56 3.Heat change of salt (J) 54.95 70.65 4.Total enthalpy change, equation 25.11 -1141.76 -1470.23 5.△Hs (J/mol salt) equation 25.12 -6514.51 -8401.07 6.Average △Hs (J/mol salt) Name of salt: Ammonium chloride -7457.79 NH4Cl Show calculations for Trial using the correct number of significant figures △ T =T final −T initial =25.00−3 2.00=−7 ° C Heat change of water ( J )=−specific heat ( H O ) x Mass of Water x △ T J x 18.81 ( g ) x (−7 ° C )=550.90 ( J ) g •° C Heat change of salt ( J )=−specific heat ( H O ) x Mass of Salt x △ T ¿−4.184 ¿−1.57 J x ( g ) x (−7 ° C )=54.95 ( J ) g •° C Total enthalpy change=−H eat change ( H O )+ △Hs ( −enenergy change salt −54.95 =−550.90+ =−1141.76 Molesalt 0.093 (−energy chang e H O ) + (−energy chang e salt ) (−550.90+ (−54.95 ) ) J J = = =−6514.51 mol salt mol e salt 0.093 mol esalt ) Trial Trial Time (NH4Cl) Temp Temp The temperature change in the Temperature change in 28 28 5s 30 27 28 26 15s25 24 27 20s25 22 26.5 21 26 Temperature Temperature (oC) (oC) 10s 20 25s minute 24 30s15 21 25 35s23 10 20.5 24 22 45s 20.5 24 20.5 24 20 24 40s 50s210 55s20 20 10 15 23.5 60s 23.5 20 first minute the following minutes 20 25 30 35 40 45 50 55 60 Time (second) 19 minutes minutes minutes minutes minutes 15s 15 21 23.5 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 Trial Trial 30s 21.5 23.5 Time (second) 45s 21.5 23.5 60s 22 22 75s 22 22 90s 22 22 105s 22 22 120s 22.5 22.5 135s 23 22.5 150s 23 22.5 165s 23 22.5 180s 23 22.5 195s 23 22.5 210s 23.5 23 225s 24 23 240s 24 23 255s 24 23.5 270s 24.5 24 285s 25 24 300s 25 24 Trial Trial Na2SO4 Trial Trial Mass of salt (g) 5.000 5.000 Moles of salt (mol)) 0.035 0.035 Mass of calorimeter (g) 207.450 207.470 Mass of calorimeter + water (g) 226.540 226.320 Mass of water (g) 19.090 18.85 33 32 o Initial temperature of water ( C) Final temperature of mixture from 33.5 32.5 graph (oC) Calculations for Enthalpy (Heat) of Solution for the dissolution of Salt 1.Change in temperature of solution, △T (℃)) Heat change of water (J) 0.5 0.5 -79.87 -39.43 3.Heat change of salt (J) -2.257 -2.257 4.Total enthalpy change, equation 25.11 144.35 103.91 5.△Hs (J/mol salt) equation 25.12 2346.40 1191.05 6.Average △Hs (J/mol salt) Name of salt: Sodium Sulphate 1768.725 Na2SO4 Show calculations for Trial using the correct number of significant figures △ T =T final −T initial =33.5−3 3.00=0.5 ° C Heat change of water ( J )=−specific heat ( H O ) x Mass of Water x △ T J x 19 09 ( g ) x ( 0.5 ° C )=−79.87 ( J ) g •° C Heat change of salt ( J )=−specific heat ( H O ) x Mass of Salt x △T ¿−4.184 ¿−0.903 J x ( g ) x ( 0.5 ° C )=−2 25 ( J ) g •° C Total enthalpy change=−Heat change ( H O ) + △Hs ( −enenergy change salt 2.25 =79.87+ =1 44 35 (J ) Mole salt 0.035 (−energy chang e H O ) + (−energy chang e salt ) ( 79.87+ ( 2.25 ) ) J J = = =234 6.40 mol salt mol e salt 0.0 35 mol e salt ) Trial Trial Time (NaOH) Temp Temp The temperature change in the first Temperature change in the 5s 32.5 32 following minutes 34 32 10s 33 34 32 15s33.5 33 Temperature Temperature (oC) (oC) 20s33.5 33 minutes minutes minutes minutes minutes 32 33 32 33 33 30s32.5 32 35s 33 32 40s 33.5 32 45s31.5 33.5 32 32 25s minute 33 32 32.5 50s 31 33.5 55s31.5 533.5 10 32 60s 32 33.5 31 33.5 15s 15 30 45 33.5 30s 33.5 45s minute 3215 20 25 30 35 40 45 50 55 60 Time (second) 32 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 Trial Trial 32 Time (second) 32 60s 33.5 32 75s 33.5 32 90s 33.5 32 105s 33.5 32 120s 33.5 32 135s 33.5 32 150s 33.5 32 165s 33.5 32 180s 33.5 32 195s 33.5 32 210s 33.5 32 225s 33.5 32.5 240s 33.5 32.5 255s 33.5 32.5 270s 33.5 32.5 285s 33.5 32.5 300s 33.5 32.5 Trial Trial NaOH Trial Trial Mass of salt (g) 5.000 5.000 Moles of salt (mol)) 0.125 0.0125 Mass of calorimeter (g) 207.530 207.440 Mass of calorimeter + water (g) 226.450 226.370 Mass of water (g) 18.920 18.84 33 32 Initial temperature of water (oC) Final temperature of mixture from 55 55.5 graph (oC) Calculations for Enthalpy (Heat) of Solution for the dissolution of Salt 1.Change in temperature of solution, △T 22 23.5 -1741.54 -1852.42 -163.9 -175.07 4.Total enthalpy change, equation 25.11 3052.74 3252.98 5.△Hs (J/mol salt) equation 25.12 15243.52 16219.92 (℃)) Heat change of water (J) 3.Heat change of salt (J) 6.Average △Hs (J/mol salt) Name of salt: Sodium hydroxide 15731.72 NaOH Show calculations for Trial using the correct number of significant figures △ T =T final −T initial =55−33.00=22 ° C Heat change of water ( J )=−specific heat ( H O ) x Mass of Water x △ T J x 18.92 ( g ) x ( 22 ° C )=−1741.54 ( J ) g •° C Heat change of salt ( J )=−specific heat ( H O ) x Mass of Salt x △T ¿−4.184 ¿−1 49 J x ( g ) x ( 22° C )=−163.9 ( J ) g •° C Total enthalpy change=−Heat change ( H O ) + △Hs ( −enenergy change salt 163.9 =1741.54+ =3052.74( J ) Mole salt 0.125 (−energy chang e H O ) + (−energy chang e salt ) ( 1741.54+ ( 163.9 ) ) J J = = =15243.52 mol salt mol e salt 125 mol e salt ) Time (Na2SO4) 5s 60 10s Trial Trial Temp Temp The temperature change in the first Temperature change in the 32 31 minute following minutes 35 32 15s50 35 35 20s60 37 37 25s 38 38 180s 53.5 58 195s 54 57.5 210s 54.7 225s 55 57.5 57 Part A.4 when a student chemist transferred the metal to the calorimeter, some water splashed out of the calorimeter Will this technique error result in the specific heat of the metal being reported as too high or too low? Explain 240s 57 255s 56 55 270s 55 56.5 285s 55 56 55 55.5 Temperature Temperature (oC) (oC) 70 40 Laboratory Answers Part A.1 The 200-nm test 40 40 minute tube also 35s40 41 43 20 contained some 44 40s 42 water (besides 30 46 45s10 42.5 the metal) that 20 was 48 50s 43.5 49 15 20 25 30 35 40 45 50 55 60 subsequently 55s10 44 10 added to the Time (second) 50 60s 44.5 calorimeter ( in 52 Part A.4) 15s 15 44.8 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 Trial Trial Considering a 45 55 30s Time (second) minutes higher specific 45 57 45s heat of water, 46 59.5 60s will the Trial Trial temperature 48 59.5 75s change in the calorimeter be higher, lower, or unaffected by this 49 59.5 90s technique error? Explain minutes 50 59 105s The presence of water in the 200-mm test tube, along with its addition 50.5 59 120s to the calorimeter, will result in a lower temperature change Water's 52 59.5 135s higher specific heat capacity causes it to absorb more heat energy, 52.5 58.5 reducing the temperature change compared to if only the metal was 150s minutes present 53 58.5 165s 50 30s30 minutes minutes 57 The presence of water in the 200-mm test tube, along with its addition to the calorimeter, will result in a lower temperature change Water's higher specific heat capacity causes it to absorb more heat energy, reducing the temperature change compared to if only the metal was present Part B The enthalpy of neutralization for all strong acid-strong base reactions should be the same within experimental error Explain Will that also be the case for all weak acid-strong base reactions? Explain 300s The enthalpy of neutralization for strong acid-strong base reactions is expected to be the same due to complete dissociation However, for weak acid-strong base reactions, the enthalpy of neutralization may vary due to incomplete dissociation and other factors affecting the reaction stoichiometry Part B Heat is lost to the Styrofoam calorimeter Assuming a 66.2 ◦C temperature change for the reaction of HCl (aq) with NaOH (aq), calculate the heat loss to the inner 2.35g Styrofoam cup The specific heat of Styrofoam is 1.34 J/g.◦C Heat loss = mass × specific heat × temperature change Given: Mass of Styrofoam cup (m) = 2.35 g Specific heat of Styrofoam (C) = 1.34 J/g•°C Temperature change (ΔT) = 6.22°CT) = 6.22°C Plugging in these values into the formula, we can calculate the heat loss: Heat loss = 2.35 g ì 1.34 J/gãC ì 6.22C Heat loss = 19.78 J Therefore, the heat loss to the inner Styrofoam cup is 19.78 Joules Part B.3 Jacob carelessly added only 40.0mL (instead of the recommended 50.0mL) of 1.1 M HCl to the 50.0 mL of 1.0M NaOH Explain the consequence of the error It will result in an insufficient amount of HCl to fully react with the NaOH This leads to incomplete neutralization and an imbalance in the stoichiometry of the reaction The consequence is that some NaOH will remain unreacted, potentially affecting the accuracy of the reaction's outcome Part B.3 The chemist used a thermometer that was miscalibrated by ℃ over the entire over the entire thermometer scale Will this factory error cause the reported energy of neutralization, △Hn to be higher, lower, or unaffected? Explain The miscalibration of the thermometer by 2°C will cause the reported energy of neutralization to be higher This is because the overestimated temperature change leads to an overestimation of the calculated energy of neutralization Part C.3 If some of the salt remains adhered to the weighing paper (and therefore is not transferred to the calorimeter), will the enthalpy of solution for the salt be reported too high or too low? Explain Because the underestimated mass of the salt used in the calculation leads to an underestimated enthalpy of solution Part C The dissolution of ammonium nitrate, NH4NO3, in water is an endothermic process Since the calorimeter is not a perfect insulator, will the enthalpy of solution, △Hs, for ammonium nitrate be reported as too high or too low if this heat change is ignored? Explain If the heat change resulting from the dissolution of ammonium nitrate in water is ignored due to the imperfect insulation of the calorimeter, the reported enthalpy of solution will be erroneously too low 10 This is because the dissolution process is endothermic and absorbs heat energy from the surroundings Inadequate insulation allows some heat to escape, leading to an underestimated enthalpy value 11 Experiment Prelaboratory Assignment Calorimetry A 20.94-g sample of a metal is heated to 99.4 oC in a hot water bath until thermal equilibrium is reached The metal sample is quickly transferred to 100.0 mL of water at 22.0oC contained in a calorimeter The thermal equilibrium temperature of the metal sample plus water mixture is 24.6oC What is the specific heat of the metal? Express the specific cheat with the correct number of significant figures Let X be the specific heat of the metal Heat lost by metal = 20.94 x X x (99.4 – 22.6) = 1566.312 cJ Mass of water = 100 x 1.00 = 100g Heat gained by water = 100 x 4.814 x (24.6 – 22.0)= 1087.84 J Heat lost = Heat gained 1566.312X = 1087.94 → X = 0.695 J/g ◦C a Experimental Procedure, Part A.1 What is the procedure for heating a metal to an exact but measured temperature? Heat treatment is a process that is used to alter the physical properties of a material in a beneficial way During a heat treatment process, a material is typically heated to a target temperature at which its physical properties change It is then cooled at a controlled rate We use heating plates to maintain proper temperature control and transmission Another way is to use boiling water, which is very effective at heating the metal b Experimental Procedure, Part A.1 How can bumping be avoided when heating water in a beaker? The most way of preventing bumping is by adding a metal rod to the reaction vessel Though, these alone cannot prevent bumping and for this reason it is advisable to boil liquids in a boiling tube, a boiling flask Experimental Procedure, Parts A.4,5 a when a metal at a higher temperature is transferred to water at a lower temperature, heat is inevitably lost to the calorimeter will this unmeasured heat loss increase or decrease the calculated value of the specific heat of metal? Explain See equation 25.5 - If you not figure out the heat loss, then you will have a higher specific heat, now you miscalculate that the metal will need less energy to change its temperature b Explain why the extrapolated temperature is used to determine the maximum temperature of the mixture rather than the highest recorded temperature in the experiment See Figure 25.5 We use extrapolation to detect other values outside the valid range of our equation 12 Experimental Procedure, Part B Three student chemists measured 50.0 mL of 1.00M NaOH in separate Styrofoam coffee cup calorimeters (Part B) Brett added 50.0 mL of 1.10 M HCl to his solution of NaOH; Dale added 45.5 mL of 1.10M HCl (equal moles) to his NaOH solution Lyndsay added 50.0 mL of 1.00M HCl to her NaOH solution Each student recorded the temperature change and calculated the enthalpy of neutralization Identify the student who observes a temperature change that will be different from that observed by the other two chemists Explain why and how (higher and lower) the temperature will be different Solution: 50 mL and 1M of NaOH → 50 x = 50 mol of NaOH a/ Brett adds → 50 x 1.1 = 55 mol of HCl b/ Dale adds → 45.5 x 1.1 = 45.5 mol of HCl c/ Lindsay adds → 50 x = 50 mol of HCl Only Date will have a limiting reactant, so the temperature is expected to be lower Experimental Procedure, Part C Angelina observes a temperature increase when her salt dissolves in water a Is the lattice energy for the salt greater or less than the hydration energy for the salt? Explain - The lattice energy for the salt is greater than the hydration energy for the salt - When salt reacts with water, the salt absorbs some of the water and turns into a hydrated salt Hydrated salts have lower energy due to the increase in temperature of the water b Will the solubility of the salt increase or decrease with temperature increases? Explain - The solubility of the salt will increase with temperature increases - The system has higher energy, lattice energy will weaker and the salt is easy to break down A 5.00g sample of KBr at 25.0 ℃ over the entire dissolves in 25.0 mL of water also at 25.0 ℃ over the entire The final equilibrium temperature of the resulting solution is 18.1℃ over the entire What is the enthalpy of solution, △Hs of KBr expressed in kilojoules per mole? See equation 25.12 • Mass of water = volume x density = 25.0 mL x 1.00 g/mL = 25.0 g • Mass of solution = mass of KBr + mass of water = 5.0 + 25.0 = 30.0 g We have, specific heat of solution = specific heat of water = 4.184 j/g·°c Heat absorbed by reaction = Heat released by solution q = mc ΔTT = 30.0 x 4.184 x (18.1 – 25.0) = -866.088 J → ΔHH(per gram) ¿ • Moles of KBr = q −866.088 J ¿ ¿−173 mass of KBr 5.0 g m 5.0 = =0.04202mol M 119.0 13 → ΔHH (per mole) ¿ q −866.088 J = =−2.06 x 104 = -20.6 kJ/mol mole of Kbr 0.04202 mol 14