Constant Pressure Calorimeter for Heat Capacity

Constant Pressure Calorimeter for Heat CapacityKanwarpal BrarPurposeTo calibrate a continuous pressure calorimeter and use it to determine the loves of the reaction and dis ascendent of different reactants and to use these wakens of the reactions to find the enthalpy of a reaction by hesss law.Analysis/ calculations coiffure the heat capacity of the coffee cup, Ccal in j*degC for all three trials and calculate the average value. Provide all these values in your report offer entireness calculations only for trial .From table 1Mass of 1.0 M NaOH solution used = 51.67gMass of 1.0 M HCl solution used = 50.85gTotal fold of final solution = 102.52gInitial temperature of reagents = 21.3 deg CFinal temperature after neutralization = 27.8 deg CHeat absorbed by calorimeter q1 = C (heat capacity)*deltaTHeat abrorbed by soluction Q2 = heat capacity (C) * throng of the solution (m)*deltaTHeat released by neutralization reaction, Q3 = heat of reaction (delta H)* gram seawallecules(n)/1mol eIn this reaction, Delta T=T2-T1=27.8degC-21.3degC=6.5degCheat capacity of the solution, C=4.02J/g degC (given) troop of the solution, m = 102.52gheat of the reaction, H = -57.3 KJ (given)= -57300 JHCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)Because HCl and NaCl react 11, any one can be used as limiting reagentMolarity of HCl = 1.0 MVolume of HCl = 50.0 ml = 0.0500 LTherefore moles of HCl, n = molarity * volume= 1.0 mol/L * 0.0500 L= 0.0500 molIt is assumed that not heat is mazed to sur round downing E system = 0 JE system = q1 + q2 + q3 = 0JQ1 = -q2 q3C1 * T = -(4.02 J/g degC * 102.52g * 6.5 deg C) (-57300 J * 0.0500 mol/1 mol)C1 * T = -2678.85 J + 2865 JC1 = 186.15 J/ TC1 = 186.15 J/ 6.5 deg CC1 = 28.64 J/ deg CTrial 1 = 28.64 J/ deg CTrial 2 = 31.09 J/deg CTrial 3 = 29.48 J/deg CAverage = 29.73 J/deg CDetermine the overall heat of reaction per mole od calcium meatl for the appendage of calcium metal.to 1.0 M HCl folloed by the addition of water and b) to water folloed by addition of 1.0 M HCl.In each case, treat the overall reaction as a single process, i.e. instead of determining a delta H value for each step, determine . atomic reactor of ca = 0.404 gmolar mass of ca = 40.08 g/molmoles of ca, n = mass/molar mass= 0.404 g/ 40.08 g/mol= 0.0100 molMass of water used, m = 50.0 g (1ml = 1g)T = Tfinal T signT = 30.5 21.4 deg C = 9.1 deg C ( table 2)Heat of the reaction per mole = -(q of reaction (Ccal * T))/moles of meatal-(Cwater*m*water*detaT(0*T) /n= -(4.184 J/ degC * 50 * 9.1 degC) /0.0100 mol= -1903.72 J/ 0.0100 mol= -190372 J/mole= -190.372 KJ/moleH = -190.372 KJ/moleb)mass of ca = 0.403gmolar mass of ca = 40.08 g/molemoles of ca = mass/ molar mass= 0.400 g/ 40.08 g/mole= 0.00998 molMass of water used = 50 g (1ml = 1 g)Temperature difference T = Tfinal TinitialT = 30.5 20.3 degC = 10.2 degC (table3)Heat of reaction per mole = -q of reaction (Ccal*T)/mole of metal= -(Cwater*mwater*T-(0*T)/n= -(4.184 J/g degC*50g*10.2 deg)/ 0.00998mole= -2133.84 J/ 0.009 98mole= -213811.62 J/mole= -213.81 KJ/moleH = -213.81 KJ/moleDetermine deltaEdissolution in J (g flavour) for the unknown sodium chloride for all three trials and calculate the average value. Provide all of these values in your report, provide full calculation only for trial 1.Unknown salt = CMass of salt = 4.013gMass of water = 100gMass of solution after reaction = 100g + 4.013g= 104.013gT = Tfinal Tinitial= 27-19.9 degC = 7.1 degCEdissolution = -q of reation= -m*C*T= -4.184 J/g degC*104.013g*7.1= -3089.85 JEdissolution/g salt = -3089.85 J/4.013g= -769.96 J/g saltTrial 1 = -769.96 J/g saltTrial 2 = -769.87 J/g saltTrial 3 = -754.18 J/g saltAverage = -764.67 J/g saltDetermine deltaEdissolution in J (g salt) for six salts in table 1. Provide all of these values in your report, provide full calculation only for LiCl.Edissolution = Elattice + Ecation hydration + Eanion hydrationEdissolution = 846 KJ/mol + (-506 KJ/mol) + (-377 KJ/mol) from tableEdissolution = -37 KJ/molEdissolution = -37000 J/molMolar mass of LiCl = 42.39 J/molEdissolution/ g of salt = Edissolution/ molar mass= -37000 J/mol/ 42.39 g/mol= -873 J/g saltEdissolution for LiCl = -873 J/g saltEdissolution of LiBr = -472 J/g saltEdissolution of NaCl = 51.3 J/g saltEdissolution of NaBr = 0 J/g saltEdissolution of KCl = 228 J/g saltDiscussion In the test, a simple constant-pressure, coffee cup calorimeter was calibrated using an acid-base neutralization reaction. the work out specific heat of calorimeter was then used to determine the heats of reactions and dissolutions of other chemical compounds.A simple constant pressure calorimeter was produced out of two styroform cups. The cups were covered with a plastic lid with a hole in centre. While erformiing the acid-base neutralization reaction, the temperature of both acid and base were measure using PH metre temperature probe. The temperature were close to each other. When HCl was added to NaOH no visible change was observed while adding the acid. Bu t the temperature of the soluction was rise after the acid was added. This showed that the reaction between HCl and NaOH was exothermic reaction. after that mass of the final solution was measured.The second objective was to find the heat of the reaction per mole of calcium metal, while following the hesss law provided in the science laboratory manual. This was done in two different trials. First the calcium metal was added to 50.0 ml of 1.0M HCl and then 50ml of water. When calcium was added to HCl it reacted vigorously creating bubbles. The highest temperature recorded was almost double the initial temperature. When water was added to this solution, no visible change was observed, but temperature was dropped by 10 degC. The overall process was still an exothermic reaction the heat of the reaction was mensurable to be -190.372 KJ/moleIn the second trial, the calcium was first added to water. This reaction was similar to the first one. Calcium reacted with the water vigorously. Th e temperature of the solution was increased showing that is was exothermic reaction. when HCl was added to this solution the temperature was dropped by 3.6 degC.Which was less than the first case. The heat of the raction waw calculated to be -213.81 KJ/mole.The closeness o fthe both results can be explained by the fact that heat of the reaction is a state function, and does not depend on the path of the reaction. this also increases the confidence in the result.The final objective of the reaction was to determine the heat of dissociation of the unknown salt, and thus find the unknown salt by comparing the heat of dissotiation to the heat dissolution of possible salts. This unknown salt tag C was white powder form. When unknown salt was added to water, temperature raise by 7.1 degC. This reaction showed that this was a exothermic reaction. the average enthalpy of dissolution of the unknown salt C was calculated to be -764.67 J/g. This value of enthalpy of dissolution corresponded to the calculated value of Lithium chloride, LiCl.A number of experiment errors could have alter the data collected, which includes the accuracy and precission of the instruments used environment conditions. The graduated cylinder was used to measure liquids was accurate to only one decimal place, or could only round the value to .0 or to .5. the measuring balance used to weigh had had high accuracy up to three decimal place, dispite that there was difference in the total weigh of the soluction in all three trials. This shows that may be weighing machine was not accurate. it is also possible that when solution was shaked to mix the reactant some of the solution lost or may be was left over on the cover lid. Or into the walls of cups and glass container. While doing the experiment some liquid was spilled that could be that reason for the difference in the weight. The volume of the solution could be measured by burettes or pipetts for higher accuracy.Overall the results of the experime nt calculations were really promising and surefooted based on the fact that they folled the theory of the experiment.ConclusionA calorimeter was prepared. The heat capacity of calorimeter was calculated to be 29.73 J/deg C. The heat reaction calcium was embed to be -190.372 KJ/mole and -213.81 KJ/mole, in the two trials. the unknown salt had -764.67 J/g salt. The unknown salt C was found to be Lithium chloride. Results of this experiment is promissign and confident.ReferencesOlmsted, john 3 Williams, greg burk Robert c. Chemistry, 1st Canadian ed john Wiley and sons ltd Mississauga, Canada, 2012, pp 511-550