Calorimetric Determination of Reaction Enthalpies

Transcription

1 H + (aq) + OH - q H 2 O Calorimetric Determination of Reaction Enthalpies Purpose: Determine the enthalpy of dissociation of CH 3 COOH CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) Techniques: Calorimetry Temperature Measurement vs Time Apparatus: Calorimeter Precision Thermometer Clock Pu rp os e: CH 3 COOH + OH - CH 3 COO - + H 2 O 1 Concepts: Strong/Weak Acids and Bases Dissociation Neutralization Limiting Reagents Heat Heat Capacity Enthalpy Calorimeter Constant Endothermic Exothermic Hess Law 0 th & 1 st Laws of Thermodynamics 2 The Measurement of Heat and Temperature q = C m T q We cannot measure heat, q, directly, but can measure its effect on substances, namely temperature * changes, T, when heat flows into, or from, a substance. For reasonable changes in temperature and a fixed mass of a substance, m, q is proportional to the mass, m and the temperature change, T. = C m T T = T final -T initial The proportionality constant, C, is called the Specific Heat of the substance. For water, C = 4.18 J / g K * Or other changes, such as melting or boiling 3 absorbed heat is positive. T > 0 ; q > 0 liberated heat is negative T < 0 ; q < 0 Calorimeter a device that measures temperature changes which accompany heat changes in its contents 4 How? Calorimeters are constructed to minimize heat exchange with environment so effects of heat are entirely in the calorimeter and its contents. MUST INSULATE! ( Use Covered, Nested Styrofoam Cups ) Must be able to separate the effects of heat on the calorimeter from the effects on its contents. MUST CALIBRATE! Environment Contents Calorimeter Determine heat absorbed by calorimeter when a known amount of heat is added to contents with known effects of heat. High precision thermometer $60 Hg I.e., determine its Calorimeter Constant 5 6 1

2 TIME CALIBRATION How do we calculate when two samples of water of known mass and temperature are mixed? E.g., suppose we add: 60 g of hot water at 50 o C to T 30 g of cold water at 25 o C mix 60 g 50 o C The Ideal Case 90 g of water at Ideally heat absorbed by cold water = - heat lost by hot water q cold = m c C c T c = - q hot = - m h C h T h 30 g *4.18 J/g * ( 25) = -60 g *4.18 J/g * ( - 50 ) 30 * ( 25) = -60 * ( 50) 30 g 25 o C q hot 90 g 42 o C = (60 * * 25) / ( ) = 42 o C 7 -q hot = q cold 8 Actual will differ from ideal because of heat exchange with container i.e., calorimeter Assume calorimeter is initially at the temperature of the cold water & it experiences same temperature change as its cold water contents. Note: The cold water should be in the calorimeter before mixing! Some heat lost by the hot water goes to increasing temperature of calorimeter as well as the cold water Now, we have q hot = q cold + q cal So, q cal = q hot q cold Calorimeter Accounting for the Calorimeter 60 g 50 o C q hot 30 g 25 o C? o C -q hot = q cold + q cal 9 10 Suppose the measured final temperature Instead of the above mixture, = 40 o C of o C The hot water loses: as we q hot = 60 g * 4.18 J/gC o * ( 50) calculated = 2510 J The cold water gains: in the q cold = 30 g * 4.18 J/gC o * ( 25) ideal case = 1880 J The calorimeter gains: q cal = q hot q cold = = 630 J T of calorimeter = T of cold water = +15 C o So, it behaves like an object with a specific heat of C cal = q cal / T cal = 630 / 15 = 42 J / C o 11 TEMPERATURE MEASUREMENT Considerations 1. Temperatures of objects not at room temperature (RT) will change in time. [ Newton s Law: dt/dt = k (T T 0 ) ] 2. When liquids of different temperatures are mixed, the final equilibrium temperature is not achieved instantaneously. 3. Some chemical reactions may occur s l o w l y How do we account for the above? CONCENTRATION CONCENTRATION vs TIME Make periodic temperature measurements and extrapolate to the time of interest. 12 2

3 55 Hot Water Temperatures vs Time This procedure enables us to tell what the temperature changes would have been IF: Temperature Time of mixing = T cal T After Mixing cold Cold Water Time (min) Cold Hot Mix 13 the mixing had been instantaneous, and the temperature of the calorimeter and its contents had equilibrated instantaneously We use time extrapolation in each of the calorimetric measurements in this exercise. 14 RELATIONSHIP BETWEEN q AND H Under the conditions of this exercise constant pressure, and assuming only PV work, heat absorbed or liberated by a process is equal to the Enthalpy change in the process: q = H fin H init = H A process for which H > 0 is called Endothermic H fin > H init - Final state has higher enthalpy A process for which H < 0 is called Exothermic We wish to Measure H of Dissociation of CH 3 COOH CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) undissociated dissociated Problems: 1. How do you make a solution of the pure undissociated starting material to put in the calorimeter? (It dissociates immediately to its equilibrium extent) 2. The ionization reaction stops when it reaches equilibrium which lies very close to the reactant side. (i.e., When it does dissociate, the reaction does not produce much product) H fin < H init - Final state has lower enthalpy 15 [ 1.0 M CH 3 COOH ionizes to produce only M CH 3 COO - ] 16 What About Neutralization Reactions? The reaction of acetic acid with a strong base does not suffer from these problems. CH 3 COOH + Na + + OH - CH 3 COO - + H 2 O + Na + H + (aq) + Cl - + OH - Cl - + H 2 O But, the reaction of NaOH with CH 3 COOH can be represented as occurring in two steps: 1. First, Dissociation: CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) 2. Then, Neutralization: H + (aq) + OH - H 2 O H neut Reaction 1 is the one in which we are interested! What about Reaction 2? 17 The neutralization reaction: H + (aq) + OH - H 2 O H neut does not involve the original acetic acid or any other specific acid in any way. Indeed, it is the net ionic equation one would write for the reaction of hydrochloric acid, or any strong, monoprotic acid, with sodium hydroxide, e.g., HBr, HNO 3, HI, HClO 3, HClO 4 So, H neut could be the H for the neutralization of hydrochloric acid with sodium hydroxide - which should be easily measurable. 18 3

4 Hess Law : H for a reaction that can What be written as a series of steps is equal From HCl to the sum of H s for the we individual seek neutralization steps. CH 3 COOH CH 3 COO - + H + H + + OH - H 2 O H neut CH 3 COOH + OH - CH 3 COO - + H 2 O From Hess Law H CH3COOH = + H neut Or, H CH3COOH From CH 3 COOH neutralization What will we Measure & Compute? 1. Determine calorimeter constant, C cal 2. Determine heat of neutralization of CH 3 COOH + OH - CH 3 COO - + H 2 O 3. Determine heat of neutralization of HCl, H CH3COOH H + (aq) + OH - H 2 O ( H HCl =) H neut 4. Compute heat of dissociaton of CH 3 COOH Weak acid Strong acid = H CH3COOH - H neut 19 = H CH3COOH - H neut 20 Procedure 1. Calibration - Determination of C cal Mix two samples of water with measured volumes (to get weights) and temperatures. Determine temperature changes at mixing, graphically [, = T cal ] Using given heat capacity of water (4.18 J / g-c o ), density, and temperature changes, calculate heat exchange [ q hot, q cold, q cal ] From these, calculate the calorimeter constant, C cal. In the example that follows, let us assume that our calculated value of C cal is 42 J / C o Determine H HCl for HCl + NaOH Mix samples of measured volumes and temperatures and given concentrations of HCl and NaOH * V HCl = 50.0 ml V NaOH = 55.0 ml M HCl = 2.20 M M NaOH = 2.30M Confirm limiting reagent mmol HCl = 50.0 * 2.20 = 110 mmol NaOH = 55.0 * 2.30 =127 Limiting Reagent Measure initial temperatures of HCl and NaOH and temperature change at mixing time, graphically. Note: We measure solution volumes instead of weights so we need densities (1.02 g/ml) 22 Temperature (ᵒC) T Temperatures vs Time C o Using the given heat capacities, densities and measured volumes of the solutions and the measured temperature change, compute the heat exchange, q, - correcting for the calorimeter Heat Capacity = 3.97 J/g C o Density = 1.02 g/ml C cal (from part 1) = 42.0 J/C o Tot Volume of Solution Tot Weight of Solution = = ml = 1.02 * = 107 g Time (minutes) NaOH HCl Mix 23 Time of mixing Our measured temperature change at mixing time (from the graph) is: T = C o 2 decimals 24 4

5 q RXN So, q SOL = 3.97 J/gC o * 107 g * C o = J q cal = 42.0 J/C o * C o = 553 J The heat absorbed by the solution and the calorimeter was liberated by the reaction = - (q SOL + q cal )= - ( ) J = J (The sign of the heat of reaction is negative!) But, how many moles of HCl were neutralized? SIG FIGS 3. Determine H CH3COOH for CH 3 COOH + NaOH The analysis for this part is identical with the strong acid case. The result is again a relatively large negative number. Suppose the result for this part of the exercise is: I.e., The neutralization of mol of HCl liberated 6147 J H CH3COOH = kj/mol H HCl = J / mol = kj/mol Now we invoke Hess Law, namely = H CH3COOH - H HCl = X (-5.24 X 10 4 ) = X 10 2 J / mol * We have determined the enthalpy of the reaction: CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) = -3.7 X 10 2 J / mol * The difference between two comparable negative numbers. Sig Figs 1. Work in assigned Pairs Notes on the Procedure 2. Do not remove Precision Thermometers from their special clamps! 3. Each pair will submit 3 plots. Put both names on all plots and attach all 3 plots to ONE of the two lab reports. No need to make copies for each report. Precision is critical in this exercise! Make sure you read thermometers to their full precision: Why you should expand the Temperature Scale Desk thermometers to nearest 0.2 o C Precision thermometers to nearest 0.02 o C ( Maximum Temperature 50 o C) In plotting, use temperature scales that provide appropriate precision. Try to read to 0.02 o C Temp Temp (oc) (oc) degree increments NaOH HCl For neutralization reactions, you should break the temperature axis to expand scale in two regions if that is helpful Read the temperatures from graphs with appropriate precision Time 1.8 (min) Time (min) NaOH HCl

6 Same Data as Previous Slide Temp (oc) degree increments Time (min) NaOH HCl 31 6