Heat of Neutralization

Heat of Neutralization 11/04/2019 Department of Chemistry and Biochemistry Heat of Neutralization 1 Abstract The purpose of this experiment was to determine the heat capacity of the calorimeter and the heatof neutralization of different reactions. We used a constructed calorimeter and thermometer tomeasure the temperature of deionized water and the heat capacity which we calculated to be45.48 J/K. We then used this calculation to find the heat of neutralization of HCl–NaOH and CH3COOH – NaOH. The heat of neutralization is the total joules released divided by the numberof moles of water produced. For the HCl- NaOH the heat capacity determined was -62.35 kJ/moland the heat of neutralization of CH3COOH- NaOH was -59.89 kJ/mol. Introduction The main goal of this procedure is to find the heat capacity of the calorimeter and the heat of neutralization of HCl – NaOH and CH3COOH – NaOH. The heat of reaction is the change in the enthalpy of a chemical reaction that occurs at a constant pressure. It is a thermodynamic unit of measurement useful for calculating the amount of energy per mole either released or produced ina reaction. Δ H = –ΔT (heat capacity of calorimeter + heat capacity of contents ) (1) Δ T represents the difference between the final and intial temperatures. ΔH represents the change in enthalpy. A positive value indicates the products have greater enthalpy, or that it is an endothermic reaction where heat is required. A negative value indicates the reactants have greater enthalpy, or that it is an exothermic reaction where heat is produced. Heat of Neutralization 2 A calorimeter is a device used to measure the heat flow of chemical reactions. The heat capacity, C, of a substance is the amount of heat required to raise the temperature of a given quantity ofthe substance by 1 degree. The heat capacity of the calorimeter is the quantity of heat absorbedby the calorimeter for each 1°C rise in temperature. The heat lost by the warm water is equal tothe heat gained by the cold water and the calorimeter. heat lost by warmer water = ( T 2 − T f) x 50g x 4.184 J/K − g (2) heated gained by cooler water = (T f − T 1) x 50g x 4.184 J/K − g (3) For instance, if T 1 equals the temperature of the calorimeter and 50mL of cooler water, if T 2 equals the temperature of 50mL of warmer water added to it, and if T f equals the temperature after mixing it, the heat lost by the warmer water is represented by equation (2). The specific heat of water is 4.184 J/K-g, and the density of water is 1.00 g/mL. The heat gained by the cooler water is represented by equation (3). Substituting equations (2) and (3) gives you the following: [( T 2 − T f) x 50 g x 4.185 J/K-g]– [( T f − T 1) x 50g x 4.184 J/K-g] = ( T f − T 1 ) x heat capacity of calorimeter (4) The heat lost to the calorimeter equals its temperature change multiplied by its heat capacity. By measuring T 1 , T 2 , and T f , the heat capacity of the calorimeter can be calculated from equation (4). The heat of neutralization (Δn H) is the change in enthalpy that occurs when one equivalent of anacid and one equivalent of a base undergo a neutralization reaction to form water and a salt. Thebalanced chemical reaction associated with the determination of heat of neutralization for thereaction between NaOH and HCl is shown in equation (5). The balanced chemical reaction Heat of Neutralization 3 associated with the determination of heat of neutralization for the reaction between NaOH and CH3COOH is shown in equation (6). H Cl(aq ) + N aOH (aq) → N aCl(aq ) + H2 O(l) + heat (5) C H3 COOH (aq) + N aOH (aq ) → CH3 COON a(aq ) + H2 O(l) + heat (6) Experimental Procedure The first step of this experiment is to construct a calorimeter. The calorimeter was already set up for us. It was two styrofoam cups covered with a lid that had a small hole in it for the thermometer. This hole prevented the thermometer from touching the bottom of the cup. The entire apparatus was placed in a 400mL beaker to provide stability. 50.0mL of deionized water was placed into the calorimeter cup and left to settle for five minutes. The temperature was then recorded to be 24.3°C. Next we placed 50.0mL of deionized water in a clean, dry 250mL beaker and heated it until the temperature reached 39.6°C. We then poured the 50.0mL of warm water into the calorimeter cup and carefully stirred the water with the thermometer. We observed the temperature for the next three minutes and recorded the temperature every 15 seconds. Next we dried the calorimeter and the thermometer with a paper towel. We measured 50.0mL of 1.0 M of NaOH and placed it in the calorimeter. Then we measured 50.0mL of 1.0 M of HCl in a dry beaker and allowed it to settle near the calorimeter for three minutes. We then measured the temperature of the acid and found it to be 24.70°C. We cleaned the thermometer then measured the temperature of the NaOH solution in the calorimeter; it was 25.02°C. Next we added the 1.0 Mof HCl into the calorimeter with the NaOH. We recorded the temperature of the solution asa function of time every 15 seconds for the next three minutes. We then calculated the heat of neutralization of HCl-NaOH per mole of water formed to be -62.35. For the last part of the Heat of Neutralization 4 experiment we emptied and dried the calorimeter and thermometer with a paper towel. We measured 50.0mL of 1.0 M of NaOH and placed it in the calorimeter. Then we measured 50.0mLof 1.0 M of CH3COOH in a dry beaker and allowed it to settle near the calorimeter for three minutes. We then measured the temperature of the acid and found it to be 24.3°C. We cleanedthe thermometer then measured the temperature of the NaOH solution in the calorimeter; it was25.09°C. Next we added the 1.0 M CH3COOH into the calorimeter with the NaOH. Werecorded the temperature of the solution as a function of time every 15 seconds for the next threeminutes. We then calculated the heat of neutralization of CH3COOH-NaOH per mole of waterformed to be -59.89. Results and Discussion Table 1. Experimental results for the determination of heat capacity of the calorimeter. Temperature of calorimeter and water, 𝑇 1 , °C 24.3 Temperature of warm water, 𝑇 2 , °C 39.6 a 31.2 Maximum temperature, 𝑇 f, °C Temperature difference, 𝑇 2 − 𝑇 f , °C b Heat lost by warm water, J Temperature difference, 𝑇 f − 𝑇 1 , °C c 8.4 1,757.28 6.9 Heat gained by cooler water, J 1,443.48 d Heat gained by calorimeter, J 313.8 c Heat capacity of calorimeter, J/K 45.48 a Determined from experimental data of temperature vs time Heat of Neutralization 5 b Eq. 2 [1] heat lost by warmer water = ( T 2 − T f) x 50g x 4.184 J/K − g heat lost by warmer water = ( 39.6°C − 31.2°C) x 50g x 4.184 J/K − g = 1,757.28J c Eq. 3 [1] heated gained by cooler water = (T f − T 1) x 50g x 4.184 J/K − g h eated gained by cooler water = (31.2°C − 24.3°C) x 50g x 4.184 J/K − g = 1, 443.48J d Eq. 4 [1] [( T 2 − T f) x 50 g x 4.185 J/K-g]– [( T f − T 1) x 50g x 4.184 J/K-g] = ( T f − T 1 ) x heat capacity of calorimeter Heat of Neutralization 6 ( 1,7 57.28J − 1 , 4 43.48J ) = 313.8J heat gained by calorimeter T f −T 1 313.8 J 6.9°C = from Eq. 4 [1] 45.48 J/K Table 2. Experimental results for the determination of heat of neutralization of HCl – NaOH. Temperature of calorimeter and NaOH, 𝑇 1 , °C 25.02 Temperature of HCl, °C 24.70 a 31.58 Maximum temperature, 𝑇 f, °C 6.72 Temperature difference, 𝑇 f − 𝑇 1 , °C b Heat gained by solution, J c Heat gained by calorimeter, J d Heat of neutralization, ∆𝐻𝑛𝑒𝑢𝑡, J e Moles of H2O produced in reaction, mol 𝐻 2,811.65 305.63 3,117.28 2 𝑂 0.05 f Heat of neutralization, ∆𝐻𝑛𝑒𝑢𝑡, kJ/mol a Determined from experimental data of temperature vs time after adding HCl to the NaOH -62.35 The temperature of HCl was 24.70℃. ΔT was determined from our curve after adding HCl to the NaOH, so we did the highest value minus the average value of NaOH and HCl to get the temperature difference. (31.58 ℃ − 24.86℃) = 6.72℃ b Eq. 3 [1]; per 100 g of solution (50 g NaOH plus 50 g of HCl) Heat of Neutralization 7 ( temperature increase x 100 g x 4.184 J/K − g) (6.72℃ x 100 g x 4.184 J/K − g)= 2,811.65 J c (𝑇 f −𝑇 1 ) × heat capacity of calorimeter (Eq. 4) ( temperature increase x heat capacity of calorimeter) 6.72℃ x 45.48 J/K = 305.63 J d heat gained by solution + heat gained by calorimeter 3 05.63 J + 2,8 11.65 J = 3, 117.28 J e Calculated stoichiometrically from the balanced chemical equation between 50 mL of 1.0 MHCl and 50 mL of 1.0 M NaOH M olarity = moles solute / liters solution 10 . M = moles / 0.05 L = 0.05 moles The number of moles of H₂O produced in the reaction of 50 mL of 1.0 MHCl and 50 mL of 1.0 NaOH was found to be 0.05 moles because they had a 1:1 ratio. f ∆H neut J mol H 2 O x 1 kJ 1000J ( total joules released / # of moles water produced) (3,117.28 J / 0.05 moles) = 62, 345.51 J This is a neutralization reaction therefore energy is lost and the △H should be a negative value making it -1(62,345.51) = -62.35 kJ/mol. Heat of Neutralization 8 Table 3. Experimental results for the determination of heat of neutralization of CH 3 COOH –NaOH. Temperature of calorimeter and NaOH, 𝑇 1 , °C 25.09 Minimum temperature, 𝑇 , °C 24.43 Temperature difference, ∆ 𝑇, °C 6.45 a Heat gained by solution, J Heat gained by calorimeter, J Heat of neutralization, ∆𝐻𝑛𝑒𝑢𝑡, J Moles of H2O produced in reaction, mol 𝐻2𝑂 Heat of neutralization, ∆𝐻𝑛𝑒𝑢𝑡, kJ/mol a 2,700.77 293.57 2,994.34 .02 -59.89 Determined from experimental data of temperature vs time after adding CH3COOH to the NaOH ΔT was determined from our curve after adding HCl to the NaOH, so we did the highest value minus the average value of NaOH and CH3COOH to get the temperature difference. (31.48 ℃ − 25.03℃) = 6.45℃ Heat gained by soultion: ( temperature increase x 100 g x 4.184 J/K − g) 31.48 ℃ − 25.025 ℃ = 6.46 (6.46℃ x 100 g x 4.184 J/K − g)= 2,700.77 J Heat gained by the calorimeter: Heat of Neutralization 9 (𝑇 f −𝑇 1 ) × heat capacity of calorimeter (Eq. 4) ( temperature increase x heat capacity of calorimeter ) 6.455℃ x 45.48 J /K = 293.57 J Heat of Neutralization in J: h eat gained by solution + heat gained by calorimeter 2 ,7 00.772 J + 293.57 J = 2, 9 94.34 J Heat of Neutralization: ( total joules released / # of moles water produced) (2, 994.34 J / .02 moles) = 59.89 This is a neutralization reaction therefore energy is lost and the △H should be a negative value making it -1( 59.89 ) = - 59.89 kJ/mol. Discussion  neutralizing the solution; by working quickly we tried to make sure that we were not losing heat   in the transfer outside of  the calorimeter. However, the transfer of solutions to e th calorimeter   was not perfect. Therefore there may have been heat lost in the transfer that was not calculated. Heat of Neutralization 10  rgest source of error in the experiment was in the calorimeter. We had to be careful while   calorimeter and calculate the heat of neutralization for each of the two neutralization reactions. The la  There were two main objectives of this experiment: determine the heat capacity of the  bases, therefore neutralizing them a bit before the thermometer was recording any possible   Another source of error could be simply not cleaning the thermometer between the acids and  neutralize the solution. Therefore it must be carefully cleaned for more accurate results. The HCl with NaOH should have a higher heat of reaction than CH3COOH because HCl is astrong acid..     measured because if it is contaminated with an acid before it goes into the ase b it could   temperature change. The thermometer has to be washed and dried after the temperature of NaOHwas The ions of HCl fully ionize compared to those of CH3COOH. This means there ismore Cl+ ions to bond with the Na+ ions of NaOH. The more bonds there are means there ismore energy released during neutralization. Heat of Neutralization 11