Heat_of_RX_Acids_Bases

Heat_of_RX_Acids_Bases - #18 annem for Acid Base Rm's. page...

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Unformatted text preview: #18 annem for Acid Base Rm's. page 1 Determination of the Enthalpy Changes for Some Acid - Base Neutralization Reactions When acids and bases combine with one another in a neutralization reaction. heat energy is produced and liberated. By using a “cuties cup“ calorimeter. it is possible to experimentally measure the amount of heat energy released per mole of acid or base reacting. to. determine the value of the “ Molar Heat of Neutralization.” or molar Aliment. An acid is any molecular species which can 'donate protons“ in solution. A strong acid is thus an acid which easily and frequently dissociates. i.e.. easily donates protonts) in solution. Hydrochloric acid is an example of a strong acid. HCI <---> H+ + Cl‘ That isto say. in the equation above. the equilibrium lies far to the right. Placed in solution. nearly every HCI molecule is dissociated into a H+ and a Cl“ at any given moment. A weak acid. on the other hand. is one which only poorly and infrequently dissociates in solution. Thus. at any given moment. there are only a few molecules dissociated. Acetic acid is an example of a weak acid. cu3coou <--—> CH3000' + H+ In the case of acetic acid. the equilibrium of the above equation can be said to lie far to the left. , In hydrochloric acid. the molecular bonding between the W” and the Cl“ is fairly weak and thus easily broken. The molecular bond between the H+ and the CH3COO' of acetic acid is relatively strong. so only a small percentage of the acetic acid molecules are ionized. i.e.. have released protons (H41 at any given time. A base is any molecular species which “accepts protons” in solution. Sodium hydroxide (lye) is a strong base. NaOH dissociates as follows: NaOH <--—> Na+ + OH’ and the hydroxyl ion (OH') is able to readily combine with the 3+ from an acid to form water. . H+ +'OH' ‘-—> HOH #18 Aline“: for Acid Base Rxn's, page 2 A strong base. like sodium hdee. as you might suspect. is one which dissociates easily and frequently. and aweak base is one which dissociates poorly and infrequently. A common example of aweak base is ammonium hydroxide. NH40H' . which we will think of dissociating as follows: NH40H <—--> NH4+,~I: OH' When a typical acid and a base are combined. the result is the formation of water and a salt: HoA + B-OH ~--> H-OH + A-B (acid) (base) (water) (salt) For example. the reaction between hydrochloric acid and sodium hydroxide. called a neutralga’ tion, can be represented as HCI + NaOH ~~> HOH + NaCl The equilibrium for the neutralization reaction. as written. lies far to the right. The reason for this far-right equilibrium is that the molecular bonds in the water molecules are quite strong. and once formed. seldom break . Breaking the molecular bonds of any substance requires an input of emery. Likewise. Whenever new molecular bonds are formed energr is released in the process For a chemical reaction then. energy is required to break the chemical bonds of the reactant(s) and energy is released when the new chemical bond(sl of the products are formed. ’ Energg "IN" to Emrgu “QUT‘ when break hoods new bonds form if the amount of energy released by the formation of "new" bonds is greater than the amount of energy consmned in breaking the “old” bonds. there is a net export of energy out of the reaction (system) to the surroundings. Such a reaction is said to be em or. if the majority of the energy is released as heat energ. as W. 11'. on the other hand. the amount of energy required to break the "old" bonds is greater than the amount of energ/ released by the formation of “new” bonds. the reaction must absorb energy from the surroundings. Such a reaction is said to be enderggnic or. if the majority of the energy is consumed as heat energr. as * NH40H is actually NH3 dissolved in H20. and the N113 accepts an H+ to form NH4+« but {01’ the sake of simplicity. we will consider the base to be NH40H. #18 Anne“; for Acid Base Rxn's. page 3 Let us apply this concept of energy release and consumption by "bond breaking 8: bond making“ to neutralization reactions between acids and bases. Consider three cases: ‘ e l. Rea n f tron ci with stron e: strong weak bond weak bond bond V weak bond \l t 3. t H-Cl + Ne-UH —-—I* H'UH + tie-Cl Notice that the sum of the energy released in the formation of "new" bonds is greater than the Sum of the energr required to break the "old" bonds. Consequently. one can expect that this reaction w0uld be moderately exothermic. i.e.. release heat energy to its surroundings. 2.R a tion ofaw aci with aw ak a : nude rate moderate 3t To "9 weak bond bond ho nd bond 2/ it it t H-CH3C00 + NH4-DH m5- H-UH + NH4-CH3COU In this case. while the net energ/ released by the formation of "new" bonds is still greater than the amount of energy required to break the "old" bonds. the difference is not so great as was the case of the strong acid and the strong base. Hence. while this neutralization reaction would be exothermic. the amount of heat energy released would be less than in the first reaction. 3. Reaction of a weal; acid with a strong base: moderate week at r0 no weak ho nd bond bond bond t i: L H-CHSCDU + Na—UH——-—Iv H—DH + Na—CH3CUD The latter neutralization should yield some amount of heat energy between that released in the first reaction and that released in the second reaction. The same result would be ex'pecteo if one reacted a strong acid with a weak base. #18 AHneut for Acid Base Rxn's, page 4 Pre-lab question it 1 (answer on answer sheet) Write a molecular. a complete ionic. and a net ionic equation for each of the following acid-base neutralization reactions: (at HN03 + NaOH --~> (b) HCI + NaOH ---> (C) CH3COOH + NaOH ---> (d) HCl + NH40H --—> (e) CH3COOH + NH40H ---> t _ In this exercise. you will be called upon to react a variety of acids and bases with one another. some weak and others strong and in the process. not only learn about calorimetry. but also to confirm the previous predictions about such reactions Pro-lab question #2 (answer on answer sheet) Based upon the notion of weak and strong acids and bases. which of the folloudng three reaction would you expect to produce the largest Aliment? The smallest Aliment? (a)HCl +NaOl-l -~—> (blNH40l-l + HNO3 ---> (c)NH40H + CH3COOH ---> You will carry out the acid base neutralizations in a Styrofoam cup with a plastic lid. Since the bottles containing the acid and base solutions will be left under the hood overnight. the initial temperatures of both solutions should be the same. Into the cup. you will place one reagent (either the acid or the base). measure the temperature. add the second reagent. and record the new temperature as the reaction concludes. ‘ #18 AKncut for Acid Base Rm’s. page 5 The reacting acid and base molecules. which are ‘the system.“ will react and “the system“ will loose heat energy to “the surroundings.“ In this case. 'the surroundings' will be the water‘ and the Styrofoam cup. Thus. the value of AHngut must be the sum of {H.Eabsorbed by the water} + (H. E. absorbed by the cup]. H.E. released _ HE. absorbed _ H.E absorbed b Rxn ‘ In Water " b Calorimeter aHmm' = [massl'isphtlh‘ll 1+ [(cal.const}(AT)] Heat Energy to Cup a... ___ ...._- As indicated in the preceding diagram. the amount of heat energy absorbed by the water can be calculated by multiplying the mass of the water by its specific heat and that value by the change in temperature: 1»; ADC gm x woe (mass) (Sp. heat) (AT0) The amount of heat energ)r absorbed by the calorimeter can be determined by multiplyin a calorimeter constant" by the Chan e in temperature; g g C31 ..._.... - O o )5. AC (cal. const) (ATOC) Since energy is neither created nor destroyed in a chemical reaction. all the heat energy produced by the neutralization reaction will be absorbed by the water and the cup. Pre~lab question #3 (answer on answer sheet) Calculate the number of calories required to change the temperature of SO-gm of water from 21.100 to 293°C ' In a “cofiee-cup' calorimeter. the solvent of the reactants is the water of the calorimeter. There is no separation as is the case in a 'bomb' calorimeter. " Also known as the “Heat Capacity" of the cup; the number of calories of heat energy absorbed per 1°C rise in temperature. ~ #18 Afineut for Acid Base Rxn's. page 6 However. before you can determine the various values for Aliment. it will be necessary to determine the value of the calorimeter constant (or heat capacity) of your coll'ee-cup calorimeter. Since no two Styrofoam cups are exactly alike. the value of this calorimeter constant must be experimentally determined for each cup. and that deter- mination will be the first task in this lab exercise. v Investigative Question #1: What is the calorimeter constant for myStyrofoarn Cup? Strategy: In essence. the calorimeter constant. with units of cal/00. is ameasure of how many calories of heat energy the cup absorbs with each 1°C rise in temperature of the cup. As the temperature of the'liquid in the Styrofoam cup increases. so too does the temperature of the cup lining rise. hence some of the heat energy from the liquid must be absorbed by the Styrofoam. if you were to mix together equal quantities of hot and cold water. you might casually expect that the resulting temperature would be midway between the two initial temperatures. However. if you were to pour 50-ml of water at 60°C into a cup containing 50-ml of cold water at 10°C. you would find that the final temperature would be less than the midpoint of 35°C. Why? Because some of the heat emery lost by the hot water would be used to warm the cup instead of being used to warm the cold water. One can visualize the situation as follows: Y ::—~—~—"' T°Hot Wate'r H.E. Lost T by Hot Water ’5' H T°Finel H.E. “stolen” H.E. Gained by by Cup ATC Cold Water _“ T°Cold Water HE. Lost by HE. Gained bg + HE. Stolen Hot Water Cold Water in Cup k mass) (sphtluTHfl = [(mass} (sp.ht)(aTc)] + [teal constXaTQ] Notice that there are eight terms in the above equation. In mixing together hot and cold water. you know. experimentally control. or can measure seven of these terms. leaving only the (cal const) as "unknown." Solving for the value of the calorimeter constant then becomes a relatively simple algebra problem. #18 AHneut for Acid Base Rxn's. page 7 Experimental Procedure: Obtainasmmfoamcup fromunderthehood. undamemmteterfrmtme stockroom. In a beaker, heat 300-400 of water to 60—65 °C. This will be the “Hot Water" Place 50ml of cold tap water into the Styrofoam cup, and measure the water temperamre. Be swe to allow sufficient time for the reading on the themmeter to but notso longatimethatooolingoowrs. Supportorholdthethemiometerinthecenterof the liquid in order to obtaina more consistent reading. Don't let the thermometer rest on the bottom of the Styrofoam cup. Measure out 50 milliliters of the hot water into a graduated cylinder and measure the ternperaaire of the hot water, once again allowing suflicient time for the reading on the thermometer to stabilize. Hold the tip of the themorneterm the center of the liquid Don't let the mennometerrestonmebottomofthegraduated cylinder. Pourthe hot waterinto thecup with the cold water. stir briefly with the thermometer, and then detem'u'ne the final temperature reached by the mixture. Ptorn the temperature data. detemtine the values of A791; and AT 03 1 for the emen'r‘nent, and from these values, calculate the the calorimeter constant far your particular mp. The spea'fic heat of water is 1 cal/gm-OC. While no two cups will have quite the same calorimeter constant, the values will generally range between 5 and 14 call 0C. Repeat this experimental detemtination until you obtain at least two fairly consistent values, La, not more than 2—3 cal/ 0C difference. Average these close values and use the result as your calorimeter constant You are now ready to experimentally detemtine the values of AHneut for various acid-base neutralizations. Investigative Question #2: I What is the values of Aline“; for various acid base neutralization reactions? Strategy: . Into the Styrofoam cup calorimeter for which you've already determined the calorimeter constant. you will place a meaSured amount of either an acid or base solution. Alter measuring the temperature of this solution. you will add an equal amoant of a reacting reagent. either acid or base. All solutions will be at the same initial temperature and the same concentrations. By measuring the final temperature of the liquid in the cup. you should be able to calculate the amount of heat energy released in the reaction: ‘ " Heat Energy released by rxn Heat Energy absorbed by + the water Heat Energy absorbed by the cu AH II as ut [(massXsphtX AU] + [(Cal. ConsOC AU] 1 Notice that the value ATC is used for both the cold water and the cup. #18 AHneut for Acid Base Rm's. page 8 Experimental Procedure: Measure outexactly 50.04731 ofthe 2.0 M acid whidayou wish toreact artdplace Uieacidmmecalorimeterlap}. Measureandrecordthetemperatureoftheadd. Place emctly50.0~mlofthe2.0Mbasewhid1youwishtoreactinagraduatedcylirider,andthen pourthebaseintotheadd. Sluflteacidaridbasetogemerwiththethemianeterandrecord fliehighesttemperamrereadiedbythenfimre. Besuretomldthetipofflzethexmmieterm thecenter of the mhmonasyoudid previously mdetemu'ntngthecalodmeterconstant. Using the data obtained, calculate the heat of neuu’ab‘zation (AHnwt) for the reaction ascanied out Tltetotalvolumeofsohltionmthecup is loo-ml. However, stnoetheliquid ts "salty water.“ the density is slightly greater than pure water, about 1.03 gm/ml. 171115 the mass ofliquid in the calorimeter will be 1013-an rather than IOOgm. Likewise, the somfic heat of a salt solution is slightly less than that of pure water, about 0.97 cal/gm OC. ~ Thus, your calailaiion for the amount of heat energy absorbed by the water should be; (vol) x (density) x (speafic heat) x (dmnge in T0), or = (100—ml)(1.03.gm/nw(o.97 cal /gm 00(13ch Pro-lab question #4 (answer on answer sheet) (a) A Styrofoam cup has a calorimeter constant of 9.8 call 0C. Ila neutralization reaction performed in this calorimeter causes the temperature to rise from 223°C to 365°C. How many calories of heat energr were absorbed by the cup? ' (b) In this same calorimeter. a neutralization reaction was carried out wherein the mass of the solution was loo-gm with a spwific heat of 1.03 call gm C’C. If the temperature rose from 22.300 to 365°C. calculate the number of calories of heat energy absorbed by the water. (c) What was the diluent of the reaction? (d) If 50.04111 of 2.0 M acid was used in the reaction. how many moles of acid were present? (e) What was the molar Aliment. in kcall mole. of the reaction ¢ Considering that one is using a Styrofoam cup as a calorimeter and a Student grade thermometer. such attention to small diii‘erences is obviously not significant - In fact. the increased density of the solution is counterbalanced by the lowered specific heat. Hawever. as a principle of good application, students should "play the game“ and incorporate these Values into their calculations. #18 Aline“: for Acid Base Rxn‘s. page 9 Carry out three different neiaralization reactions so as to react a [‘strong" + “strong’Y, a ['strong" + "weak'l. and a [‘weak’ + “weak? combination , Thus, you should ezpect to obtain three graduated values for the Mm. The strong acids available are: HCl (hydrochan 82; HNO3 (nitric). The weak acid available is CH3COOH, {acetic}. The strong base available is NaOH (sodium hydroxide) The weak base amilable is NH40H (ammonium hydroxide). (which is NH3 gas dissolved in water). \ Repeat each measurement until you obtain reasonably consistent results in each case. - . Once you have calailated the AHneutfor the reaction, then you want to calculate the mm W of reactant. Heats of neutralization are usually recorded as the amount of heat energy released mr mole of reactants in order to make broad cmzparisons. To make this calculation, you need only recall the following relationship; M: {71; . which can be rearranged to n = MVL Since you used 0.050-liters of 2.0 M solution in each case, the rumber of moles of acid would always be (0.050) x (2.0) = 0.1 moles, I which means that fined. _ W 0.1mole " 10me By subsfimo‘ng the values in this raaoproportion, you can calculate the molar Anew: for each of your esperimentally detem1ined reactions. Arrange these data into an organized table for easy visual comparison. Have you confimzed the original predictions? #18 Aanut for Acid Base Rm's. page 10 #18 Aline“; for Acid Base Rm‘s. page 1 1 PRE- LAB ANSWER SIEET: Pre-lab Question #1 Write a molecular. a complete ionic. and a net ionic equation for each of the renewing acid-base neutralization reactions: (a) HN03 + NaOH —--> (b)HCl + NaOH -—-> (C) CH3COOH + NaOH ---> (d)HC1 +NH401—I --.> (e) CH3COOH + NH40H ---> #18 AHneut for Acid Base Rxn's, page 12 Pre-lab Question #2: Based upon the notion of weak and strong acids and bases. which of the following reactions would you expect to produce the largest AHneut‘? The smallest Aliment? What kind of value of AHneut would you expect of the remaining reaction? (a)HC1 + NaOH ---> (b) NH40H + HNO3 --—> (C) NH40H + CH3COOH ---> Pro-Lab Question #3: Calculate the number of calories of heat energy required to change the temperature of 5.0-gm of water from 21.1 to 29.300. Pre-Lab Question #4: (a) A Styrofoam cup has a calorimeter constant'of 9.8 call 0C. If a neutralization reaction is performed in this calorimeter causes the temperature to rise from 223°C to 365°C. how many calories of heat energy were absorbed by the cup? _ #18 Aflncut for Acid Base Rxn's. page 13 (b) In this same calorimeter. a neutralization reaction was carried out wherein the mass of the solution was loo-gm with a specific heat of 1.03 call ngC. lithe temperature rose from 223°C to 365°C. calculate the number of calories of heat energy absorbed by the solution. to) What was the AHneut of the reaction? (at) If 50.0-ml of 2.0 M acid was used in the reaction. how many moles of acid were present? (6) What was the molar Aliment. in kcall mole. of the reaction? ...
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This note was uploaded on 02/06/2012 for the course BIO 210 taught by Professor Tricca during the Fall '11 term at Canada College.

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Heat_of_RX_Acids_Bases - #18 annem for Acid Base Rm's. page...

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