Chem 115 POGIL Worksheet - Week #7 First Law, Enthalpy, Calorimetry, and Hess s Law

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1 Chem 115 POGIL Wrksheet - Week #7 First Law, Enthalpy, Calrimetry, and Hess s Law Why? In additin t mass changes, chemical reactins invlve heat changes assciated with changes in the substances internal energy. Like mass-based stichimetry, these changes are quantitative. One f the mst imprtant physical relatinships gverning energy change is the First Law f Thermdynamics. Mst ften we will cnsider this in terms f a thermdynamic functin called enthalpy. Knwing abut the First Law and enthalpy is essential t understanding the relatinship between heat change and chemical reactins. Knwing hw much heat is gained r lst in a chemical r physical prcess is imprtant in many real-life applicatins, such as determining the calric cntent f fds, the heat ptential f fuels, and the heat required r liberated t prduce useful materials. The heat assciated with a prcess is an extensive prperty (related t amunt) and ften can be measured with an apparatus called a calrimeter. But in many cases, it is impractical (r dwnright dangerus!) t measure the heat f a certain prcess in a calrimeter. Frtunately, internal energy and enthalpy are state functins. This means their values depend slely n the physical state f the system, and nt n hw that state was reached. As a result, we can determine the enthalpy f a prcess frm the values f any series f steps that add t give the desired verall prcess. In ther wrds, we d nt always need t measure heat change directly. Learning Objectives Knw the First Law f Thermdynamics Understand the relatinships between heat, wrk, internal energy, and enthalpy Understand the cncepts f heat capacity, mlar heat capacity, and specific heat Understand the principles f calrimetry Understand Hess s Law and its use fr calculating reactin enthalpies Success Criteria Be able t calculate heat and temperature changes Be able t apply the First Law f Thermdynamics Be able t calculate heat r enthalpy frm calrimeter data Be able t use Hess s Law t calculate reactin enthalpies Prerequisite Have read sectins 5.1 thrugh 5.6 in the text

2 Infrmatin (First Law f Thermdynamics) Energy is the ability t d wrk r transfer heat. Wrk is the transfer f energy frm ne bdy t anther. In a sense, wrk is energy in the prcess f transfer. This assciatin between wrk and energy allws us t define a unit f energy as that quantity transferred when a unit f wrk is dne. In ther wrds, energy and wrk have the same units. The SI unit f energy is the jule (J),which is a derived unit defined as fllws: 2 2 jule (J) kg m s Because the SI unit f frce, the newtn, is defined as 2 newtn (N) = kg m s we can als think f the jule as being a newtn-meter: jule (J) = kg m s (kg m s )(m) = (newtn)(m) Systems f chemical interest usually cnsist f certain amunts f substances underging physical r chemical change in a cntainer f sme type. The substances themselves cnstitute the system. Everything else, such as the cntainer r the immediate labratry envirnment, is cnsidered part f the surrundings. The energy f either the system r surrundings can take many frms: thermal (heat), radiant (light), chemical, mechanical, electrical. But all f these frms f energy can be viewed in terms f equivalent amunts f kinetic energy (K, energy f mtin) and ptential energy (U, energy f psitin r cmpsitin). Thus, the ttal energy f a system may be defined as the sum f its kinetic and ptential energies. E t = K + U This energy is called the internal energy, E, f the system. As the system underges chemical r physical change, the frms f its energy may change, and sme energy may be gained frm r lst t the surrundings, but energy is neither created nr destryed. This idea is the essence f the First Law f Thermdynamics: Energy can be transferred frm ne bject t anther, and its frms can be intercnverted, but energy can neither be created nr destryed. Cnsistent with the First Law f Thermdynamics, the energies f the system and surrundings can change nly if they exchange energy with each ther. Thus, if the energy f ne increases, the energy f the ther decreases by the same amunt, r vice versa, maintaining a cnstant ttal energy.

3 In general, the ptential and kinetic energies f a system, which make up its internal energy, cannt be evaluated n an abslute scale. Hwever, we can bserve and measure changes in the magnitude f the internal energy, ÄE, defined as 1 ÄE = E final E initial = E f Ei A system underging chemical r physical change can change its internal energy by transferring heat (q), ding wrk (w), r bth: ÄE = q + w If the system receives heat frm its surrundings, the sign n q is psitive (q > 0); if it gives ff heat the sign n q is negative (q < 0). Likewise, if the surrundings d wrk n the system, the sign n w is psitive (w > 0); if the system des wrk n the surrundings, the sign n w is negative (w < 0). An example f a system ding wrk wuld be the expansin f a gas against a cnstant external pressure. This might be the result f a gas-prducing chemical reactin. A simple example f heat transfer ccurs when a ht bject, such as a piece f metal, is placed in rm-temperature water. If we cnsider the metal t be the system, as it transfers heat t its surrundings, the water, the temperature f the metal will g dwn and the temperature f the water will g up. After sme time, a cnditin f thermal equilibrium will ccur, at which pint the temperature f bth the metal and the water will be the same. This illustratin raises the fllwing general pints abut heat flw: (a) The sense f heat flw is always frm the htter bject t the cler bject. (b) Thermal equilibrium ccurs when bth system and surrundings are at the same temperature. (c) The quantity f heat lst frm the htter bject is equal t the heat gained by the cler bject. The heat transferred between the system and surrundings represents a change in heat cntent f the system. If a system takes up heat frm its surrundings, its heat cntent will be higher at the end f the prcess. Cnsequently, q will be psitive. Such a prcess is endthermic. If the system gives ff heat t its surrundings, its heat cntent at the end f the prcess will be less. 1 The symbl Ä (Greek delta) indicates a change in sme prperty. We will see ther changes in prperties, indicated by delta functins, which will always be defined as the difference between the final value minus the initial value; e.g., Äx = x f x i. This rder f cmparisn (final minus initial) is crucial t imparting the crrect sign n the value f the change.

4 Cnsequently, q will be negative. Such a prcess is exthermic. If a prcess is endthermic in ne directin, it is exthermic in the ppsite directin, and vice versa. Key Questins 1. What are the tw majr ways in which the internal energy f an bject can be categrized? Hw d these ways differ frm ne anther? 2. If a system lses heat, where des it g? 3. Describe the fllwing prcesses as exthermic r endthermic: a. an ice cube melts n a warm surface b. water freezes c. calcium chlride is mixed with water, resulting in a very ht slutin d. ammnium nitrate is mixed in water, resulting in a very cld slutin Infrmatin (Enthalpy) We will ften be interested in the heat cntent under cnstant pressure cnditins, called the enthalpy, H, f the system. Like internal energy, H cannt be measured directly, but we can measure the change in H, defined as ÄH H f H i = qp where H f and H i are the final and initial heat cntent f the system, respectively, under cnstant pressure; and q P is the heat transferred under cnstant pressure cnditins. Under cnstant pressure cnditins, we can rewrite the defining equatin fr internal energy as frm which it fllws ÄE = q P + w = ÄH + w ÄH = ÄE w Nte that if n wrk is dne (w = 0), ÄH = ÄE. This means that the enthalpy change is the internal energy change when n wrk is dne. The amunt f heat transferred in a chemical r physical prcess depends upn (1) physical cnditins, (2) amunts f substances, and (3) directin f change. The physical cnditins n which enthalpy depends are temperature (T), pressure (P), and physical state (i.e., slid, liquid, gas, aqueus slutin). Therefre, we must be careful when quting values f q r ÄH t be sure t specify these cnditins. T avid cnfusin, we define Standard cnditins: T = 25 C; P = 1 atm = 760 mm Hg = in. Hg; all substances in their usual states fr these cnditins (the standard state).

5 Inasmuch as physical state is imprtant, it is cmmn practice t nte the states (s, l, g, aq) alng with the frmulas f each chemical cmpund in a thermchemical equatin. A thermchemical equatin is an rdinary chemical equatin written in cnjunctin with a thermchemical value (e. g., q, ÄH), where the reactant and prduct cefficients are taken as numbers f mles. When the value f the enthalpy is quted fr the reactin ccurring under standard cnditins, it is called the standard enthalpy, symblized ÄH. Nte the differences in the fllwing thermchemical reactins, due t subtle changes in cnditins: Standard cnditins: H 2(g) + ½O 2(g) H2O(l) ÄH = 286 kj/ml Nn-standard cnditins, T = 25 C: H 2(g) + ½O 2(g) H2O(g) ÄH = 242 kj/ml Nn-standard cnditins, T = 50 C: H (g) + ½O (g) H O(g) ÄH = 241 kj/ml Enthalpy depends upn the amunt f substance; i.e., it is an extensive prperty. Since thermchemical equatins are read with the understanding that the cefficients are numbers f mles, changing the cefficients by multiplying thrugh the equatin als multiplies the value f the enthalpy. Cmpare the fllwing: H 2(g) + ½ O 2(g) H2O(l) ÄH = 286 kj 2 H (g) + O (g) 2 H O(l) ÄH = 572 kj Reversing the directin f a reactin reverses the sense f the heat transfer. As we have seen, an exthermic prcess in ne directin will be an endthermic prcess in its reverse directin. The heat transferred is the same in bth cases, but the directin changes, as indicated by a sign change n the value f q r ÄH. Fr example H 2(g) + ½ O 2(g) H2O(l) H O(l) H (g) + ½ O (g) ÄH = 286 kj/ml ÄH = +286 kj/ml Exercises 4. The thermchemical equatin fr the burning f ne mle f benzene under standard cnditins is C H (l) + 15/2 O (g) 6 CO (g) + 3 H O(l) H = kj 6 6 cmb a. Is this reactin exthermic r endthermic? b. Hw much heat is released when a 5.00-g sample f benzene is burned in excess xygen under standard cnditins? (m.w. C H = u) Infrmatin (Heat Capacity, Mlar Heat Capacity, and Specific Heat) 6 6 The heat capacity, C, is the amunt f heat, q, required t raise the temperature, ÄT, f an bject by 1 C. The three variables are related by the equatin q = CÄT

6 The value f C in this equatin, and likewise the magnitudes f q and ÄT, pertain t a certain sample and depend n the amunt. Fr example, if we cmpare a teaspn f water with a swimming pl f water, the heat capacity f the swimming pl water is vastly larger. Hwever, n a per mle r per gram basis, the heat capacities are the same. Thus, fr pure substances, the heat capacity is usually defined n the basis f a ne mle r ne gram sample. The mlar heat capacity, C m, is the heat capacity per mle f substance, and the defining equatin can be written q = ncmät where n is the number f mles in the sample. Likewise, the specific heat, C s, is the heat capacity per gram f substance, and the defining equatin can be written q = mcsät where m is the mass f substance in grams. In ther wrds, fr a specific sample f a pure substance, C = nc m = mc s. In using these relatinships, realize that a change in temperature expressed in C is the same value if expressed in K units. Exercises 5. The specific heat f irn (at. wt. = u) is J/g K. What is the heat capacity f a 23.5-g blck f irn? What is the value f the mlar heat capacity f irn? 6. Hw many jules f heat are required t raise the temperature f a 23.5-g blck f irn frm 25.0 C t 96.2 C? Infrmatin (Calrimetry) Heat absrbed r liberated by a chemical reactin can be determined with a calrimeter. The heat f the reactin is nt directly bserved, but rather its effect n an amunt f water surrunding the reactin vessel in the calrimeter is measured. The heat capacity f the calrimeter, C cal, can be experimentally determined r calculated, and with this and the bserved temperature change the heat change f the calrimeter can be calculated: q = C ÄT cal The heat change experienced by the calrimeter is equal in magnitude but ppsite in sign t that fr the reactin that caused it; i.e., q = q rxn Thus, if the calrimeter experiences an endthermic heat change, q cal > 0, characterized by a rise in its temperature, this must have been the cnsequence f an exthermic reactin, fr which q rxn < 0 (negative). Cnversely, if the calrimeter experiences an exthermic heat change, q < 0, cal cal cal

7 characterized by a fall in its temperature, this must have been the cnsequence f an endthermic reactin, fr which q > 0 (psitive). rxn th A simple calrimeter cnstructed frm Styrfam cffee cups (see Fig in the 12 editin f yur bk), such as yu will use in the labratry, measures reactin heats under cnstant pressure cnditins; thus, q rxn = ÄH rxn, the change in enthalpy f the reactin. This is ften used t measure the heat change f a slutin frmed in the inner cup. The specific heat f the slutin is generally assumed t be the same as that f pure water, J/g K. The heat capacity f the calrimeter is calculated as the prduct f the mass f the slutin times th J/g K.. A bmb calrimeter (cf. Fig in the 12 editin f yur bk) measures heats under cnstant vlume cnditins; thus, q rxn = ÄE rxn, the change in internal energy f the reacin. In mst cases, the difference between ÄH and ÄE is small and is ften ignred. Exercises 7. A 2.58-g sample f KNO 3 (f.w. = u) was added t g f water in a cffee-cup calrimeter. The initial temperature f the water was 22.5 C, and the temperature f the slutin after mixing was 20.4 C. On the basis f this experiment, what is the heat f slutin per mle f KNO [f.w. = u]? The specific heat f water is J/g K The cmbustin f 1.50 g f glucse, C6H12O 6 (m.w. = 180.0), caused the temperature f a bmb calrimeter t rise frm C t C. The calrimeter had a heat capacity f 4.42 kj/k. Calculate the heat f cmbustin f ne mle f glucse: C H O (s) + 6 O (g) 6 CO (g) + 6 H O(l) q =? rxn Infrmatin (Standard Cnditins, State Functins and Hess's Law) Remember that the measured value f ÄH depends n the states f all reactants and prducts (s, l, g, aq) and the temperature and pressure under which the reactin ccurs. Thus, as nted abve, it is useful t define a set f standard cnditins, defined as T = 25 C, P = 1 atm, and all substances in their usual states fr these cnditins (the standard state). The standard state f an element is its mst stable state under standard cnditins; e.g., H 2(g), C(s) graphite, S 8(s), P 4(s). Fr cmpunds, the standard state is the mst prevalent state under standard cnditins; e.g., H O(l), CO (g), C H (g), C H (l) Enthalpy is a state functin, which nly depends upn current cnditins (the state f the system) fr its value, nt n hw the current state was reached. As applied t ÄH = H f H i, the value f the enthalpy change fr any prcess depends nly n the difference between the final and initial states, nt n the path taken t get frm the initial t the final state. This means that any set f steps, whether real r imagined, that takes the system frm the initial state t the final state f interest will have a sum f ÄH values fr all the steps that is identical t the value f ÄH fr the verall prcess if dne directly. This principle, called Hess's Law f Cnstant Heat Summatin, was first established by G. H. Hess in 1840: The enthalpy change fr a reactin is independent f path.

8 In applying Hess's Law, a set f given thermchemical equatins is manipulated such that they add t give a balanced thermchemical equatin fr the prcess f interest (the target equatin). In ding this, whenever a given thermchemical equatin is multiplied (usually by an integer, but smetimes by a ratinal fractin such as ½, a, etc.), its ÄH is likewise multiplied. Whenever the directin f a given thermchemical equatin is reversed, its ÄH value changes sign. Fr example, suppse we wish t calculate ÄH fr the reactin, H2O(l) + CO(g) CO 2(g) + H 2(g) given the fllwing tw thrmchemical equatins: (a) H 2(g) + ½ O 2(g) H2O(l) (b) CO(g) + ½ O (g) CO (g) 2 2 ÄH = kj ÄH = kj If we reverse equatin (a) and keep equatin (b) as given, we can add t get the desired verall reactin (the target equatin): Exercises ( a) H2O(l) H 2(g) + ½O 2(g) ÄH = kj (b) CO(g) + ½O 2(g) CO 2(g) ÄH = kj H O(l) + CO(g) CO (g) + H (g) ÄH = +2.9 kj 9. Calculate ÄH fr the reactin, Given: CS 2(l) + 2 H2O(l) CO 2(g) + 2 H2S(g) (a) CS 2(l) + 3 O 2(g) CO 2(g) + 2 SO 2(g) (b) H S(g) + 3/2 O (g) H O(l) + SO (g) 2 ÄH = kj ÄH = kj 10. Calculate ÄH fr the reactin, Given: 2 NH 3(g) + 3 N2O(g) 4 N 2(g) + 3 H2O(l) (a) 4 NH 3(g) + 3 O 2(g) 2 N 2(g) + 6 H2O(l) (b) N2O(g) + H 2(g) N 2(g) + H2O(l) (c) H (g) + ½ O (g) H O(l) ÄH = kj ÄH = kj ÄH = kj

9 11. The heat f cmbustin fr a cmpund cmpsed f C and H, pssibly with either r bth O and N, is the heat liberated when ne mle f the substance is burned with the stichimetric amunt f O (g) t prduce, CO (g), H O(l), and N (g), as may be apprpriate. 2 i. Write the balanced equatin fr the cmbustin f ne mle f C2H 6(g), which wuld be the basis fr the thermchemical equatin defining the heat f cmbustin f ethane. ii. Calculate the heat f cmbustin f C2H 6(g) frm the fllwing data: (a) C2H 2(g) + 2 H 2(g) C2H 6(g) (b) H 2(g) + ½ O 2(g) H2O(l) (c) 2 C(s) + H 2(g) C2H 2(g) (d) C(s) + O (g) CO (g) 2 2 ÄH = kj ÄH = kj ÄH = kj ÄH = kj