c) This is a problem about the enthalpy change in a chemical reaction. We use

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1 Chapter 2 Prblems Page 1 f (a) A hiker absrbs 1.00 L f water. It evaprates at 20 ºC. Hw much heat is required fr this prcess? (b) What wuld the temperature drp fr the hiker be (assuming the clthes and hiker have a mass f 60 kg and a heat capacity f water)? (c) Hw many grams f sucrse is needed t be metablized t accunt fr this amunt f heat? a) Evapratin is a phase change (liq gas), s we use the latent heat f vaprizatin f water at 20 ºC frm able 2.2. his is a prcess at cnstant pressure, s q = ΔH = where L V is the (latent) heat f vaprizatin and m is mass. P ml V he mass m = (1 L)( g/cm 3 )(1000 cm 3 /L) = g = kg hus the required heat is q P = ml V = ( kg)(2447 kj/kg) = 2443 kj b) his is a prblem in heat capacity, s qp = C P Δ = mcp Δ where C P is the heat capacity f the whle system and the value with the bar ver it is the specific heat capacity, that is the per unit mass r per mle value. Rearranging this equatin leaves qp 2443kJ Δ = = = 9.73K mcp (60kg)(4.184kJ / kg K) where the minus sign fr q P cmes frm the fact that heat is being remved frm the system. c) his is a prblem abut the enthalpy change in a chemical reactin. We use ΔH = ΔH f ΔH where the sums are f the enthalpies f frmatin f prducts prd react f and reactants. Fr the sucrse reactin, taking int accunt the numbers f mles f prducts and reactants, the cmbustin f ne mle f sucrse has an enthalpy change f ΔH = ( 11ml)( 241.8kJ + (12ml)( 393.5kJ (1ml)( 2222kJ (12ml)(0kJ = 5160kJ his is an exthermic reactin. prduce 2442 kj, the hiker wuld need t burn (2442 kj)/(5160 kj/ml) = 0.47 ml f sucrse. Given that the mlar mass f sucrse is g/ml, this wuld amunt t a mass f m = (#ml) X M.W. = (0.47 ml)(342.3 g/ml) = 162 g

2 Chapter 2 Prblems Page 2 f (a) As heat is added t ice at cnstant P, increases until the ice melts. After melting, cntinues t increase. Cmplete the plts fr q vs. and C P vs.. (b)as heat is added t a weak inic slutin, q is linear with. If DNA is added t the buffer, the DNA strands separate. Draw the q vs. and C P vs. curves. a) Until the ice melts, the heat added will be prprtinal t the f the ice. his is because C P is (very nearly) a cnstant and q = C P Δ. When the ice melts, the will arrest at the melting (0 ºC) until the sample is all liquid. hen the relatinship will be linear again, with a larger slpe since C P f water is larger than C P f ice. hat is enugh t describe the q vs. plt. Nte that C P = dq/d = slpe f the q vs. plt. hat is enugh t describe the C P vs. plt: q C P b) here is a phase change as hydrgen and ther nncvalent bnds are brken in the DNA. his ccurs ver a range f since nt every bnd has the same strength. his means that the heat required (beynd heating the slutin) will increase while the DNA is unwinding. he phase transitin will be brad, nt sharp as in a). he C P vs. plt is the derivative f the q vs. plt, s it will have a brad peak centered at the melting. We culd measure the heat f melting f DNA by integrating the C p vs. curve ver the range where the phase transitins ccurs, and subtracting the same integral fr pure buffer with n DNA: In the plts, the DNA slutin is shwn as slid, the plain buffer alne as dashed. q C P

3 Chapter 2 Prblems Page 3 f One ml f an ideal gas at 27 ºC and 1 atm is heated and expands reversibly at cnstant P t = 327 ºC. Given that C V = 20.8 J/K ml, calculate (a) the wrk dne n the gas, (b) ΔE and ΔH and (c) the heat absrbed. a) he expansin is at cnstant pressure, s w = -P ext ΔV. We need t cmpute V i and V f. V i = nr i /P = (1 ml)( L atm/ml K)(300 K)/(1 atm) = 24.6 L V f = nr f /P = (1 ml)( L atm/ml K)(600 K)/(1 atm) = 49.2 L S ΔV = V f V i = 49.2 L 24.6 L = 24.6 L and thus w = -P ext ΔV = -(1 atm)(24.6 L)(101.3 J/L atm) = kj, i.e. wrk is dne by the gas as it expands. b) Recall that, fr an ideal gas, ΔE = q P + w P = C P Δ + w p and C P = C V + nr (see eq. 2.31b and the discussin abve it). Frm this we can calculate C P and, with the answer t a) the value f ΔE. If yu lk at p. 43, eq. 2.32, yu see that this is all cmbined, with sme algebra, int a simpler relatinship: ΔE = C V Δ = (1 ml)(20.8 J/K ml)(300 K) = 6.24 kj. Nw, mving t ΔH, we recall that ΔH = ΔE + Δ(PV) = ΔH = ΔE + PΔV at cnstant P. his is easy t cmpute: ΔH = 6.24 kj kj = 8.73 kj Nte that bth ΔH and ΔE increased as expected (the temperature is higher). c) he heat int the system (the gas) fr this cnstant pressure situatin is q = ΔH = 8.73 kj. Again, heat must flw in t raise the and increase the V f the gas it makes sense.

4 Chapter 2 Prblems Page 4 f Indicate fr the fllwing prcesses whether q, w, ΔE and ΔH are psitive, negative r zer. (a) w cpper bars, ne at 80 ºC and ther at 20 ºC are brught int cntact in an insulated cntainer; (b) a sample f liquid in an insulated cntainer is stirred by a mechanical linkage t an external mtr; (c) a sample f H 2 gas is mixed with an equimlar amunt f N 2 gas at cnstant and P where n chemical reactin ccurs. a) he system is the tw bars. Since the system is in a thermally islated cntainer, n heat flws t r frm the system, s q = 0. he vlume f the ht cpper bar will decrease and that f the cl bar increase, s the net change will be zer, thus w = 0. Because q and w are zer, ΔE is zer. Since ΔE = 0 and PΔV = 0, it must be that ΔH = ΔE + PΔV = 0. b) As in a), the system is thermally islated, s q = 0. he mechanical linkage allws wrk (stirring) t be dne n the liquid, s w > 0. ΔE = q + w > 0 since q = 0 and w > 0. his is a liquid, s Δ(PV) is negligibly small. hus, ΔH = ΔE + Δ(PV) > 0. c) he wrding suggests that the tw gases are ideal. hus, when they mix there is n change, and thus q = C P Δ = 0. here is n vlume change (the sum f the vlumes des nt change), s there is n wrk dne. hat means that ΔE = q + w = 0. Likewise, it must be that ΔH = ΔE + PΔV = 0.

5 Chapter 2 Prblems Page 5 f g f liquid water at 55 C is mixed with 10 g f ice at -10 C. P is cnstant and n heat leaves the system. Calculate the final temperature. here is much mre ht water than cld ice, s it is likely that the ice will melt cmpletely. S let s assume that and calculate the heat needed t raise the temperature f the ice t 0 C. his is a simple heat capacity prblem: q icewarm = mcδ = ( 10g)(2.113kJ / K kg)(273k ( 263k))(1kg /1000g) = kJ Remving that much heat frm the 55 C liquid water will change its temperature (anther heat capacity prblem) by Δ = qcl / mc = kJ /(0.1kg)(4.18kJ / K kg) = K. hat is, the new liquid water temperature is lwer by degrees, r it is C. Nte that the heat added t the ht water is negative. It flws int the clder ice t melt it. S nw we have 10 g f ice at 0 C and 100 g f liquid water at C. he ice will melt, and t d that requires heat fr melting, fr which we need t use the mass f ice and its latent heat f melting: qmelt = mlm = ( 10g)(333.4kJ / K kg)(1kg /1000g) = kJ As befre, this heat must cme frm the warm water, s its temperature will drp nce again: Δ = qcl / mc = 3.334kJ /(0.1kg)(4.18kJ / K kg) = K S the temperature f the ht water drps an additinal degrees t C. At this pint we have tw masses f liquid water at tw different temperatures. hey will reach the same final temperature, and t d s they the ht batch will give heat t the cld batch. hus, we can equate (with a minus sign) the heat capacity heat fr each pl f water: m100 CΔ 100 = m10 CΔ10 r ( 100g) C( f C) = (10g) C( f 0 C) he specific heat capacity cancels n bth sides, s we slve fr the final : ( 100g)( f ) + (10g)( f ) = (10g)(0 C) + (100g)(46.52 C) r f = ( 100g)(46.52 C) /(110g) = C. his makes sense: the final result is fairly clse t the riginal temperature f the much mre massive liquid water.

6 Chapter 2 Prblems Page 6 f Crps can yield 20 kg f sucrse per hectare per hr. in sunlight f 1 kw/m 2. (a) Using standard enthalpies f frmatin, calculate the enthalpy f prductin f sucrse frm CO 2 (g) and H 2 O(l), (b) calculate the energy needed t synthesize 20 kg f sucrse and (c) cmpute the efficiency f cnversin f sunlight t sucrse. a) Lk up the enthalpies f frmatin f the reactin cmpnents frm the appendices. Be careful abut the phases (liq, gas, slid). CO 2 (g): kj/ml H 2 O(l): kj/ml Sucrse(s): kj/ml O 2 (g): 0 kj/ml he enthalpy change in ne mle f the reactin is ΔH = ( 1ml)( 2222kJ + (12ml)( 0kJ (11ml)( 285.8kJ (12ml)( 393.5kJ = 5644kJ Nte that the reactin requires energy t prceed. b) he number f mles f sucrse in 20 kg is n = (20 kg)(1000 g/kg)/(342.3 g/ml) = 58.4 ml, s the ttal enthalpy change fr the prductin f 58.4 ml f sucrse is ΔH = (58.4 ml)(5644 kj/ml) = 330 MJ. his is the amunt prduced in ne hectare in ne hr. c) In ne hectare (10,000 m 2 ), the energy that is available frm sunlight in ne hur is E = (1 kw/m 2 )(1 hr)(10 4 m 2 )(1000 W/kW)(1 J/sec W)(60 sec/min)(60 min/hr) = 3.6 X J = 36,000 MJ. S the efficiency e = (energy captured)/(energy input) = (330 MJ)/(36,000 MJ) = , pretty lw efficiency.