Sessin #22: Hmewrk Slutins Prblem #1 (a) In the cntext f amrphus inrganic cmpunds, name tw netwrk frmers, tw netwrk mdifiers, and ne intermediate. (b) Sketch the variatin f mlar vlume with temperature fr pure silica. Shw glass frmatin at tw different cling rates. Als shw the crystallizatin prcess. On each cling curve, label the melting pint r the glass transitin temperature. (c) What are tw key factrs that determine whether a material will slidify as a glass r a crystal? Slutin (a) frmers: SiO 2, P 2 O 5, B 2 O 3, GeO 2 mdifiers: CaO, MgO, Na 2 O intermediates: Al 2 O 3, TiO 2, ZnO (b) (c) One factr is hw extreme is the change in atmic arrangement required upn transfrming liquid int slid, i.e., cmplexities f bth structures. Anther is the cling rate/viscsity f the melt. Prblem #2 (a) Draw the netwrk structure f a brate glass. (b) Explain hw the additin f Na 2 O t B 2 O 3 decreases viscsity f the glass melt. (c) T raise the glass transitin temperature f the brate glass, d yu increase r decrease the cling rate? Explain. (d) Describe ne surface treatment methd by which yu can strengthen a brate glass. (e) Explain why brate glass is transparent t visible light.
Slutin (a) Curtesy f Jhn Wiley & Sns. Used with permissin. (b) Na 2 O is a mdifier. It dissciates t frm Na + catins and O 2- anins. The xide anins attack the bridging -O- xygens in the brate netwrk thereby disrupting it. The resulting smaller brate units are able t mve mre easily that the antecedent larger units. This is manifest in the lwer value f viscsity fr the mdified melt. (c) Increase. As high cling rates there is nt enugh time fr the melt cnstituents t rearrange themselves befre the temperature has fallen t the pint at which the existing structure is rendered immbile. (d) One methd is in exchange. Immerse the glass in a mlten salt cntaining alkali catins that are larger than thse present in the glass. The cncentratin gradient in such catins will drive them t diffuse int the glass where they will substitute fr smaller ins. This results in the generatin f cmpressive stress in the surface and raises the level f stress needed t break the glass A secnd methd is tempering. Subject the glass t surface cling by air jet. This leads t differential cling rates: the surface regin is cling at a higher rate than the interir. If yu lk at a plt f V vs. T, the final vlume f the fast-cled surface layer will be greater than that f the slw-cled interir. This results in the generatin f cmpressive stress in the surface and raised the level f stress needed t break the glass. (e) Sme reasns fr transparency: cvalent bnds; tightly bund electrns; high band gap energy, ~5 ev; n dpants that wuld intrduce dnr r acceptr levels, e.g. the Hpe diamnd. Prblem #3 The decay rate f 14 C in living tissue is 15.3 disintegratins per minute per gram f carbn. Experimentally, the decay rate can be measured t ±0.1 disintegratins per minute per gram f carbn. The half-life f 14 C is 5730 years. (a) What is the maximum age f a sample that can be dated and what is the uncertainty assciated with this measurement? (b) What is the minimum age f a sample that can be dated and what is the uncertainty assciated with this measurement?
Slutin (a) Radiactive decay is a 1 st rder reactin which can be mdeled as: dc = kc r c = c e -kt With a little algebra we can get an expressin fr the relatinship between time, t, and the instant value f the decay rate. At any time, t, we can write and at time zer, dc -kt = kc = kc e (1) dc = kc (2) Divide eq. 1 by eq. 2 t get r r t = e dc where t reduce clutter let r =. ln 2 Take the lgarithm f bth sides f eq. 3 and substitute k =. t With rearrangement, this gives -kt 1/2 (3) t1/2 rt t = - ln (4) ln 2 r S, fr the ldest specimen we wuld measure the minimum instant decay rate f 0.1 ± 0.1 disintegratins per minute per gram. Set this equal t r t in eq. 4 and slve fr t t get 41585 ± 5730 years. (b) Fr the yungest specimen we wuld measure the maximum instant decay rate f 15.2 ± 0.1 disintegratins per minute per gram. Set this equal t r t in eq. 4 and slve fr t t get 54 ± 54 years. Prblem #4 Slutin Nte that althugh the experimental uncertainty remains cnstant at 0.1 disintegratins per minute per gram, because this is cmpared against a measurement that can vary frm 15.3 dwn t undetectable, the derived uncertainty in time varies greatly frm 54 years t 5730 years, respectively. What is the activatin energy f a prcess which is bserved t increase by a factr f three when the temperature is increased frm rm temperature (20 C) t 40 C? 1 -E A RT 1 k = Ae ; 2 1 -E A RT 2 k = 3k = Ae 1 = 3 e EA 1 1 - R T 1 T 2
EA 1 1 ln 3 = R T1 T2 R ln 3 4 E A = = 4.19 10 = 41.9 kj/mle 1 1 293 313 Prblem #5 A first-rder chemical reactin is fund t have an activatin energy (E A ) f 250 kj/mle and a pre-expnential (A) f 1.7 x 10 14 s 1. (a) Determine the rate cnstant at T = 750 C. (b) What fractin f the reactin will be cmpleted at 600 C in a perid f10 minutes? (c) At what temperature will the reactin be three times as fast as at 750 C? Slutin 2.5 10 5 E - A - 14 23-1 (a) k = Ae RT = 1.7 10 e 8.31 10 = 28.8 s (b) Requires knwledge f k 600 : 600 2.5 10 5-14 8.31 873 k = 1.7 10 e = 0.184 c c kt 0.184 600-48 = e = e = 1.3 10 0 c = 0 means the reactin is essentially 100% cmplete. (c) k = 3 k = 3 28.8 = Ae 1 E A - RTx 14 EA ln 86.4 = ln(1.7 10 ) - RT A x 14 T = E ( ) R ln 86.4 - ln(1.7 10 ) x x Prblem #6 T = 1063 K = 790 C Whiskey, suspected t be f the mnshine variety, is analyzed fr its age by determining its amunt f naturally ccurring tritium (T) which is a radiactive hydrgen istpe ( 3 H) with a half-life f 12.5 years. In this shine the activity is fund t be 6% f that encuntered in fresh burbn. What is the age f the whiskey in questin?
Slutin c = e kt c c ln kt 0.06c = ; c = 0.06c x ln 0.06 = -kt x ln 0.06 ln 0.06 t x = - = = 50.7 years ln 2 0.693 t 12.5 1/2
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