Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia) The pressure may be as high as 9 bar. Express the pressure f 9 bar as: A. KPa B. Atm C. mm Hg D. PSI 2. A. (6 pints)i have a rigid vessel filled with gas at a 1 atm pressure and a temperature f 25 C, what is the pressure in the vessel when the temperature is drpped t -25 C? P and T change T must be in K s 25+273 = 298K, -25+273=248K B. (6 pints) I have a rigid vessel filled with a gas at a pressure f 1.6 atm what percent f the mlecules in the vessel have t remved t get the pressure t drp t 0.25 atm. n and P change Let s make n 1 a nice easy number like 100% S if we start with 100% f mlecules, 15.6% remain, and we had t remve 100-15.6 r 84.4% f the mlecules. 1
3. (12 pints)what s the equatin...? t get mlar mass f a gas frm its density t get the vlume f a gas at STP At STP, 1 mle f any gas has a vlume f 22.42 L t get energy frm heat and wrk E=q+w t get the change in enthalpy frm heat released at a cnstant pressure ÄH=q P t get enthalpy f a reactin frm enthalpy f frmatin values 0 0 ÄH = (n ÄH ) - (n ÄH ) RXN P F Prduct R F Reactant t get enthalpy f a reactin frm bnd energies ÄH = Bnds Brken in reactants - Bnds frmed in prducts RXN 4. I have islated a new element called Zecfusdium. It has an atmic mass f 125 g/mle. If it takes 1000 J f heat energy t raise the temperature f 1 gram f Zecfusdium by 5K A. (6 pints)what is the Specific Heat Capacity f Zecfusdium? Energy = Specific Heat Capacity (SHC) grams ÄT SHC = Energy /(grams ÄT) = 1000J/(1 gram 5K) =200 J/g K B. (6 pints)what is the Mlar Heat Capacity f Zecfusdium? Energy = Mlar Heat Capacity (MHC) mles ÄT MHC = Energy /(mles ÄT) Mles = 1 g / (125 g/ml) =.008 mle MHC= 1000J/(.008 mle 5K) =25000 J/ml K r 25 kj/ml K 2
5.(12 pints) What s the difference between An extensive and an intensive prperty? An extensive prperty depends n the amunt f material is a system, but an intensive prperty is independent f the amunt f material. A path functin and a State functin? A path functin depends n the path taken between tw state, a state functin nly depends n the difference between the tw states. E and H? E=q+w, H = E +PV A calrimeter used t measure ÄH a calrimeter used t measure Ä E? ÄH is measured with a cffee cup calrimeter, where yu the pressure is cnstant ÄE is measured with a bmb calrimeter where the vlume is cnstant ÄH fus and ÄH vap ÄH fus is the energy required t melt 1 mle f material at its melting pint ÄH vap is the energy required t vaprize 1 mle f material at its biling pint The numerical value f a pressure measured in trr and ne measured in mm Hg. The numerical values f trr and mmhg are the same! 6. (12 pints) One f the ways methanl is prduced cmmercially is t reduce carbn mnxide with H 2 in the reactin: CO(g) + 2H 2(g) CH3OH(g) (Methanl - CH3OH has a central C surrunded by 3H s and 1 O. The remaining H is then attached t the O) Determine the ÄH fr this reactin using the fllwing bnd energies: Bnd Bnd Energy(kJ/ml) H C-H 414 O-H 464 C O + 2H-H H-C-O-H C-C 347 C=C 615 H C C 881 C-O 351 ÄH Rxn = Bnds brken in reactants - Bnds frmed in prducts C=O 730 = [940 + 2(435)]-[3(414)+351+464)] C O 940 = 1810-2057 H-H 435 = -247 kj/ml H=H 620 3
7. (12 pints) Hw much heat energy will it take t change 100 grams f slid sdium at 25 C t a liquid at 200 C? Sme cnstants that yu might need: ÄH fus =2.60 kj/ml ÄH vap =97.4 kj/ml Melting pint 97.8 C Biling pint 883 C Mlar heat capacity f slid Na = 28.2 J/ml K Mlar heat capacity f liquid Na = 30.8J/ml K Mlar heat capacity f gaseus Na = 20.8 J/ml K Three step prcess Step 1: Heat sdium slid frm 25 t 97.4 using mlar heat capacity f slid Step 2: Melt sdium at 97.4 using ÄH fus Step 3: heat liquid sdium at 97.4 up tp 200 using mlar heat capacity f liquid Mles f sdium = 100g/ 23g/ml =4.35 ml Step 1: ÄT = 97.4-25 = 72.4; Step 2: E= mles ÄH Fus Step 3: ÄT = 200-97.4 = 102.6; E = MHC ÄT ml = 28.2 72.4 4.35 =8881J =4.35 2.60 =11.31 kj E = MHC ÄT ml = 30.8 102.6 4.35 = 13750 J Cnverting t kj and adding 8.88 + 11.31 + 13.75 = 33.94 kj 8. (12 pints) Help me befre Mrs. Z finds ut! I drpped a 1L bttle f water in a rm in the basement. The rm is 10 ft x 15 ft by 8 ft and has a vlume f 1200 cu ft r abut 0 34,000 liters. If the temperature is 72F (~ 22 C) and the equilibrium vapr pressure f water is 19.8 trr at this temperature, will all the water evaprate int the air, r will sme water remain as a liquid (and a wet spt n the flr). Yu can assume that the air was dry in the rm at the start, and that I clsed the dr s Mrs.Z culdn t see (and the water vapr can t leave the rm). Yu can t guess yes r n. Yu have t calculate the amunt f water in the air f the rm and shw me that it is mre r less than 1 kg (the mass f 1 liter f water). Calculate the maximum water vapr that culd be in the air using PV=nRT n=pv/rt T in K and P in atm T = 273+22 = 295K, P = 19.8 trr 1atm/760 trr =.026 atm n=(.026 atm 34,000L)/(.08206 l atm/k ml 295K =36.52 ml cnvert t g 36.52 ml 18 g/ml = 657 g f water vapr in the air Unfrtunately I spilled 1 liter f water which wuld weigh 1000g, s there is still 1000-657 = 343 grams f water n the flr! 4
Extra Credit - up t 5 pints In Chapter 13 we had. In Chapter 14 we used R again, but n said that. While they may lk like different numbers, they actu express the same relatinship, just with different units. Prve that t me nw. Cnvert. 2 2 I will give yu a few hints. Sme f the cnversins yu need include: 1J=1 kg m /s, 1 2 3 Pa= 1 kg/(m s ), and 1 liter = (.1m). Needless t say, yu have t shw me explicitly each cnversin yu use, yu cannt simply say.08206 l atm/k ml = 8.314 J/K ml. Dn t be afraid if yur final answer is nt exactly 8.314 because there will be sme small rundff errrs. I wanted yu t see hw the cnversin use the abve basic SI units, s I nly gave partial credit if yu used the 1l atm = 101.325 J frm the cnversin sheet. One f the difficulties is cnverting frm liters t m 3 3 3 1 liter = (.1m) =.1m.1m.1m =.001m =8.314 J/K ml 5