REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj

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1 Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml) H = H (Prducts) - H (Reactants) H (Prducts) = (1 x -125) + ( x -271) = -98 kj H (Reactants) = (1 x -46.1) + ( x 0) = kj H = (-46.1) = kj 2. Given H= kj r the reactin shwn belw, calculate the H r CO (g). Cr 2 O (s) + CO (g) 2 Cr (s) + CO 2 (g) H (kj/ml) 119.7??? H = H (Prducts) - H (Reactants) = [( x -9.5)] - [(1 x ) + H (CO)] - H (CO) = = +1.5 kj -1.5 kj H (CO) = = kj. The air within a pistn equipped with a cylinder absrbs 565 J heat and expands rm an initial vlume 0.10 L t a inal vlume 0.85 L against a pressure 1.0 atm. What is the change in the internal energy the air within the pistn? E = q + w q = J 101. J w = -P V = (1.0 atm)(0.75 L)( ) = -76 J (wrk is dne by the system) 1 Latm E = +565 J - 76 J = 489 J 0 1

2 4. Hw much wrk (in J) is required t expand the vlume a pump by 2.5 L against an external pressure 1.1 atm? 101. J w = P V = (1.1 atm)(2.5 L)( ) = -280 J 1 Latm (negative sign indicates wrk is dne by the system) 5. Calculate the wrk assciated with the llwing reactin at 1.00 atm and 25 C. Is the wrk dne by the system r n the system? 2 H 2 (g) + O 2 (g) 2 H 2 O (l) w = P V V= vlume changed caused by change in mles gas nrt ( ml)(0.0821)(298 K) V= = 7.4 L P 1 atm 101. J w = (1.00 atm)(7.4 L)( ) = 7440 J 1 Latm w = J since wrk is dne n the system due t cmpressin 6. The enthalpy change r vaprizatin methanl (CH OH) at 25 C is 8.0 kj/ml. I the entrpy methanl vapr at 25 C is 255 J/mlK, what is the entrpy the liquid methanl at this temperature? CH OH (l) CH OH (g) H = H = +8.0 kj/ml vaprizatin vap sys H T 298 K S = S - S = J/K surr -(-8.0x10 ) J S vap = = = J/K vap gas liq S = S - S = 255 J/K -(127.5 J/K) = 127 J/K liq gas vap 2

3 7. A reactin has H rxn = 107 kj and S rxn = 285 J/K. At what temperature is the change in entrpy r the reactin equal t the change in entrpy r the surrundings? - Hsys S surr = S surr = S sys = 285 J/K T - Hsys -(-107x10 J) T = = = 75 K S 285 J/K sys 8. Determine the entrpy change ( S ) in J/K r the reactin shwn belw, given the standard entrpies r each: 2 SO 2 (g) + O 2 (g) 2 SO (g) S (J/K ml) S = S (Prducts) - S (Reactants) S (Prducts) = (2 x 256.6) = 51.2 J/K S (Reactants) = (2 x 248.1) + (1 x 205.0) = J/K S = = J/K 9. Given the llwing thermdynamic data, estimate the temperature ( C) at which the reactin shwn belw becmes spntaneus. CaCO (s) CaO (s) + CO 2 (g) H = +184 kj S = +166 J/K G =+00 kj G = H - T S Reactin reaches equilibrium when G = 0 Therere, T S = H = +184 kj +184 kj +184 kj S kj/k T = = = 1108 K = 85 C Temperature must be greater than 85 C r reactin t becme spntaneus

4 10. Calculate the ree energy ( G ) in kj r the reactin shwn belw, given the G values r each substance: 4 NH (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) G (kj/ml) G = G (Prducts) - G (Reactants) G = [(4 x (86.67) + 6 x ( )] - [4 x (-16.48) + 5 x (0)] G = ( ) = kj 11. Methanl can be prduced by the reactin shwn belw, with the llwing thermdynamic data given at 25 C. CO (g) + 2 H 2 (g) CH OH (l) H (kj/ml) G (kj/ml) S (J/ml K) ??? a) Calculate G and H r this reactin G = G (Prducts) - G (Reactants) G = (1 x ) - [(1 x -17.) + 0] = kj H = H (Prducts) - H (Reactants) H = (1 x -28.6) - [(1 x ) + 0] = kj b) Calculate S (in J/K) r this reactin. G = H - T S H - G kj - (-28.9 kj) S = = = -0. kj/k = - J/K T 298 K c) Calculate S r hydrgen. S = S (Prducts) - S (Reactants) S = (1 x 126.8) - [(1 x 197.9) + (2 x S S H = = 11 J/ml K 2 2 H2 ) = - J/K 4

5 12. At 25 C the equilibrium cnstant, K p, r the reactin belw is atm. Br 2 (l) Br 2 (g) a) What is G 298 r this reactin? K = K p G = -RT ln K = -(8.14 J/ml K )(298 K )(ln 0.281) 298 p G 298 =.14 x10 J/ml b) I requires 19 J t vaprize 1.00 g liquid brmine at 25 C and 1.00 atm. Calculate H and S at 25 C r this reactin. 19 J g 1 g 1 ml 4 H = x =.084 x10 J/ml H - G ( ) J/ml S = = = 92.9 J/ml K T 298 K c) Calculate the nrmal biling pint brmine. Assume H and S are nt aected by temperature. (Hint: At the nrmal biling pint liquid and vapr are in equilibrium) At nrmal biling pint G = 0 G = H T S = 0 H 0840 J/ml T = = = 2 K S 92.9 J/ml K 5

6 1. ClF can be prepared by the reactin shwn belw: Cl 2 (g) + F 2 (g) 2 ClF (g) Fr ClF, H = 16.2 kj/ml and G = 12.0 kj/ml a) Calculate the value the equilibrium cnstant r this reactin at 25 C. G = 2(-12 kj) - 0 = -246 kj G = -RT ln K G -246x10 J ln K = = = RT -(8.14 J/K)(298 K) K = e = 1.x b) Calculate S r this reactin at 25 C. G = H - T S H = 2(-16.2 kj) - 0 = kj H - G (-26.4x10 J) - (- 246x10 J) S = = = -270 J/K T 298 K c) I ClF prduced were a liquid instead a gas, hw wuld the S r the reactin be dierent (sign and magnitude) than calculated abve? Explain. I ClF prduced was a liquid, S wuld be a larger negative number, since liquid is mre rdered than gas and has less entrpy. 6

7 14. Prductin ammnia rm nitrgen and hydrgen gases is an imprtant industrial reactin shwn belw: N 2 (g) + H 2 (g) H = 92.8 kj 2 NH (g) S = 198. J/K a) Calculate G r this reactin at 500 C. Assume H and S are nt temperature dependent. G = H - T S - G 500 = kj - (77 K)(-198.x10 kj/k) = 60.9 kj b) Calculate G at 25 C r this reactin i the reactin mixture cnsists 1.0 atm N 2,.0 atm H 2 and 1.0 atm NH. G = G + RT ln Q P (1.0) Q = = =.7x10 P P (1.0)(.0) 2 2 NH -2 N2 H2 - G 298 = H - T S = kj - (298 K)(-198.x10 kj/k) = -. kj G = -. kj + -2 (8.14 J/K)(298 K)(ln.7x10 ) 10 J/kJ = kj 15. What are the signs H, S and G r the sublimatin dry ice (slid CO 2 ) at 25 C? Since sublimatin dry ice is endthermic, H = psitive (+) Since gas has greater entrpy than slid, S = psitive (+) Since dry ice sublimes at 25 C spntaneusly, G = negative ( ) 7

8 16. Using the llwing data, calculate the value K sp r Ba(NO ) 2, ne the least sluble the cmmn nitrate salts. Species Ba 2+ (aq) NO Ba(NO ) 2 (s) G -561 kj/ml -109 kj/ml -797 kj/ml Ba(NO ) 2 (s) Ba 2+ (aq) + 2 NO (aq) G = G (Prducts) - G (Reactants) G = [(2 x (-109) + 1(-561)] - [1 x (-797)] = (- 797) = + 18 kj G 18x10 J G = -RT ln K ln K = = = RT -(8.14 J/K)(298 K) K = e = 7.0x Shw that hydrgen cyanide (HCN) is a gas at 25 C by estimating its nrmal biling pint rm the llwing data: H (kj/ml) S (J/mlK) HCN (l) HCN (g) The nrmal biling pint HCN is the temperature at which liquid and gaseus HCN are in equilibrium: HCN (l) HCN (g) H = = 26.2 kj S = = 89 J/K At equilibrium C = 0 = H T S ΔH 26.2x10 J/ml T = = = 294 K= 21 C ΔS 89 J/ml K Since 21 C is less than 25 C, HCN wuld exist as a gas at this temperature 8