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Practice test _3 - 1 10 points Given the following...

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Unformatted text preview: 1) 10 points. Given the following thermodynamic data, calculate the boiling point of phosphorous trichloride, PC13. AH°f (25°C) kJ/mol. AS° (25°C) meol. AG°f (25°C) kJ/mol PC13 (1) -3197 217.18 -2723 PC13(g) —287.0 311.78 -267.8 Assume that AH and AS do not change With temperature. 2) 10 points. Calculate the amount of work done When 100 liters of an ideal gas originally at a pressure of 1 atm is compressed to a final volume of 10 liters and a final pressure of 10 atm. 3) 10 points. When a 5 g sample of aluminum powder (MW : 27gfmol) reacts With iron oxide, FezOs, in a bomb calorimeter according to the reaction: 2 Al (8) + Fe203 (s) ——. A1203 (s) + 2 Fe (8) the temperature rises by 8.3 °C. The heat capacity of the calorimeter is 22.3 kJ/°C. a) Calculate AE for the reaction of 1.00 mol of aluminum powder in kJ/mol. b) Calculate AH for this reaction at 298 K in kamol. 4) 10 points. Calculate the equilibrium constant for the following reaction at 298 K and 1 atm of pressure. 2 H20 (1) >>> H3o+ (aq) + OH' (aq) AH°f (25° C) kJ/mol. AS° (25°C) J frnol. AG°f (25° C) kJ/mol H20 (1) —285.83 69.91 —237. 18 H301 (aq) —285.83 69.91 —237. 18 OH' (aq) —229. 99 — 10.75 — 157.32 5) 8 points. For the reaction: 2 NO (g) + Cl2 >>> 2NOCl (g) the NOCl concentration increases at a rate of 0.030 mol/L-s. a) What is the rate of the disappearance of C12 in molfL-s ? b) What is the rate of the overall reaction in rnol/L-s ? 6) 10 points Calculate AG“ for the following reaction at 298 K and 1 atm of pressure. Circle your answer for AG“ and indicate if the reaction is spontaneous as written. AH°f (25°C) kJ/rnol. AS° (25°C) meol. AG°f (25°C) kJ/mol H2 (g) 0 +130 0 02 (g) 0 +205 0 H20 (g) -241 +188 —228 2 H2(s) + 02(g) >>> 2 H20 (g) 6b) To what temperature would you have to heat water to drive the reaction toward H2 and 02? (one suggested route for generating hydrogen for the hydrogen economy). 7) 8 points. Technetium 99mTc is used as a medical diagnostic agent. The half life of 99mTc is only 6 hours. How many hours does a patient have to wait until the 9ngc has decayed to less than 1 % of the original dose? 8) 10 points. A reaction rate triples when the temperature is increased from 50°C to 100 °C. What is the activation energy in U? 9) 14 points. The following reaction, N2 (g) + 2 H2 (g) >>> N2H4 (g) is believed to occur by the following elementary steps: Step 1. N2 + H2 >>> NZH2 Step 2. NZH2 + H2 >>> N2H4 The experimental rate law is : Rate = K[N2][H2]2 a) Write the rate law for step 1 assuming that step 1 is the rate determining step. b) Write the rate law for step 2 assuming that step 2 is the rate determining step. NOTE: The rate law expression should NOT contain an intermediate in it. 0) Compare your answers for parts a and b to the experimental rate law and decide which step is the rate determining step for this reaction. 10) 10 points. Use the following experimental initial rate data obtained at some particular fixed temperature to determine the rate law and rate constant for the reaction of nitrogen With ozone: ICl (g) + H2(g)>>>12(g) + 2 HCl (g) Initial Concentration Initial Concentration Initial Rate of (M) (M) Reaction in moHL-s [1011 -EE_— 3-7 X 10’s 7-4 X 10'5 14-8 X 10.5 a) What is the rate law? b) What is the rate constant? c) What are the units of the rate constant? Henry’s Law: C : kP Roult’s Law: Psolmt = P° o solvent Xsolvent Boiling Point Elevation: ATB = kBrn M = mies m: mgles g Freezing Point Depression: ATF = kFrn Osmotic Pressure: H = MRT 760 Torr = 1 atm R = 0.0821 L'Eltm/le'K R = 8.314 J/m01'K C]C[D]d aA+bB —’CC+dD =[— ‘— KeQ [AiarBib K1) 2 KART)An _ 2 _ ay2+by+C20 y=w 2a [H301 [OH'] = 1 X 10'14 2 kW pH + pOH =14 = pkW pH = -10g[H3O+] [H3O+] = 10'pH pOH = -10g[OH'] [OH'] = 10'pOH + _ k :[H3O ][A ] k :[BH+][0rr] a [HA] b [B] kakb : kw pka + pkb : pkw “b pH: pka +10g— “21 W = —PAV 1 L0 atrn = 101.3 J qSUIT 2 CAT qSUIT = —qSyStEm AE=q+ W AH =AE+ PAV AH =AE+AnRT AG 2 AH — TAS AG 2 AG° + RTan AG° = —RT1nKeq Keq = e’AGWRT AS° AH° aneq 2 Y‘ RT rate=—fl=_fl:+A[q =+fl aAt bAt CAt dAt rate = 1([A]"[B]y rate = 1([A]0 [A] = [A]0 —kt rate = 1([A]1 [A] = [A]0 e'kt 1n[A] =111[A]0 — kt t1,2 = 0.693/k 2 1 1 rate : k[A] — = + kt [A] [A]0 k : AeXp(—Ea/RT) 111k : lnA — Ea/R (1/T) 111i: _Ea [1_1] 10 Answers: T = 345.6 K 91,117] 1 2 3 —999.5 kJ/mol same answer a and b — no gases involved 4 1 x 10'14 5 —0.015 mol/Ls and 0.015 molfLs 6 (also says problem 5) -456 KT , T = 5118 K 7 39.9 hrs 8 22 k] 9 rate : K[N2] [H2], b) k[N2] [H2]2 0) step 2 is rate limiting 10 rate = k[ICl] [H2] b 1.23X10'3 c L/mol-s 11 ...
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