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### homework1solutions

Course: ENGR 423, Fall 2009
School: Wisconsin
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Word Count: 581

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1 Homework Problem 1 Problem 1: Calculate the standard free energy change of the reaction: &lt;Ni&gt; + (O2) = &lt;NiO&gt; at 327oC from the following data: Ho298, NiO = -57,500 cal/mole So298, Ni = 7.12 cal/deg/mole So298,O2 = 49.02 cal/deg/mole So298, NiO = 9.10 cal/deg/mole Cp,Ni = 6.03 + 10.44 x 10-6 T2 - 2.5 x 10-3 T cal/deg/mole Cp,O2 = 7.16 + 1 x 10-3 T 0.4 x 105 T-2 cal/deg/mole Cp,NiO =...

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1 Homework Problem 1 Problem 1: Calculate the standard free energy change of the reaction: <Ni> + (O2) = <NiO> at 327oC from the following data: Ho298, NiO = -57,500 cal/mole So298, Ni = 7.12 cal/deg/mole So298,O2 = 49.02 cal/deg/mole So298, NiO = 9.10 cal/deg/mole Cp,Ni = 6.03 + 10.44 x 10-6 T2 - 2.5 x 10-3 T cal/deg/mole Cp,O2 = 7.16 + 1 x 10-3 T 0.4 x 105 T-2 cal/deg/mole Cp,NiO = 12.91 cal/deg/mole 1 NEEP 423 Lesson 13 Homework 1 Problem 1 Reaction: <Ni> + (O2) = <NiO> Steps 1: Calculate CP (as a function of T) Step 2: Calculate So298 Step 3: From Ho298, (given), calculate H600, Step 4: From So298 (step 2), calculate S600 Step 5: G600 = H600 TS600 Step 1: CP, 298 = CP, NiO, 298 CP, Ni, 298 (CP,O2, 298) Cp, = 3.30 - 10.44 x 10-6 T2 + 2.0 x 10-3 T + 0.2 x 105 T-2 cal/deg Step 2: S298 = S298, NiO S298, Ni (S298,O2) S298 = 9.10 7.12 24.51 = - 22.53 cal/deg. 2 NEEP 423 Lesson 13 Homework 1 Problem 1 Step 3: H0298 = - 57,500 cal/mole (given) H O T 2 = H O T 1 + CpdT T1 T2 H O 600 = H O 298 + (3.30 - 10.44 x 10-6 T 2 + 2.0 x 10 -3 T + 0.2 x 105 T -2 )dT 298 600 = -56,858 cal 3 NEEP 423 Lesson 13 Homework 1 Problem 1 Step 4: Cp ST 2 = ST 1 + dt T T1 600 T2 (3.30 - 10.44 x 10-6 T 2 + 2.0 x 10 -3 T + 0.2 x 105 T -2 ) S 600 = -22.53 + dt T 298 = -20.96 cal/deg 4 NEEP 423 Lesson 13 Homework 1 Problem 1 Step 5: G 600 = H 600 - TS 600 G 600 = -56,858 - 600 * (-20.96) = -44, 282 cal. 5 NEEP 423 Lesson 13 Homework 1 Problem 2 Problem 2: Chromium plates are bright annealed at 727oC in hydrogen at 1 atm. pressure. Calculate the permissible moisture content in the hydrogen if there is to be no oxidation during the annealing process. Neglect the possibility of dissolution of hydrogen in chromium. Given that for the reaction: 2Cr + 3H2O= Cr2O3 + 3H2, G0 = -91,050 + 22.8 T cal/mole G0 = -91,050 + 22.8 T = - 68,250 G0 = -RT ln K (R = 1.987 cal/deg/mole, T = 1000) -68, 250 = -1987 ln K ln K = 34.35 K = 8.27x 1014 6 NEEP 423 Lesson 1 13 Homework Problem 2 aCr 2O 3.a 3 H 2 K= 3 2 a H 20.a Cr Assuming aCr2O3 and aCr are unity and a = pressure for ideal gas p H2 K= 3 p H 20 p3H 2 8.27 x1014 = 3 p H 20 pH 20 = 1.065 x10 -5 pH 2 3 The moisture content should not be greater than 1.065x10-3 by volume to prevent Cr oxidation 7 NEEP 423 Lesson 13 Homework 1 Problem 3 Problem 3: A diffusion couple is formed by welding together of two alloys of elements A and B. The alloy to right of the weld contains 40% of element A and the alloy to the left contains 50% of element A. The diffusion couple is heated quickly to a temperature of 800oC and held there for 40 hours and then cooled to room temperature. Chemical analysis of the diffusion couple showed that at a distance of 0.20cm to the right of the weld the composition of element A is 42.5%. Calculate (i) the diffusion coefficient of A (assume diffusion coefficient to be independent of composition over the composition range of the alloys) and (ii) for the same diffusion temperature calculate the length of time required to achieve a composition of 42.5% A at a distance of 0.40 cm to the right of the weld. 8 NEEP 423 Lesson 13 Homework 1 Problem 3 C1 = 0.5 C2 =0.4 C1 + C 2 C1 - C 2 x Cx , t = - erf 2 2 2 Dt x C1 + C 2 C1 - C 2 Cx , t = - erf 2 2 2 D *t 0 .2 0.5 + 0.4 0.5 - 0.4 0.425 = - erf 2 2 2 D *144,000 C1 = 50% (0.5), C2 = 40% (0.4), Cx,t = 42.5 (0.425), x = 0.2cm t = 40 x 3600 = 144,000 sec. Since, erf (0.5)~0.5 0.2 0.5 = erf 2 D *144,000 0.2 0.5 = 2 D *144,000 D = 2 x 10-7 cm2/sec 9 NEEP 423 Lesson 13 Homework 1 Problem 3 0.2 0.5 + 0.4 0.5 - 0.4 0.425 = - erf 2 2 2 D *144,000 0.5 + 0.4 0.5 - 0.4 0.4 0.425 = - erf 2 2 2 D *t 0.2 0.4 erf = erf 2 D *144,000 2 D *t x1 x2 = t1 t2 2 2 t = 4 x 40 = 160 hours 10 NEEP 423 Lesson 13 Homework 1 Problem 4 11 NEEP 423 Lesson 13 Homework 1 Problem 4 12 NEEP 423 Lesson 13
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