Question

material science and engineering

copper oxidizes to Cu2O exhibiting parabolic behavior at temperatures above 500 degree C. the activation energy for diffusion is 163kJ/mol. Of .011 cl of Cu2O forms after exposure to air after 10 minutes at 625 degree C, what will the oxide thickness be after exposure of a clean copper surface to air at 680 degree C for 18 minutes?

Copper oxidizes to Cu20 exhibiting parabolic behavior at temperatures above 500 °C. The activation energy for diffusion is 163 kJ/mol. If 0.011 cm of Cu2O forms after exposure to air after 10 minutes at 625 °C, what will the oxide thickness be after exposure of a clean copper surface to air at 680 °C for 18 minutes?

Answer 1

The problem involves the concept of parabolic rate law and temperature dependence of rate constant on temperature. A detailed explanation is given in below images.

porabolic Law 2012 Lince oxidation follow 500°c. which be above con as. dx Twhere x il X at x oxide thickness ² da K [k - Conssent or rate dt X constant kot ² x dx = Integrating both sides. fxdx Lkoll 2 ke tc X 3 x ² 2 Kt to 7-0 ce doni fond -taht of integration. where ke kob ko constant facter Activation Energy 4 T = Temperature (k) Ea

Se we get X = 2 K L tc New udly at t=0 x=0 a = LKlol to c=0 So Ix²= 2K17- the required porabolic röke lower on Step? Calculation of rate constant for T = 625 =898k Since K is defendent Hemfperchuse. aven At At T = 898 K for t= 10 min X = 0.0llcm ta - 1631677 mol So Putting values in a (0.011)' 2 (K 1 ) x ( 101 ki = 0.011 X 0.011 0.00000605 스 2 x 10 cm / min

Step 2. Also calculation of ko - Ea Also RT We know that k= ko e - Eu RTI K = Kol -163 0.00000 do 5 = kal RX898 RE 0.008314 K1 /molok) 163 So FI 0100000 665 = ko e 2.008314 X898. -163 0.0.0.0.0.0 Gos = ko e 7:465 - 6 Oto 6:05 x 10 =ko 3,289 X lo to? ko = Gios 3.289 * ó 10 -10 to 4 ko 11839 X 10 cm² l min cans tant (K2) at Calcul ation of sale 7 = 6900 953K - Ea K₂ = ko e RT2 109 X e K2 = (1.839X10 kes 18 19 x 50 x L -163 60.008314) X 953) t168 7,92

K = 1.839 X 10" X (1.153 x 10 2/min ܘܐ ܐ ܘ 2 ܐ ܝܐ - ܨܠ ܐ ܐ Stepy. Colau lahin of for ta thickness of oxide at 6806 If minutes. Using 2 = 2K E Here K = K2= 2.120 x 10 cm /min t=18 min X 18 So x ² = 2x 2.120xlo 4 x² = 7.632 x 10 Ст x = d. 16 x 10 cm or 0276 cm TXE thickness of oxide weuld 10.0 276 cm at 650c f for 18 minute

Hope it would be helpful for you.