che294f04hmwrk14soln - CHE/MAE 294 F04 Homework 14 Solution...

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CHE/MAE 294 F04 Homework 14 Solution 3.Dec.01 > restart: Problem 1 Linear and Parabolic Rate Laws One of the constants in these two equations is equal to the initial film thickness (thickness at t=0). > EqnLineRate := y = c1*t + yo: EqnParaRate := y^2 = c2*t + yo^2: Contant Solutions Solve for constants and redisplay rate law formulations. > c1 := solve(subs({yo = 50,y = 150, t = 10},EqnLineRate),c1); c2 := solve(subs({yo = 50,y = 150, t = 10},EqnParaRate),c2); := c1 10 := c2 2000 > EqnLineRate; EqnParaRate; = y + 10 t yo = y 2 + 2000 t yo 2 Growth Times Determine growth times to reach 200 nm > tLine := solve(subs({y=250,yo=50},EqnLineRate),t); tPara := solve(subs({y=250,yo=50},EqnParaRate),t); := tLine 20 := tPara 30 Linear Rate: 250 nm is reached in 20 min Parabolic Rate: 250 nm is reached in 30 min
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-- Problem 2 Logarithmic Rate Law Expression > EqnLogRate := y = kr*ln(cr*t + 1): Subsitutions The above expression has two unknowns. We have two (time, thickness) points, so we can determine the two unknowns. Create two algebraic expressions and solve ... > EqnLogRate1 := subs({y=50,t=2},EqnLogRate); EqnLogRate2 := subs({y=150,t=10},EqnLogRate); := EqnLogRate1 = 50 kr ( ) ln + 2 cr 1 := EqnLogRate2 = 150 kr ( ) ln + 10 cr 1 > solnset := solve({EqnLogRate1,EqnLogRate2},{kr,cr}); solnset = kr 50 1 ( ) ln ( ) RootOf , + - _Z 2 _Z 4 = label _L1 , { := = cr - 1 2 ( ) RootOf , + - _Z 2 _Z 4 = label _L1 1 2 } Assign the values and show the rate law expression.
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