Prob_6.6_sol_rev - Thermodynamics of Materials HW 2 Problem...

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Thermodynamics of Materials © John Kieffer, University of Michigan (2009) 1 HW 2 Problem 6.6 1000 g of Al are pre-heated to 700˚C and combined with Cr 2 O 3 . The reaction enthalpy for 2Al(s) + Cr 2 O 3 (s) = Al 2 O 3 (s) + 2Cr(s) H R = H(Al 2 O 3 ) – H(Cr 2 O 3 ) = – 541000 J/mol at room temperature. There are two paths in the enthalpy vs. temperature diagram that we can pursue to solve the problem. The first is to consider the reaction taking place at room temperature, and all reactants plus ballast needs to be heated up from room temperature to 1600 K. 1000 g of Al correspond to 37.07 mol. Let x = 37.07/2 = 18.52 mol. The process can be written as 2x Al + ( x + y ) Cr 2 O 3 = x Al 2 O 3 + 2 x Cr + y Cr 2 O 3 (1) In this molar balance equation, x mol of Cr 2 O 3 react and y mol are carried along as bal- last to prevent overheating. The corresponding enthalpy balance is 2 xc p ,Al l ( ) dT 973 934 2 x H ,Al + 2 xc p ,Al s ( ) dT 934 298 + x H R + 2 xc p ,Cr s ( ) + xc p ,Al 2 O 3 + yc p ,Cr 2 O 3 ( ) dT 298 1600 = 0 .
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Prob_6.6_sol_rev - Thermodynamics of Materials HW 2 Problem...

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