Horvath Practice Problems 35

Horvath Practice Problems 35 - This is hard to solve...

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This is hard to solve algebraically but you can use your calcu- lators to get three solutions: x =0 . 459 ,x . 550 . 992 . Only one of these solutions can be the real answer. We can f gure that out by inspection. Since the f nal concentration of B is 1 . 00 2 x, only the x = 0 . 459 solution makes sense. The other solutions give a negative concentration for B. Thus the f nal equilibrium conditions are [ A ]=1 . 00 0 . 459 = 0 . 541 M( 8 . 4 ) [ B . 00 2 × 0 . 459 = 0 . 082 M [ C ]=0 . 459 M Example 2 : Consider the reaction A ( g ) ­ B ( g ) + C ( g ) , (8.5) where K p =4 . 23 × 10 1 . The initial partial pressure of A is 0 . 333 atm . Determine the concentration of reactants and
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