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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 = 0 . 550 , x = 0 . 992 . Only one of these solutions can be the real answer. We can fi gure that out by inspection. Since the fi 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 fi nal equilibrium conditions are [ A ] = 1 . 00 0 . 459 = 0 . 541 M (8.4) [ B ] = 1 . 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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