HW Key _12 s problems part b

HW Key _12 s problems part b - S49. Use information from...

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S49. Use information from the appendices to predict whether diamond or graphite is the stable form of carbon at 1 atm pressure. If the density of graphite is 2.26 g/cm 3 and that of diamond is 3.51 g/cm 3 , to what value must you change the pressure to make diamond the stable form of carbon at room temperature? You may assume that both diamond and graphite are incompressible. To determine whether graphite or diamond is more stable at 1 atm and 298 K we need to calculate G for the process C (graphite) C (diamond) This can be found using the data in the back of the book and the expression Δ G ° = n Δ G f ° (diamond) n Δ G f ° (graphite) = (1 mole)(2.900 kJ/mole) (1 mole)(0 kJ/mole) = + 2.900 kJ Because G is positive, we know that the conversion from graphite to diamond is not spontaneous, but the conversion of diamond to graphite is. This means that graphite is the more stable form. Alternatively we can see this from the fact that Δ G f ° (graphite)=0 , which says it is the naturally occurring form of carbon. To find the pressure at which diamond is the stable form of carbon, we need to find the
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This note was uploaded on 02/01/2011 for the course CHEM 111 taught by Professor Owens,g during the Fall '08 term at University of Utah.

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HW Key _12 s problems part b - S49. Use information from...

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