Question

Problem. 4·The extent of reaction generally depends on pressure as well as temperature. For the reaction C (graphite)C (diamond) the standard state free energy change at 25 C is 2866 J/mol. The density of graphite is 2.25 g/cc and that of diamond is approximately 3.51 g/cc; both solids may be considered to be incompressible. To help chemical engineering students pay their tuition, we are thinking of setting up equipment to run this reaction in the senior laboratory. Estimate what pressure, at room temperature (-25 °C), must be used to convert old pencil leads' (graphite) into diamonds.

Answer 1

From equation $\left [ \partial \Delta G/\partial P \right ]$ _T =$\Delta$V

d$\Delta$G=$\Delta$Vdp at Constant T.

Therefore ,

$\int_{1}^{2}$d$\Delta G$=$\Delta G\int_{P1}^{P2}$$\Delta Vdp=\Delta G$₂ -$\Delta G1=\Delta V(P2-P1)$

P2=$\Delta G2-\Delta G1/\Delta V +P1$   --------------Equation 2

The initial state is graphite and the final state is diamond .Now the $\Delta G$ given in the above equation are the free energy changes for the conversion of graphite to diamond ,which takes +2,900 j mol^-1 at 25 c and 1 bar pressure .

$\Delta G=\Delta f G$_diamond-$\Delta G$_graphite =2900-0=2900 j mol^-1

To find the pressure at which the free energy change for this conversion becomes 0,so that diamond and graphite will be in equilibrium and the conversion can take place,Therefore our initial state correspondents to $\Delta G$1=2900 j mol^-1while the final state corresponds to $\Delta G$2=0 .

$\Delta V$=12.011 g mol^-1(1/3.51-1/2.25) cm^3/g*10^-6 m³/cm³

        =-1.915*10^-6 m³mol^-1

Using the above equation 2

p2=0-2900 j mol^-1/-1.92*10^-6m³ mol^-1   + 10⁵ pa

   =1.51*10⁹ pa

   =1.51 * 10⁴ bar

So above pressure is required to convert old pencil leads(graphite ) into diamonds .