Diamonds are formed from graphite under high pressure in coal mines. Calculate the equilibrium pressure (in atm) at which graphite is converted to diamonds at ${25}^{\circ }C$ (assumed constant) given densities of ${\rho }_{\text{graphite}}=2g/cc\&{\rho }_{\text{diamond}}=3g/cc(\Delta G_f^{\circ })$ for diamonds is ${\text{3 k J m}}^{-1}$ from graphite

For conversion $C$ (graphite)$\to C$ (diamond) the $\Delta S$ is

For conversion $C$ (graphite)$\to C$ (diamond) the $\Delta S$ is

(i) Determine ${\Delta }_f{H}^{\circ }(NO,g)$ at ${25}^{\circ }C$. Using the following information
${\Delta }_f{H}^{\circ }(C{O}_2,g)=-393.5$ kJ/mol
$\begin{array}{cccc}2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)\left(g\right)& & & {\Delta }_r{H}^{\circ }=-114.0\text{kJ/mol}\\ 2CO\left(g\right)+O_2\left(g\right)\to 2C{O}_2\left(g\right)& & & {\Delta }_r{H}^{\circ }=-566.0\text{kJ/mol}\\ 4CO\left(g\right)+2N{O}_2\left(g\right)\to 4C{O}_2\left(g\right)& +& N_2\left(g\right)& {\Delta }_r{H}^{\circ }=-1198.4\text{kJ/mol}\end{array}$
(ii) Calculate the equilibrium pressure (in Pascal) for the conversion of grapgite to diamond at ${25}^{\circ }C$. The densities of graphite and diamond may be taken to be 2.20 and 3.40 g/cc respectively independent of pressure.
Given : $\Delta G^{\circ }(C\left(\text{graphite}\right)\to C\left(\text{diamond}\right))=2900$ J/mol.

Densities of diamond and graphite are $3.5$ and $2.3gm{L}^{-1}$, respectively. The increase of pressure on the equilibrium $C_{\text{diamond}}\iff C_{\text{graphite}}$

Calculate the enthalpy change accompanying the transformation of C(graphite) to C(diamond). Given that the enthalpies of combustion of graphite and diamond are 393.5 and 395.4 KJ respectively

The enthalpies of combustion of $C_{\left(\text{graphite}\right)}\text{and}{C}_{\left(\text{diamond}\right)}$ are $-393.5\text{and}-395.4kJ/mol$ respectively. The enthalpy of conversion of $C_{\left(\text{graphite}\right)}$ to $C_{\left(\text{diamond}\right)}$ in kJ/mol is:

Why is $C-C$ bond length in graphite shorter than in $C-C$ bond length in diamond?

The standard state Gibbs free energies of formation of ) C(graphite and C(diamond) at T = 298 K are
${\Delta }_f{G}^{\circ }\left[\text{C(graphite}\right)]=0kJmo{l}^{-1}$
${\Delta }_f{G}^{\circ }\left[\text{C(diamond}\right)]=2.9kJmo{l}^{-1}$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [ ) C(graphite ] to diamond [C(diamond)] reduces its volume by $2\times {10}^{-6}{m}^{3}mo{l}^{-1}.$ If ) C(graphite is converted to C(diamond) isothermally at T = 298 K, the pressure at which ) C(graphite is in equilibrium with C(diamond), is
$[\text{Useful information:}1J=1kg{m}^2{s}^{-2},1Pa=1kg{m}^{-1}{s}^{-2},1\text{bar}={10}^{5}Pa]$

A : $C_{\text{diamond}}\to C_{\text{graphite}}$
$\Delta H\text{and}\Delta U$ are same for this reaction.
R: Entropy increases during the conversion of diamond to graphite.

Calculate the enthalpy change accompanying the conversion of 10 g of graphite into diamond if the heats of combustion of C (graphite) and C (diamond) are -94.05 and -94.50 kcal respectively.