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Question: The densities of graphite and diamond at \[298K\]are 2.25 and \(3.31gc{{m}^{-3}}\) respectively. If ...

The densities of graphite and diamond at 298K298Kare 2.25 and 3.31gcm−33.31gc{{m}^{-3}} respectively. If the standard free energy difference is1895Jmol−11895Jmo{{l}^{-1}}, the pressure at which graphite will be transformed into diamond is:
A. 9.92×108Pa9.92\times {{10}^{8}}Pa
B. 9.92×107Pa9.92\times {{10}^{7}}Pa
C. 9.92×106Pa9.92\times {{10}^{6}}Pa
D. None of these

Explanation

Solution

As we know that both graphite and diamond are all allotropes of carbon. When the graphite is converted into the diamond and thermal reaction takes place. When the graphite is converted into diamond then high temperature and high pressure is required.

Formula used:
ΔG=−PΔV\Delta G=-P\Delta V
Where ΔG\Delta G- Standard free energy
P - Pressure
ΔV\Delta V -Volume

Complete step by step answer:
Given that density of diamond =3.31gcm−33.31gc{{m}^{-3}}
Density of graphite = 2.25gcm−32.25gc{{m}^{-3}}
Standard free energy difference =1895Jmol−11895Jmo{{l}^{-1}}
Energy = ΔG\Delta G
P =?
As we know that Density = MassVolume\dfrac{Mass}{Volume}
Mass of graphite & diamond = 12
(Because both are allotropes of carbon)
Volume of diamond = Mass of carbonDensity of diamond\dfrac{Mass\,of\,carbon}{Density\,of\,diamond}
Vd{{V} {d}} =123.31×10−3L=\dfrac{12}{3.31}\times {{10} ^ {-3}} L
=3.6253×10−3L=3.6253\times {{10} ^ {-3}} L
Volume of graphite = Mass of carbonDensity of graphite\dfrac{Mass\,of\,carbon}{Density\,of\,graphite}
Vg{{V} {g}} =122.25×10−3L=\dfrac{12}{2.25}\times {{10} ^ {-3}} L
=5.33×10−3L=5.33\times {{10} ^ {-3}} L
ΔV=Vd−Vg\Delta V= {{V} {d}}-{{V} {g}}
=(3.625−5.33)×10−3L=(3.625-5.33)\times {{10}^{-3}}L
=−1.71×10−3L=-1.71\times {{10} ^ {-3}} L
Now ΔG=−PΔV\Delta G=-P\Delta V
1895=−P(−1.71×10−3)1895=-P (-1.71\times {{10} ^ {-3}}) (1atm=101.3Pa)(1atm=101.3Pa)
1895=P×101.3×1.7×10−3 P=1895101.3×1.7×10−3=10.93×102atm      =11.07×108Pa 1895=P\times 101.3\times 1.7\times {{10} ^ {-3}} \\\ P=\dfrac{1895}{101.3\times 1.7\times {{10} ^ {-3}}} = 10.93\times {{10} ^ {2}}atm \\\ \,\,\,\,\,=11.07\times {{10} ^ {8}}Pa \\\

So, the correct answer is “Option D”.

Note:
Convert volume into litres from centimetres to use it in the density formula. Standard free energy is the energy change during the formation of a substance from its elements in their most stable state.
The expression for standard free energy must be known.
The knowledge of how the unit Pascal is converted to atm & atm is converted to Pascal must be remembered.