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Grade 12th passPhysical Chemistry

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm–3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is

Profile image of Yogiraj
10 Years agoGrade 12th pass
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3 Answers

Profile image of Sakshi
10 Years ago
just refer to the definations of heat enthalipies of transformation and other enthalipies to get the answer.
u also need to use hess law for this question
Profile image of Omgirish
9 Years ago
Free energy = useful work done∆G = p∆v1895= P(v2 - v1)1895= P(12/2.25 - 12/3.31) × 10^-6(m) × . 10^-3(kg) .......v=m/d1895= P(5.33 - 3.62) × 10^-91895= P × 1.71 × 10^-9 P= 1895/1.71×10^-9 P= 1108 × 10^9 P= 1.1 × 10^6
Profile image of Kushagra Madhukar
5 Years ago
Dear student,
Please find the solution to your problem below.
 
Volume of Graphite, Vg = 12/2.25 cm3 = 12/2.25 x 10-3 L
Volume of diamond, Vd = 12/3.31 cm3 = 12/3.31 x 10-3 L
We know ΔG = – PΔV = – P x (Vg – Vd) x 101.3     [ Since, 1 L-atm = 101.3 J]
Hence, 1895 = – ( – 1.71 x 10-3) x 101.3 x P
on simplifying we get, P = 10.93 x 103 atm = 11.08 x 108 Pa
 
Hope it helps.
Thanks and regards,
Kushagra

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