For reactions in condensed phases (liquids and solids), the difference between A
ID: 870904 • Letter: F
Question
For reactions in condensed phases (liquids and solids), the difference between All and A U is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that All and AU can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth?s surface. Calculate triangle H - triangle U for the conversion of 1 mole of graphite to 1 mole of diamond at a pressure of 4.00 X 10^4 atm. The densities of graphite and diamond are 2.25 g/cm^3 and 3.52 g/cm^3, respectively. -0. 0722 kJ/molExplanation / Answer
we know that
dH - dU = pdV
given
1 mole of graphite and diamond
we know that
1 mole of graphite and diamond weighs 12 g
we know that
volume = mass / density
so
volume of graphite = mass of graphite / density of graphite
volume of graphite = 12 / 2.25
volume of graphite = 5.333 cm3
similarly
volume of diamond = 12 / 3.52
volume of diamond = 3.409 cm3
now the reaction is
graphite ----> diamond
so
dV = volume of diamond - volume of graphite
dV = 3.409 - 5.333
dV = -1.924 cm3
dV = -1.924 x 10-6 m3
now
given
pressure = 4 x 10^4 atm
pressure = 4 x 10^4 x 101325 pa
pressure (p)= 4.053 x 10^9 Pa
so
now
dH - dU = pdV
dH - dU = 4.053 x 10^9 x (-1.924 x 10-6)
dH - dU = -7797.97 J
dH - dU = -7.80 kJ
so
the dH - dU value is -7.8 kJ /mol
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