Question

Given the *temperature*-dependent (molar) heat capacity
of graphite:

Cp,n= 16.86( J K^-1 mol^-1) +4.77*10^-3 JK^-2mol^-1)T-8.54*10^5(JKmol^-1)T^-2

calculate the change in the molar enthalpy of graphite in going from T = 298 K to

T = 348K.

Answer #1

molar enthalpy dH = CpdT

Given,

T1 = 298 K

T2 = 348 K

The above equation can also be rearranged as,

dH = a(T2-T1) + b(T2^2-T1^2)/2 + c/(T2^-1-T1^-1)

with a = 16.86, b = 4.77 x 10^-3, c = 8.54 x 10^5

Feeding values for T1 and T2

dH = 16.86(348-298) + 4.77 x 10^-3(348^2-298^2) + 8.54 x 10^5(348^-1-298^-1)

= 843 + 154.071 - 411.75

= 585.32 J/mol = 5.85 kJ/mol

the molar enthalpy as calculated would be 5.85 kJ/mol

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