What conclusions can you derive from the size and the signs of the enthalpy and

Question

What conclusions can you derive from the size and the signs of the enthalpy and entropy of fusion of naphthalene. Compare them to those for water, which you can find or calculate using an appropriate reference (cite your source). Use your data to predict when the standard free energy change, Delta G degree is zero for the process: C_10H_8 C_10H_8 (l) Explain your answer. Explain why using the relation Delta S degree = Delta H degree/T might be more accurate than extrapolation to 1/T_int = 0, where the intercept Delta S degree/R. List the measurements that would contribute to the overall uncertainty in the enthalpy and entropy changes found in this experiment Make an estimate of the uncertainty in the enthalpy change you found (you may need to consider different linen through the data points in your graph).

Explanation / Answer

1)

Molar enthalpy of fusion

fus H0 (water)= 6.01 kJ/mol

fus H0 (naphthalene)= 18.22 kJ/mol

Standard entropy of fusion

fus S0 (water)= 22.0 J/mol.K

fus S0 (naphthalene)= 51.57 J/mol.K

Enthalpy of fusion for naphthalene is greater than water means more heat is required to melt naphthalene.

Entropy of fusion of naphthalene is greater than water means by melting naphthalene changes to more ordered state . That is is in solid state naphthalene is not much ordered as ice.It might contain free space.That is water has more compact structure.

2)

C10H8(s) C10H8(l)

G0 = H0 - TS0

If G0 =0

H0 should be equal to TS0

fus H0 (naphthalene)= 18.22 kJ/mol

fus S0 (naphthalene)= 51.57 J/mol.K

T0fus = 353.4 K

TS0 = 353.4 K x 51.57 J/mol.K =18224.838 J/mol = 18.22 kJ/ mol

So it is confirmed G0 =0

3)

Since G0 = H0 - TS0

and  G0 =0

0 = H0 - TS0

H0 =  TS0

S0 = H0 /T

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