Calculate the enthalpy of fusion of naphthalene (C10H8) given that its melting point is 128 Celsius and its entropy of fusion is 47.7 J (k*mol) .
dG = dH-TdS
dG = 0 at the phase change.
dHfusion = Tfusion x dSfusion
dHfusion = 128 x 47.7 = ?
To calculate the enthalpy of fusion (ΔHfusion) of naphthalene (C10H8), we can use the relationship between enthalpy, entropy, and temperature:
ΔHfusion = ΔSfusion × T
Given:
Melting point (T) = 128 °C = 128 + 273 = 401 K
Entropy of fusion (ΔSfusion) = 47.7 J/(K*mol)
Now, substitute the values into the equation:
ΔHfusion = 47.7 J/(K*mol) × 401 K
Calculating this, we get:
ΔHfusion = 19114.7 J/mol
Therefore, the enthalpy of fusion of naphthalene is approximately 19114.7 J/mol.
To calculate the enthalpy of fusion of naphthalene, we need to use the equation:
ΔH = TΔS
where ΔH is the enthalpy of fusion, T is the temperature in Kelvin, and ΔS is the entropy of fusion in Joules per Kelvin (J/K). We are given that the melting point of naphthalene is 128 degrees Celsius, so we need to convert this to Kelvin.
To convert Celsius to Kelvin, we use the formula:
T(K) = T(°C) + 273.15
Therefore, the melting point of naphthalene in Kelvin is:
T = 128 + 273.15 = 401.15 K
Now, we substitute the values into the equation:
ΔH = 401.15 K * 47.7 J/K
Calculating this, we get:
ΔH = 19,127.455 J
Therefore, the enthalpy of fusion of naphthalene is approximately 19,127.455 J.