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I guess Cp = dH/dT= dq/dT=dS Thus SdT = integral of CpdT. From that ∆H is given and we can calculate ∆G. However, entropy for this reaction should be -7.4J/molK and I am getting like 26J/molK.

Also S is not ∆S which is a big flag in this working.

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Entropy change is

$$ \Delta S = \int \frac{\delta q_{rev}}{T} $$

Under an assumption that the phase transformation is reversible, this becomes

$$ \Delta_{pt} S = \frac{\Delta_{pt} H}{T_{pt}}$$

Enthalpy and entropy depend on temperature as

$$ \Delta H(T_2) - \Delta H(T_1) = \int C_p dT $$

$$ \Delta S(T_2) - \Delta S(T_1) = \int \frac{C_p}{T} dT $$

The Gibbs energy change is

$$ \Delta G = \Delta H - T \Delta S $$

The Gibbs energy change of the phase transformation at the transformation conditions is calculated directly from the enthalpy change and transition temperature.

To calculate the value at a different temperature, you must first calculate the enthalpy and entropy changes at the different temperature.

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  • $\begingroup$ It's probably worth noting that $ \Delta S = \int \frac{\delta q_{rev}}{T}$, rather than $\int \frac{\delta q}{T}$. However because, exceptionally, we can consider phase changes as reversible (I say exceptionally because this is not the case for most real-world macroscopic processes), we have $\delta q= \delta q_{rev}$ and thus $ \Delta S_{phase \,change} = \int \frac{\delta q}{T} $. $\endgroup$
    – theorist
    Apr 25, 2022 at 6:28
  • $\begingroup$ @theorist Amended as such. Thanks. $\endgroup$ Apr 25, 2022 at 13:35

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