Answer:
[tex]s_{Pb(l),500\°C}=100.83\frac{J}{mol*K}[/tex]
Explanation:
Hello,
In this case, for the calculation of the standard entropy of liquid lead at 500 °C (773.15 K), starting by solid lead 298.15 K we need to consider three processes:
1. Heating of solid lead at 298.15 K to 600.55 K (melting point).
2. Melting of solid lead to liquid lead.
3. Heating of liquid lead at 600.55 K (melting point) to 773.15 K.
Which can be written in terms of entropy by:
[tex]s_{Pb(l),500\°C}=s_{Pb(s),298.15K}+s_1+s_2+s_3[/tex]
Whereas each entropy is computed as follows:
[tex]s_1=\int\limits^{600.55K}_{298.15K} {\frac{22.13 + 0.01172 T + 1.00 x 10^{-5} T^2}{T} } \, dT =20.4\frac{J}{mol*K}\\\\\\s_2=\frac{4770\frac{J}{mol} }{600.55K}= 7.94\frac{J}{mol*K}\\\\\\s_3=\int\limits^{773.15K}_{600.55K} {\frac{32.51-0.00301T}{T} } \, dT=7.69\frac{J}{mol*K}[/tex]
Therefore, the standard entropy of liquid lead at 500 °C turns out:
[tex]s_{Pb(l),500\°C}=64.80+20.4+7.94+7.69\\\\s_{Pb(l),500\°C}=100.83\frac{J}{mol*K}[/tex]
Best regards.
