Question

The standard entropy of Pb(s) at 298.15 K is 64.80 J K–1 mol–1. Assume that the heat capacity of Pb(s) is given by: CP,m(Pb, s) J K−1mol−1 = 22.13 + 0.01172 T K + 1.00 x 10−5 T 2 K2 The melting point is 327.4 ℃ and the heat of fusion is 4770 J mol-1. Assume that the heat capacity of Pb(l) is given by: CP,m(Pb, l) J K−1mol−1 = 32.51 − 0.00301 T K Calculate the standard entropy of Pb(l) at 500 ℃

Answer 1

Answer:

$s_{Pb(l),500\°C}=100.83\frac{J}{mol*K}$

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:

$s_{Pb(l),500\°C}=s_{Pb(s),298.15K}+s_1+s_2+s_3$

Whereas each entropy is computed as follows:

$s_1=\int\limits^{600.55K}_{298.15K} {\frac{22.13 + 0.01172 T + 1.00 x 10^{-5} T^2}{T} } \, dT =20.4\frac{J}{mol*K}\\\\\\s_2=\frac{4770\frac{J}{mol} }{600.55K}= 7.94\frac{J}{mol*K}\\\\\\s_3=\int\limits^{773.15K}_{600.55K} {\frac{32.51-0.00301T}{T} } \, dT=7.69\frac{J}{mol*K}$

Therefore, the standard entropy of liquid lead at 500 °C turns out:

$s_{Pb(l),500\°C}=64.80+20.4+7.94+7.69\\\\s_{Pb(l),500\°C}=100.83\frac{J}{mol*K}$

Best regards.