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

`s_(Pb(l),500\°C)=100.83(J)/(mol*K)`

Step-by-step 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) {(22.13 + 0.01172 T + 1.00 x 10^(-5) T^2)/(T) } \, dT =20.4(J)/(mol*K)\\\\\\s_2=(4770(J)/(mol) )/(600.55K)= 7.94(J)/(mol*K)\\\\\\s_3=\int\limits^(773.15K)_(600.55K) {(32.51-0.00301T)/(T) } \, dT=7.69(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(J)/(mol*K)`

Best regards.