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21-Mat-A1 Thermodynamics · December 2017

Question 2 of 7: Entropy and Enthalpy of Lead Through Its Melting Point

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2017 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).

Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — source of the first-law/second-law relations used throughout and of the attached Ellingham diagram (Fig. 9-3) and its underlying 298 K formation data used in Question 7; supporting 298 K entropies for Ca(s), CaO(s), CO(g), CO₂(g), H₂(g) and H₂O(g) from the NIST–JANAF Thermochemical Tables.

Question 2: Entropy and Enthalpy of Lead Through Its Melting Point (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $S^\circ_{298}(\text{Pb,s})=64.8$ J K−1mol−1; $T_m=327.4\,{}^\circ\text{C}=600.4$ K; $\Delta H_{fus}=4770.0$ J mol−1; $C_p(\text{Pb,s})=22.13+0.012T+1.0\times10^{-5}T^2$; $C_p(\text{Pb,l})=32.51-0.003T$ (J K−1mol−1, T in K). Path: 298 K (solid) → 600.4 K (solid, heating) → 600.4 K (fusion) → 873 K (liquid, heating).

Find. (a) $S^\circ_{873\text{K}}(\text{Pb})$; (b) $\Delta H$ for Pb from 298 K to 873 K.

Approach. Build the path solid(298 K)→solid($T_m$)→liquid($T_m$)→liquid(873 K); integrate $C_p/T$ for entropy and $C_p$ for enthalpy over each sensible-heat leg, and add $\Delta H_{fus}/T_m$ (entropy) or $\Delta H_{fus}$ (enthalpy) for the phase change.

  1. (a) Entropy — heating the solid, 298 K→600.4 K. $\Delta S=\displaystyle\int_{298}^{600.4}\dfrac{C_p(\text{Pb,s})}{T}\,dT=22.13\ln\!\left(\dfrac{600.4}{298}\right)+0.012(600.4-298)+\dfrac{1.0\times10^{-5}}{2}(600.4^2-298^2)$ $$=15.50+3.63+1.36=\boxed{20.49\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$
  2. Entropy of fusion. $\Delta S_{fus}=\dfrac{\Delta H_{fus}}{T_m}=\dfrac{4770.0}{600.4}=\boxed{7.94\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$
  3. Entropy — heating the liquid, 600.4 K→873 K. $\Delta S=\displaystyle\int_{600.4}^{873}\dfrac{C_p(\text{Pb,l})}{T}\,dT=32.51\ln\!\left(\dfrac{873}{600.4}\right)-0.003(873-600.4)=12.17-0.82=\boxed{11.35\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$
  4. (a) Total. $$S^\circ_{873\text{K}}=S^\circ_{298}+\Delta S_{\text{heat solid}}+\Delta S_{fus}+\Delta S_{\text{heat liquid}}=64.8+20.49+7.94+11.35=\boxed{104.6\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$
  5. (b) Enthalpy — heating the solid. $\Delta H=\displaystyle\int_{298}^{600.4}C_p(\text{Pb,s})\,dT=22.13(600.4-298)+0.006(600.4^2-298^2)+\dfrac{1.0\times10^{-5}}{3}(600.4^3-298^3)$ $$=6692.1+1630.1+633.2=\boxed{8955.4\ \text{J}\,\text{mol}^{-1}}.$$
  6. (b) Enthalpy — heating the liquid. $\Delta H=\displaystyle\int_{600.4}^{873}C_p(\text{Pb,l})\,dT=32.51(873-600.4)-0.0015(873^2-600.4^2)=8862.2-602.5=\boxed{8259.8\ \text{J}\,\text{mol}^{-1}}.$
  7. (b) Total. $$\Delta H_{298\rightarrow873}=8955.4+4770.0+8259.8=\boxed{21{,}985.2\ \text{J}\,\text{mol}^{-1}\approx22.0\ \text{kJ}\,\text{mol}^{-1}}.$$
QuantityValue
ΔS (heat solid, 298→600.4 K)20.49 J K−1mol−1
ΔSfus7.94 J K−1mol−1
ΔS (heat liquid, 600.4→873 K)11.35 J K−1mol−1
(a) $S^\circ_{873\text{K}}$104.6 J K−1mol−1
(b) $\Delta H_{298\rightarrow873\text{K}}$21,985 J mol−1 (≈22.0 kJ mol−1)