Chapter 5: Problem 45
The standard entropy of \(\mathrm{Pb}(s)\) at \(298.15 \mathrm{K}\) is \(64.80 \mathrm{J}\) \(\mathrm{K}^{-1} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(s)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, s)}{\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=22.13+0.01172 \frac{T}{\mathrm{K}}+1.00 \times 10^{-5} \frac{T^{2}}{\mathrm{K}^{2}} \\] The melting point is \(327.4^{\circ} \mathrm{C}\) and the heat of fusion under these conditions is \(4770 . \mathrm{J} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(l)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, l)}{\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}}=32.51-0.00301 \frac{T}{\mathrm{K}} \\] a. Calculate the standard entropy of \(\mathrm{Pb}(l)\) at \(725^{\circ} \mathrm{C}\). b. Calculate \(\Delta H\) for the transformation \(\mathrm{Pb}\left(s, 25.0^{\circ} \mathrm{C}\right) \rightarrow\) \(\mathrm{Pb}\left(L, 725^{\circ} \mathrm{C}\right)\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.