91Ó°ÊÓ

Use the following data to calculate the standard molar entropy of \(\mathrm{CO}(\mathrm{g})\) at its normal boiling point. Carbon monoxide undergoes a solid-solid phase transition at \(61.6 \mathrm{K}\). Compare your result with the calculated value of \(160.3 \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\). Why is there a discrepancy between the calculated value and the experimental value? $$\begin{array}{r}\bar{C}_{p}\left[\mathrm{CO}\left(\mathrm{s}_{1}\right)\right] / R=-2.820+\left(0.3317 \mathrm{K}^{-1}\right) T-\left(6.408 \times 10^{-3} \mathrm{K}^{-2}\right) T^{2} \\\\+\left(6.002 \times 10^{-5} \mathrm{K}^{-3}\right) T^{3} \\\10 \mathrm{K} \leq T \leq 61.6 \mathrm{K}\end{array}$$ $$\begin{array}{r}\bar{C}_{P}\left[\mathrm{CO}\left(\mathrm{s}_{2}\right)\right] / R=2.436+\left(0.05694 \mathrm{K}^{-1}\right) T \\\61.6 \mathrm{K} \leq T \leq 68.1 \mathrm{K}\end{array}$$ $$\begin{array}{r}\bar{C}_{P}[\mathrm{CO}(1)] / R=5.967+\left(0.0330 \mathrm{K}^{-1}\right) T-\left(2.088 \times 10^{-4} \mathrm{K}^{-2}\right) T^{2} \\\68.1 \mathrm{K} \leq T \leq 81.6 \mathrm{K}\end{array}$$ and \(T_{\mathrm{trs}}\left(\mathrm{s}_{1} \rightarrow \mathrm{s}_{2}\right)=61.6 \mathrm{K}, T_{\mathrm{fus}}=68.1 \mathrm{K}, T_{\mathrm{vap}}=81.6 \mathrm{K}, \Delta_{\mathrm{fus}} \bar{H}=0.836 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\) \(\Delta_{\mathrm{trs}} \bar{H}=0.633 \mathrm{kJ} \cdot \mathrm{mol}^{-1}, \Delta_{\mathrm{vap}} \bar{H}=6.04 \mathrm{kJ} \cdot \mathrm{mol}^{-1}, \Theta_{\mathrm{D}}=79.5 \mathrm{K},\) and the correction for \\[\text { nonideality }=0.879 \mathrm{J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\\]