Question 3.7: A 143.0 g sample of C(s) in the form of graphite is heated f...
A 143.0 g sample of C\left( s \right) in the form of graphite is heated from 300. to 600. K at a constant pressure. Over this temperature range,C_{P,m} has been determined to be
\frac{C_{P, m}}{J k^{-1}mol ^{-1}}=-12.19+0.1126\frac{T}{K} -1.947 × 10^{-4}\frac{T^{2}}{K^{2}}+1.919 ×10^{-7}\frac{T^{3}}{K^{3}}-7.800× 10^{11}\frac{T^{4}}{K^{4}}
Calculate \Delta H and q_{P}. How large is the relative error in \Delta H if we neglect the temperature-dependent terms in C_{P,m} and assume that C_{P,m} maintains its value at 300. K throughout the temperature interval?
=\frac{143.0g}{12.00 g mol^{-1}}\frac{J}{mol}\int_{300.}^{600.}\left( ^{-12.19+0.1126\frac{T}{K}-1.947 ×10^{-4}\frac{T^{2}}{K^{2}}+19.19}_{×10^{-7}\frac{T^{3}}{K^{3}}-7.800× 10^{-11}\frac{T^{4}}{T^{4}}}\right)d\frac{T}{K}
=\frac{134.0}{12.00}×\left[ ^{-12.19\frac{T}{K}+0.0563\frac{T^{2}}{K^{2}}-6.49×10^{-5}\frac{T^{3}}{K^{3}}+4.798}_{×10^{-8}\frac{T^{4}}{K^{4}}-1.56 ×10^{-11}\frac{T^{5}}{K^{5}}} \right]^{600.}_{300.}J=46.9 KJ
From Equation (3.28), \Delta H =q_{P}.
\left( U_{f}+P_{f}V_{f} \right)-\left( U_{i}+P_{i}V_{i} \right)q_{p} or \Delta H =q_{P} (3.28)
If we had assumed C_{p,m}=8.617 J mol^{-1} K^{-1}, which is the calculated value at 300.K, \Delta H =143.0 g/12.00 g mol^{-1} × 8.617 J K^{-1} mol^{-1} × \left[ 600. k -300. K \right]=30.8 KJ. The relative error is 100 × \left( 30.8 KJ-46.9 KJ \right)/46.9 KJ =-34.3\%.In this case, it is not reasonable to assume that C_{p,m} is independent of temperature.