Question

2.9 The standard enthalpy of formation of the metallocene bis(benzene)chromium was measured in a calorimeter. It was found for the reaction Cr(C6H6)2(s) →
Cr(s) + 2 C6H6(g) that . Find the corresponding reaction enthalpy and estimate the standard enthalpy of formation of the compound at 583 K. The constant-pressure molar heat capacity of benzene is 136.1 J K−1 mol−1 in its liquid range and 81.67 J K−1 mol−1 as a gas.

Answer 1

Using Kirchoffs law and a temperature difference of Δ T = 583 − 298 = 285 K ΔT=583−298=285K, I calculated Δ C p = ∑ ( p r o d u c t s ) C p , m − ∑ ( r e a c t a n t s ) C p , m ΔCp=∑(products)Cp,m−∑(reactants)Cp,m for the reaction C 6 H 6 ( l ) → C 6 H 6 ( g ) C6H6(l)→C6H6(g), where the product molar heat capacity is that given for the gas, the reactant is that for the liquid. I then solved Δ f H ( T 2 ) = Δ f H ( T 1 ) + ∫ Δ C p ΔfH(T2)=ΔfH(T1)+∫ΔCp using Δ f H ( T 1 ) = 49.0 k J / m o l ΔfH(T1)=49.0 kJ/mol. Since there is a phase change, I wasn't sure how to include that information so I solved it the same as above, using Δ v a p H = 30.8 k J / m o l ΔvapH=30.8 kJ/mol for T 1 = 298 K T1=298 K, then added it to the above amount. Using the reaction enthalpy of 17.7 k J / m o l 17.7 kJ/mol and Hess's Law, I solved for the enthalpy of formation at 583 K for C r ( C 6 H 6 ) 2 Cr(C6H6)2 . Now, I understand some problems with this approach by I'm not sure how to solve it otherwise with the given information. The temperature range is massive, so Kirchoff's Law is expected to give a poor answer. I'm also not sure how to incorporate a phase change into Kirchoff's Law. Any help or guidance is appreciated.