Thermodynamics question....please solve the second part. i know the reaction ent

Question

Thermodynamics question....please solve the second part. i know the reaction enthalpy is 17.694 kj/mol..i dont know how to solve for the standard enthalpy of formation of the compound bis(benzene)chromium at 583 K. I am pretty sure you have to use all the data given...?

You are interested in determining the standard enthalpy of formation of the compound called bis(benzene)chromium [Cr(C6H6)2 (s)] at 583 K.

Through the use of a calorimeter, it was found that for the reaction

Cr(C6H6)2 (s) Cr(s) + 2 C6H6 (g)        rU (583 K) = + 8.0 kJmol-1.

Find the corresponding reaction enthalpy, and estimate the standard enthalpy of formation of the compound bis(benzene)chromium at 583 K.

The constant-pressure molar heat capacity of liquid benzene [Cp,m (l)] is 136.1 JK-1mol-1 , and Cp,m (g) = 81.67 JK-1mol-1 for gaseous benzene; both Cp,m are considered constant over the temperature range considered. The boiling point of benzene is 353 K.

Values for constant-pressure molar heat capacities at 298 K for carbon is 12.011M/g/mol) (graphite) and H2 (g) (2.016 M/g/mol) values for the enthalpy of formation of benzene is (49.2 kj mol^-1) at 298K, and that for the enthalpy of vaporization of benzene (87.19 JK^-1mol^-1) at 353K.

Explanation / Answer

H=deltaU+nRT

The first part is easy, H=+17.694kJ/mol.

Using Kirchoffs law and a temperature difference of T=583298=285K,

I calculated Cp=(products)Cp,m(reactants)Cp,m for the reaction 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

fH(T2)=fH(T1)+Cp using fH(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 vapH=30.8 kJ/mol for T1=298 K, then added it to the above amount.

Using the reaction enthalpy of 17.7 kJ/mol and Hess's Law, I solved for the enthalpy of formation at 583 K for Cr(C6H6)2 .
fH=49.0 kJ/mol for 6C(graphite)+3H2(g)C6H6(l).

6C(gr)+3H2(g)C6H6(l)

6C(gr)+3H2(g)C6H6(g)
Using the values Cp,m for hydrogen, carbon, and benzene in liquid form at 298 K, solve for fH at 583 K.

Then add in the molar enthalpy of vaporization to find fH for gaseous benzene?
If I solve it as above, I get an enthalpy of formation of the metallocene equal to 116.0 kJ/mol

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