You are interested in determining the standard enthalpy of formation of the comp

Question

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 (12.011M/g/mol) (graphite) and H2 (g) (2.016 M/g/mol) are found in data tables at the end of the textbook, as well as values for the enthalpy of formation of benzene (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

it was found that for the reaction

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

so enthalpy of formation of Cr(C6H6)2 (s) = - 8 KJ / mole as the two reactions are reverse of each other

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.