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 (graphite) and H2 (g) are found in data tables at the end of the textbook, as well as values for the enthalpy of formation of benzene at 298K, and that for the enthalpy of vaporization of benzene at 353K. (I don't have my textbook)

Explanation / Answer

dH = dU + dn RT

dH = 8 * 10^3 J / mol + (2-0) * 8.314 * 583

dH = 17.694 kJ /mol

Cr(C6H6)2 (s) Cr(s) + 2 C6H6 (g) dH = 17.694 kJ/mol

Cr(s) + 2 C6H6 (g) Cr(C6H6)2 (s) dH= -17.694 kJ/mol

dH formation = dH products - dH reactants

dH f = (-17694 J) - (2*81.67 J + 0)

dH f = -17.857 kJ

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