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Consider the three compounds cyclohexane, C_6H_12, cyclohexene, c_6H_10 and benz

ID: 946656 • Letter: C

Question

Consider the three compounds cyclohexane, C_6H_12, cyclohexene, c_6H_10 and benzene, C_6H_6 Calculate the enthalpy change for the hydrogenation reaction of gaseous cyclohexene to form gaseous cyclohexane. C_6H_10(g) + H_2(g) rightarrow C_6H_12(g) Both cyclohexene and cyclohexane are liquids at standard conditions. The enthalpies of vaporization of cyclohexene and cyclohexane are +33.5 kJ mol^-1 and 32.0 kJ mol^-1 respectively. Use these values to calculate the enthalpy of hydrogenation of cyclohexene, C_6H_10(I) + H_2(g) rightarrow C_6H_12(I) and compare your value with the literature value of -121.9 kJ mol^-1. Assuming that beneze consists of alternate C=C and C-C bonds, use average bond enthalpies to calculate the enthalpy of hydrogen of beneze, C_6H_6(I) + 3H_2(g) rightarrow C_6H_12(I). (Benzene is a liquid at standard conditions and the enthalpy of vaporization of benzene is +33.9 kJ mol^-1) It might be expected that the enthalpy of hydrogenation of benzene would be three times the value of the enthalpy of hydrogenation of cyclohexene as three double bonds are hydrogenated. In facts the value is -210 kJ mol^-1. The reason for the difference is that the alternative double and single bonds give increased stability to the benzene ring through a process known as delocalization. Estimated the value of the delocalization energy of benzene is kJ mol^-1.

Explanation / Answer

(a) Enthalpy change for hydrogenation of cyclohexene

dHo = dHo(products) - dHo(reactants)

       = (-123) - (-5.4)

       = -117.6 kJ/mol

(b) Enthalpy of hydrogenation of cyclohexene

= (-156.2) - (-38.8)

= -117.4 kJ/mol

The value is slightly lower than literature value of -121.9 kJ/mol

(c) Enthalpy of hydrogenation using bond energies for benzene to cyclohexane

in benzene we are breaking three C=C bonds and forming 6 C-H bonds

enthalpy change = (3 x 614) - (6 x 413) = -636 kJ/mol

(d) delocalisation energy of benzene

= -636 - (-210)

= -426 kJ/mol