Answer to Question #208909 in Physical Chemistry for Nurudeen

Question

Calculate the heat of combustion of ethane if heat of formation is

33kJmol'

5. The heat of combustion of benzene, CoHscn and cyclohexane, CsHiżın are-

-3268 and - 3920kJmoirespectively. Calculate:

(a) the heat of hydrogenation of benzene to cyclohexane.

(5) the heat required to hydrogenate a double bond in benzene.

Answer 1

Combustion of ethane is described by the equation:

C2H6(g) + 3.5 O2(g) = 2 CO2(g) + 3 H2O(l)

the heat of combustion is equal to enthalpy change in this reaction:

$\Delta$_cH = - $\Delta$_fH(C2H6(g)) + 2 $\Delta$_fH(CO2(g)) + 3 $\Delta$_fH(H2O(l))

given that formation enthaplies are $\Delta$_fH(C2H6(g)) = 33 kJ/mole; $\Delta$_fH(CO2(g)) = -394 kJ/mole; $\Delta$_fH(H2O(l)) = -286 kJ/mole, the heat of combustion of ethane is:

Q = $\Delta$_cH = -33 - 2*394 - 3*286 = -1679 kJ/mole.

1) Combustion reaction of benzene is:

C6H6(l) + 7.5 O2(g) = 6 CO2(g) + 3 H2O(l)

combustion enthaply

$\Delta$_c1H = - $\Delta$_fH(C6H6(g)) + 6 $\Delta$_fH(CO2(g)) + 3 $\Delta$_fH(H2O(l))

2) Combustion reaction of cyclohexane is:

C6H12(l) + 9 O2(g) = 6 CO2(g) + 6 H2O(g)

combustion enthaply

$\Delta$_c2H = - $\Delta$_fH(C6H12(l)) + 6 $\Delta$_fH(CO2(g)) + 6 $\Delta$_fH(H2O(l))

3) Hydrogenation of benzene is:

C6H6(l) + 3 H2(g) = C6H12(g)

(a) heat of hydrogenation of benzene to cyclohexane is equal to enthalpy in reaction 3):

$\Delta$_h3H = - $\Delta$_fH(C6H6(l)) + $\Delta$_fH(C6H12(l)), which can be expressed from combustion enthalpies:

$\Delta$_h3H = - $\Delta$_fH(C6H6(l)) + $\Delta$_fH(C6H12(l)) = $\Delta$_c1H - $\Delta$_c2H - 3 $\Delta$_fH(H2O(l))

given that $\Delta$_c1H = -3268 kJ/mole and $\Delta$_c2H = - 3920 kJ/mole, and $\Delta$_fH(H2O(g)) = -286kJ/mole, the hydrogenation enthalpy of benzene is:

$\Delta$_h3H = 3268 - 3920 + 3*286= 206 kJ/mole

(b) Hydrogenation of one double bond in benzene is

C6H6(g) + H2(g) = C6H8(g)

Since the enthaply of hydogenation of benzene to cyclohexane corresponds to hydrogenation of three double bonds, the heat of hydagentaion of one bond is one-third of heat of benzene hydrogenation:

$\Delta$_hH = 1/3 $\Delta$_h3H = 206/3 = 68.67 kJ/mole