Calculate the standard heat of reaction for the following reaction: the hydrogenation of benzene to cyclohexane. (1) C6H6(g) + 3H2(g) → C6H12(g) (2) C6H6(g) + 7 1/2 O2(g) → 6CO2(g) + 3H2O(l) ΔH°c = -3287.4 kJ (3) C6H12(g) + 9 O2 → 6CO2(g) + 6H2O(l) ΔH°c = -3949.2 kJ (4) C(s) + O2(g) → CO2(g) ΔH°c

Question

Calculate the standard heat of reaction for the following reaction: the hydrogenation of
benzene to cyclohexane.
(1) [latex]C_{6}H_{6}\left(g\right)+3H_{2}\left(g\right)\longrightarrow C_{6}H_{12}\left(g\right) [/latex]
(2) [latex]C_{6}H_{6}\left(g\right)+7\frac{1}{2}O_{2}\left(g\right)\longrightarrow 6CO_{2\left(g\right) }+3H_{2}O[/latex] (1)                [latex]\Delta x^{°}_{c}[/latex] = -3287.4 kJ
(3) [latex]C_{6}H_{12}\left(g\right)+9O_{2}\longrightarrow 6CO_{2\left(g\right) }+6H_{2}O[/latex] (1)                                       [latex]\Delta x^{°}_{c}[/latex] = -3949.2 kJ
(4) [latex]C\left(s\right)+O_{2} \left(g\right)\longrightarrow CO_{2} [/latex] [latex]\Delta x^{°}_{c}[/latex] = -393.12 kJ
(5) [latex]H_{2}\left(g\right)+\frac{1}{2} O_{2}\left(g\right)\longrightarrow H_{2}O[/latex] (1)           [latex]\Delta x^{°}_{c}[/latex] = -285.58 kJ

Note: unlike heats of formation, the standard state of water for heats of combustion is
liquid. Standard pressure and temperature are the same 25°C, 1 atm.

Accepted Answer

Method 1
Using the more general equation 3.26
[latex]\Delta H^{°}_{r} =\sum{\Delta H^{°}_{⨍}}[/latex] , products - [latex]\sum{\Delta H^{°}_{⨍}} reactants[/latex]

the enthalpy of formation of [latex]C_{6}H_{6}[/latex] and [latex]C_{6}H_{12}[/latex] can be calculated, and from these values the heat of reaction (1).

From reaction (2)
[latex]\Delta H^{°}_{c}\left(C_{6}H_{6} \right)[/latex] = 6× [latex]\Delta H^{°}_{c}\left(CO_{2} \right)[/latex] + 3 × [latex]\Delta H^{°}_{c}\left(H_{2}O \right)[/latex] - [latex]\Delta H^{°}_{⨍}\left(C_{6}H_{6} \right)[/latex]
3287.4 = 6(-393.12) + 3(-285.58) - [latex]\Delta H^{°}_{⨍}\left(C_{6}H_{6} \right)[/latex]

[latex]\Delta H^{°}_{⨍}\left(C_{6}H_{6} \right)[/latex] =-3287.4 - 3215.52 = [latex]\underline{\underline{71.88 kJ/mol} }[/latex]

[latex]\Delta H^{°}_{c}\left(C_{6}H_{12} \right)[/latex] =-3949.2 = 6(-393.12) + 6(-285.58)-[latex]\Delta H^{°}_{⨍}\left(C_{6}H_{12} \right)[/latex]
[latex]\Delta H^{°}_{⨍}\left(C_{6}H_{12} \right)[/latex] =3949.2 - 4072.28 = [latex]-\underline{\underline{123.06 kJ/mol} }[/latex]

[latex]\Delta H^{°}_{r}[/latex] = [latex]\Delta H^{°}_{⨍}\left(C_{6}H_{12} \right)[/latex] - [latex]\Delta H^{°}_{⨍}\left(C_{6}H_{6} \right)[/latex]
[latex]\Delta H^{°}_{r}[/latex] = (-123.06) - (71.88) = - [latex]\underline{\underline{195 kJ/mol} } [/latex]

Note: enthalpy of formation of [latex]H_{2}[/latex] is zero.

Method 2
Using equation (3.27)

[latex] \Delta H^{°}_{r} =\sum{\Delta H^{°}_{c}}[/latex], reactants - [latex]\sum{\Delta H^{°}_{c}}[/latex]

[latex]\Delta H^{°}_{r}[/latex]=([latex]\Delta H^{°}_{c}\left(C_{6}H_{6} \right)[/latex] ) + 3 × [latex]\Delta H^{°}_{c}\left((H_{2}) \right)[/latex] - [latex]\Delta H^{°}_{c}\left(C_{6}H_{12} \right)[/latex]

= (-3287.4 + 3(-285.88)) - (-3949.2) = -[latex]\underline{\underline{196 kJ/mol} }[/latex]
Heat of reaction - [latex]\Delta H^{°}_{r}[/latex]= [latex]\underline{\underline{196 kJ/mol} }[/latex]

Question 3.9: Calculate the standard heat of reaction for the following re...

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