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Question: Calculate standard enthalpy of formation of benzene if, $\Delta_c$H°(C$_6$H$_6$)$_{(l)}$ = -3267 kJ,...

Calculate standard enthalpy of formation of benzene if, Δc\Delta_cH°(C6_6H6_6)(l)_{(l)} = -3267 kJ, Δf\Delta_fH°(CO2_2)(g)_{(g)} = -393.5 kJ mol−1^{-1} and Δf\Delta_fH°(H2_2O)(l)_{(l)} = -285.8kJ mol−1^{-1}.

A

-679.3 kJ mol−1^{-1}

B

-38.6 kJ mol−1^{-1}

C

48.6 kJ mol−1^{-1}

D

+32.67 kJ mol−1^{-1}

Answer

48.6 kJ mol−1^{-1}

Explanation

Solution

The combustion reaction for benzene is:

C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)

Using Hess’s law:

ΔcH∘=∑n ΔfH∘(products)−ΔfH∘(benzene)\Delta_cH^\circ = \sum n\,\Delta_fH^\circ(\text{products}) - \Delta_fH^\circ(\text{benzene})

Given:

ΔcH∘(C6H6)=−3267 kJ\Delta_cH^\circ(C_6H_6) = -3267\text{ kJ} ΔfH∘(CO2)=−393.5 kJ/mol,ΔfH∘(H2O)=−285.8 kJ/mol\Delta_fH^\circ(CO_2) = -393.5\text{ kJ/mol}, \quad \Delta_fH^\circ(H_2O) = -285.8\text{ kJ/mol}

Standard enthalpy of formation of O2O_2 is zero.

Calculate the total enthalpy of products:

6(−393.5)+3(−285.8)=−2361−857.4=−3218.4 kJ6(-393.5) + 3(-285.8) = -2361 - 857.4 = -3218.4\text{ kJ}

Now, apply Hess’s law:

−3267=−3218.4−ΔfH∘(C6H6)-3267 = -3218.4 - \Delta_fH^\circ(C_6H_6)

Solve for ΔfH∘(C6H6)\Delta_fH^\circ(C_6H_6):

ΔfH∘(C6H6)=−3218.4+3267=48.6 kJ/mol\Delta_fH^\circ(C_6H_6) = -3218.4 + 3267 = 48.6\text{ kJ/mol}