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Question: Calculate the standard enthalpy of formation of benzene from following data. $\Delta_fH^{\circ}(C_6...

Calculate the standard enthalpy of formation of benzene from following data.

ΔfH∘(C6H6)=−3200\Delta_fH^{\circ}(C_6H_6) = -3200 kJ,

ΔfH∘(CO2)=−400\Delta_fH^{\circ}(CO_2) = -400 kJ mol−1^{-1},

ΔfH∘(H2O)=−300\Delta_fH^{\circ}(H_2O) = -300 kJ mol−1^{-1}

A

50 kJ mol−1^{-1}

B

-70 kJ mol−1^{-1}

C

150 kJ mol−1^{-1}

D

-100 kJ mol−1^{-1}

Answer

-100 kJ mol−1^{-1}

Explanation

Solution

For the combustion of benzene:

C6H6+152O2→6 CO2+3 H2OC_6H_6 + \frac{15}{2} O_2 \to 6\,CO_2 + 3\,H_2O

Using Hess's law:

ΔcH∘=∑n ΔfH∘(products)−ΔfH∘(C6H6)\Delta_cH^\circ = \sum n\,\Delta_fH^\circ(\text{products}) - \Delta_fH^\circ(C_6H_6)

Given:

ΔcH∘=−3200 kJ,ΔfH∘(CO2)=−400 kJ/mol,ΔfH∘(H2O)=−300 kJ/mol\Delta_cH^\circ = -3200\ \text{kJ},\quad \Delta_fH^\circ(CO_2) = -400\ \text{kJ/mol},\quad \Delta_fH^\circ(H_2O) = -300\ \text{kJ/mol}

Calculate the total for products:

6(−400)+3(−300)=−2400−900=−3300 kJ6(-400) + 3(-300) = -2400 - 900 = -3300\ \text{kJ}

So,

−3200=−3300−ΔfH∘(C6H6)-3200 = -3300 - \Delta_fH^\circ(C_6H_6)

Rearrange to solve for ΔfH∘(C6H6)\Delta_fH^\circ(C_6H_6):

ΔfH∘(C6H6)=−3300+3200=−100 kJ/mol\Delta_fH^\circ(C_6H_6) = -3300 + 3200 = -100\ \text{kJ/mol}