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Question: Calculate standard enthalpy of formation for benzene from the following data. \[{{\text{C}}_{6}}{{...

Calculate standard enthalpy of formation for benzene from the following data.
C6H6(l)+152O2(l)→6CO2(g)+3H2O(l) ΔH∘=-3267KJ{{\text{C}}_{6}}{{\text{H}}_{6(l)}}\text{+}\dfrac{15}{2}{{\text{O}}_{2(l)}}\to \text{6C}{{\text{O}}_{2(g)}}\text{+3}{{\text{H}}_{2}}{{\text{O}}_{(l)}}\text{ }\Delta {{\text{H}}^{\circ }}\text{=-3267KJ}

& {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(C}{{\text{O}}_{2}}\text{)= -393}\text{.5 KJ/mole} \\\ & {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(}{{\text{C}}_{2}}\text{O)= -285}\text{.8KJ/mole} \\\ \end{aligned}$$
Explanation

Solution

We calculate the calculate the enthalpy of the benzene by the formula as ΔH∘=ΣΔfH∘(products) -ΣΔfH∘(reactants)\Delta {{\text{H}}^{\circ }}=\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(products) -}\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(reactants)}, here ΔH∘ \Delta {{\text{H}}^{\circ }}~is the total enthalpy of the reaction and whose value is given as 3267 KJ mole−13267\text{ }KJ\text{ }mol{{e}^{-1}} and ΔfH∘CO2(g){{\Delta }_{f}}{{\text{H}}^{\circ }}\text{C}{{\text{O}}_{2(g)}}=−393.5 KJ mole−1-393.5\text{ }KJ\text{ }mol{{e}^{-1}} , ΔfH∘H2Ol{{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{H}}_{2}}{{\text{O}}_{l}}= −258.8 KJ mole−1-258.8\text{ }KJ\text{ }mol{{e}^{-1}} and ΔfH∘O2(g){{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{O}}_{2(g)}}=KJ mole−1KJ\text{ }mol{{e}^{-1}} . Now calculate its enthalpy.

Complete step by step answer:
First of all, what is the enthalpy of formation? From the enthalpy of formation, we simplify the total change in the enthalpy of the reaction when 1mole of the compound is formed from its constituents’ elements.
- We can easily calculate the enthalpy of benzene in the following reaction as:
C6H6(l)+152O2l→6CO2(g)+3H2O(l) ΔH∘=-3267KJ{{\text{C}}_{6}}{{\text{H}}_{6(l)}}\text{+}\dfrac{15}{2}{{\text{O}}_{2l}}\to \text{6C}{{\text{O}}_{2(g)}}\text{+3}{{\text{H}}_{2}}{{\text{O}}_{(l)}}\text{ }\Delta {{\text{H}}^{\circ }}\text{=-3267KJ}
ΔH∘=ΣΔfH∘(products) -ΣΔfH∘(reactants)\Delta {{\text{H}}^{\circ }}=\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(products) -}\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(reactants)}---------(A)
ΣΔfH∘(products)\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{(products)}= 6×ΔfH∘CO2(g)6\times {{\Delta }_{f}}{{\text{H}}^{\circ }}\text{C}{{\text{O}}_{2(g)}} + 3×ΔfH∘H2Ol3\times {{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{H}}_{2}}{{\text{O}}_{l}}---(1)
As we know that, ΔfH∘CO2(g){{\Delta }_{f}}{{\text{H}}^{\circ }}\text{C}{{\text{O}}_{2(g)}}=−393.5 KJ mole−1-393.5\text{ }KJ\text{ }mol{{e}^{-1}} (given)
ΔfH∘H2Ol{{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{H}}_{2}}{{\text{O}}_{l}}= −258.8 KJ mole−1-258.8\text{ }KJ\text{ }mol{{e}^{-1}}
Put these values in equation(1), we get:

\Delta {{\text{H}}^{\circ }}~=6\text{ }\left( -393.5 \right)\text{ }+\text{ }3\left( -258.8 \right)\text{ }KJ\text{ }mol{{e}^{-1}} \\\ ~~~~~~~~=~-2361\text{ }+-\text{ }857.4\text{ }KJ\text{ }mol{{e}^{-1}} \\\ ~~~~~~~~=\text{ }-3218.4\text{ }KJ\text{ }mol{{e}^{-1}} \\\ \end{array}$$ \- Now, we will calculate the enthalpy for the reactants $$\Sigma {{\Delta }_{f}}{{\text{H}}^{\circ }}(\text{reactants)}$$ = $6\times {{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{C}}_{6}}{{\text{H}}_{6(l)}}$ + $\dfrac{15}{2}\times {{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{O}}_{2(g)}}$ As we know that, ${{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{C}}_{6}}{{\text{H}}_{6(l)}}$= $$x\text{ }KJ\text{ }mol{{e}^{-1}}$$ (given) ${{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{O}}_{2(g)}}$= $$0KJ\text{ }mol{{e}^{-1}}$$ \- Put all the values in equation(A), we get: $$\Delta {{\text{H}}^{\circ }}~$$=$1012.5\text{ }+{{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{C}}_{6}}{{\text{H}}_{6(l)}}$ KJ/ mole \- As we know that ,$$\Delta {{\text{H}}^{\circ }}~$$= $$-3267\text{ }KJ\text{ }mol{{e}^{-1}}$$, then; $-3267\text{ }=\text{ }-3218.4\text{ }+{{\Delta }_{f}}{{\text{H}}^{\circ }}{{\text{C}}_{6}}{{\text{H}}_{6(l)}}\text{ }KJ\text{ }mol{{e}^{-1}}$ $$-3267\text{ }=\text{ }-3218.4\text{ }+\text{ }x\text{ }KJ\text{ }mol{{e}^{-1}}$$ $$x=\text{ }-\text{ }3267\text{ }+\text{ }3218.4\text{ =}-48.6~KJ\text{ }mol{{e}^{-1}}$$ \- Thus, the standard enthalpy of formation for benzene from the reaction; $${{\text{C}}_{6}}{{\text{H}}_{6(l)}}\text{+}\dfrac{15}{2}{{\text{O}}_{2l}}\to \text{6C}{{\text{O}}_{2(g)}}\text{+3}{{\text{H}}_{2}}{{\text{O}}_{(l)}}\text{ }\Delta {{\text{H}}^{\circ }}\text{=-3267KJ}$$ is: $$-48.6~KJ\text{ }mol{{e}^{-1}}$$. **Note:** The enthalpy of formation of oxygen is taken as zero in the above reaction because when the elements are present in their molecular form like oxygen gas, or in any solid form etc. their standard enthalpy of formation is always taken as zero as they undergo no change in their formation.