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Question: At \( 300K \) the standard enthalpies of formation of \( {{C}_{6}}{{H}_{5}}COO{{H}_{(s)}},C{{O}_{2(g...

At 300K300K the standard enthalpies of formation of C6H5COOH(s),CO2(g){{C}_{6}}{{H}_{5}}COO{{H}_{(s)}},C{{O}_{2(g)}} and H2O(l){{H}_{2}}{{O}_{(l)}} are −408,−393-408,-393 and −286kJmol−1-286kJmo{{l}^{-1}} respectively. Calculate the heat of combustion of benzoic acid at constant volume.
(A) ΔH=−4891kJmol−1,ΔU=−3199.75kJmol−1\Delta H=-4891kJmo{{l}^{-1}},\Delta U=-3199.75kJmo{{l}^{-1}}
(B) ΔH=−3751kJmol−1,ΔU=−1000.75kJmol−1\Delta H=-3751kJmo{{l}^{-1}},\Delta U=-1000.75kJmo{{l}^{-1}}
(C) ΔH=−4501kJmol−1,ΔU=−3199.75kJmol−1\Delta H=-4501kJmo{{l}^{-1}},\Delta U=-3199.75kJmo{{l}^{-1}}
(D) ΔH=−3201kJmol−1,ΔU=−3199.75kJmol−1\Delta H=-3201kJmo{{l}^{-1}},\Delta U=-3199.75kJmo{{l}^{-1}}

Explanation

Solution

The heat of combustion of benzoic acid can be calculated using the reaction of the combustion of one mole of benzoic acid. The enthalpy of a reaction can be determined by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products. If a certain amount of heat is absorbed during the course of the reaction, the enthalpy of the reaction is said to be positive. If a certain amount of heat is released during the course of the reaction, the enthalpy of the reaction is said to be negative.

Complete step by step solution:
The chemical reaction for the combustion of benzoic acid can be written as:
C6H5COOH+152O2→7CO2+3H2O{{C}_{6}}{{H}_{5}}COOH+\dfrac{15}{2}{{O}_{2}}\to 7C{{O}_{2}}+3{{H}_{2}}O
Now we know that the enthalpy of a reaction is given by the formula, ΔrH=(ΔfH)products−(ΔfH)reactan⁡ts{{\Delta }_{r}}H={{\left( {{\Delta }_{f}}H \right)}_{products}}-{{\left( {{\Delta }_{f}}H \right)}_{reac\tan ts}}
Substituting the values we can
ΔrH=7ΔfH(CO2)+3ΔfH(H2O)−ΔfH(C6H5COOH){{\Delta }_{r}}H=7{{\Delta }_{f}}H\left( C{{O}_{2}} \right)+3{{\Delta }_{f}}H\left( {{H}_{2}}O \right)-{{\Delta }_{f}}H\left( {{C}_{6}}{{H}_{5}}COOH \right)
Substituting the values: ΔH=7×(−393)+3×(−286)−(−408)=−3201kJ\Delta H=7\times (-393)+3\times (-286)-(-408)=-3201kJ
⇒ΔH=−3201kJ\Rightarrow \Delta H=-3201kJ
We know that the exchange of heat at constant volume is equal to the change in internal energy of the system. So we can write the formula: ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta {{n}_{g}}RT
On solving we get value for ΔU\Delta U ;
ΔU=−3201+(7−152)×8.3141000×300\Delta U=-3201+\left( 7-\dfrac{15}{2} \right)\times \dfrac{8.314}{1000}\times 300
⇒ΔU=−3199.75kJ\Rightarrow \Delta U=-3199.75kJ
Therefore, correct answer is option D ΔH=−3201kJmol−1,ΔU=−3199.75kJmol−1\Delta H=-3201kJmo{{l}^{-1}}, \Delta U=-3199.75kJmo{{l}^{-1}} .

Note:
The enthalpy of formation of any element present in its standard state is always zero. In this reaction oxygen is present as a dioxygen molecule in the gaseous state. Hence, its enthalpy of formation is zero.