Question

The combustion of benzene(l) gives carbon dioxide(l) and water(l). Given that the heat of combustion of benzene at constant volume is -3263.9KJ per mol at \[{25^ \circ }C\],heat of combustion in KJ per mol of benzene at constant temperature will be :
(R=\[8.314\]J per Kg)

Answer 1

Hint: For the process of combustion of benzene to gaseous state we need to make sure that\[\Delta \]H should always be more than \[\Delta \]U. WE will further apply the following formula:
\[\Delta \]H= \[\Delta \]U+ \[\Delta \]\[ng\]RT
Where, \[\Delta \]H=Heat at constant pressure
\[\Delta \]U= heat at constant volume

Complete step-by-step answer:We will now try to write the given equation in the equation as shown below:
\[C6H6\left( l \right) + \dfrac{{15}}{2}O2\left( g \right) \to 6CO2\left( g \right) + 3H2O\left( l \right)\]
After writing this equation we will find \[\Delta \] \[ng\]
\[\Delta \]\[ng\]= \[6 - 7.5 = - 1.5\]
HERE ,it is the difference between the number of gaseous molecules of product and the gaseous molecules of reactants.
\[\Delta \]U=\[ - 3263.9\]\[\Delta \]H= \[\Delta \]U+ \[\Delta \]\[ng\]RT
Here,we will now find the difference between the number of gaseous molecules of a product and the gaseous molecules of reactants.
\[\Delta \]\[ng\]=\[ - 1.5\]
Now we are given, the value of R:
R=\[8.314\]J perK per mol
We will now put the values in the equation,and we get,
\[\Delta H= - 3263.9 + \left( { - 1.5} \right)8.314 \times \dfrac{1}{{1000}} \times 298 \\
  \Delta H = - 3267.6KJ \]
\[\Delta \]H=Heat at constant pressure
\[\Delta \]U= heat at constant volume

Note: $ {C_6}{H_6}$ is the chemical formula of benzene. It is a molecule consisting of 6 carbon atoms and 6 hydrogen atoms.The products of complete combustion are carbon dioxide and water, as is the case for all organic molecules.