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

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 $$$$ngRT
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 $$$$ng= $6 - 7.5 = - 1.5$
HERE ,it is the difference between the number of gaseous molecules of product and the gaseous molecules of reactants.
$\Delta$U=- 3263.9$$$$\DeltaH= $\Delta$U+ \Delta $$$$ngRT
Here,we will now find the difference between the number of gaseous molecules of a product and the gaseous molecules of reactants.
\Delta $$$$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 = - 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.