Question

What is the value of n for combustion of $1$ mole of benzene , when both the reactants are products of the gases ?

Answer 1

To proceed with this question we need to know the combustion reaction. So, the combustion process is the reaction between the hydrocarbons like benzene with the oxygen , and the products formed out of this reaction are the carbon dioxide and the water.

Complete answer:
Combustion is a clean reaction i.e. it does not leave any residue behind it and results in the formation of the carbon dioxide and the water.
Now, this information will help us to write the balanced chemical equation for the given combustion reaction with benzene and then to calculate the $\Delta n$ .
$\Delta n$ is the difference in the number of moles between the product and the reactant.
Basically $\Delta n$ is the difference between the coefficients of gaseous products and the coefficients of the gaseous reactants.
So, let's write the balanced chemical equation of combustion of 1 mole of benzene, here the number of moles of reactants and the product plays an important role:
${C_6}{H_6}(g) + 7\dfrac{1}{2}{O_2}(g) \to 6C{O_2} + 3{H_2}O$
So, assuming as given in question that all products formed are in the gaseous phase, lets calculate the $\Delta n$ :
$\Delta n$ $=$ Number of moles of product – number of moles of reactant
\Delta n$$$$ = 6 + 3 - 1 - 7\dfrac{1}{2}
$\Delta n$ $= \dfrac{1}{2}$
Hence , the value of $\Delta n$ for the combustion of $1$ mole of benzene is $\dfrac{1}{2}$.
Therefore, the correct answer is option D.

Note:
In the above reaction, we have first added the coefficients or the number of moles of the product i.e. six moles of carbon=dioxide and the three moles of water. Whereas on the reactant side , first we will add one mole of benzene and $7\dfrac{1}{2}$ moles of the oxygen. Then we will proceed as the Number of moles of product – number of moles of reactant.