Question

The difference between $\Delta {\text{H}}$ and $\Delta {\text{U}}$ for the combustion reaction of benzene at ${127^\circ }{\text{C}}$ is:
(A) -200R
(B) -600R
(C) 200R
(D) 600R

Answer 1

Combustion is a high temperature exothermic chemical reaction involving a hydrocarbon or organic molecule reacting with oxygen to give carbon dioxide, water and evolution of heat. For example, the combustion of carbon can be expressed as:
${\text{C(s) + }}{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{C}}{{\text{O}}_{\text{2}}}{\text{(g)}}$
The change in enthalpy when one mole of a substance undergoes complete combustion is termed as enthalpy of combustion.
The enthalpy change $\Delta {\text{H}}$ of a chemical reaction is related to the internal energy change $\Delta {\text{U}}$ at constant pressure P with volume change $\Delta {\text{V}}$ by the relation:
$\Delta {\text{H = }}\Delta {\text{U + P}}\Delta {\text{V}}$

Complete step by step answer:
We need to find out the difference between $\Delta {\text{H}}$ and $\Delta {\text{U}}$ for the combustion reaction of benzene at ${127^\circ }{\text{C}}$ .
For an ideal gas, ${\text{PV = nRT}}$ , R is the gas constant and T is the temperature.
For reactants, ${\text{P}}{{\text{V}}_{\text{r}}}{\text{ = }}{{\text{n}}_{\text{r}}}{\text{RT}}$ at constant T and P.
For products, ${\text{P}}{{\text{V}}_{\text{p}}}{\text{ = }}{{\text{n}}_{\text{p}}}{\text{RT}}$ at constant T and P.
Here, ${{\text{n}}_{\text{r}}}$ and ${{\text{n}}_{\text{p}}}$are the number of moles of reactants and products respectively and ${{\text{V}}_{\text{r}}}$ and ${{\text{V}}_{\text{p}}}$ are the volume of reactants and products respectively.
Therefore, by subtraction we will get,

$${\text{P}}\left( {{{\text{V}}_{\text{p}}}{\text{ - }}{{\text{V}}_{\text{r}}}} \right){\text{ = }}\left( {{{\text{n}}_{\text{p}}} - {{\text{n}}_{\text{r}}}} \right){\text{RT}} \\\ \Rightarrow {\text{P}}\Delta {\text{V = }}\Delta {{\text{n}}_{\text{g}}}{\text{RT}} \\\$$

Here, $\Delta {{\text{n}}_{\text{g}}}$ is the difference in the number of moles of gaseous products and reactants.
Therefore, $\Delta {\text{H = }}\Delta {\text{U + }}\Delta {{\text{n}}_{\text{g}}}{\text{RT}}$
Now, the reaction for combustion of benzene $\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}} \right)$ is:
${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( {\text{l}} \right) + 15{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to 12{\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) + 6{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)$
Or, ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( {\text{l}} \right) + 7.5{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to 6{\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) + 3{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)$
Here,
$\Delta {{\text{n}}_{\text{g}}} = 6 - 7.5 \\\ \Rightarrow \Delta {{\text{n}}_{\text{g}}} = - 1.5 \\\$
Temperature,
${\text{T}} = {127^\circ }{\text{C}} \\\ \Rightarrow {\text{T}} = \left( {127 + 273} \right){\text{K}} \\\ \Rightarrow {\text{T}} = 400{\text{K}} \\\$
Using the relation $\Delta {\text{H = }}\Delta {\text{U + }}\Delta {{\text{n}}_{\text{g}}}{\text{RT}}$ , we have
$\Delta {\text{H = }}\Delta {\text{U + }}\left( { - 1.5} \right) \times {\text{R}} \times 400 \\\ \Rightarrow \Delta {\text{H - }}\Delta {\text{U = - 1}}{\text{.5}} \times {\text{400R}} \\\ \Rightarrow \Delta {\text{H - }}\Delta {\text{U = }} - 600{\text{R}} \\\$

So, the correct answer is B.

Note: When the change in volume is zero i.e., $\Delta {\text{V}} = 0$ or when none of the reactants or products are in gaseous states or when the number of moles of reactants is equal to the number of moles of products $\left( {\Delta {{\text{n}}_{\text{g}}} = 0} \right)$ , then change in enthalpy is equal to change in internal energy, i.e., $\Delta {\text{H = }}\Delta {\text{U}}$ . When $\Delta {{\text{n}}_{\text{g}}}$ is positive, $\Delta {\text{H > }}\Delta {\text{U}}$ and when $\Delta {{\text{n}}_{\text{g}}}$ is negative, $\Delta {\text{H < }}\Delta {\text{U}}$ .