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# For the combustion of benzene to gaseous carbon dioxide and liquid water, $\Delta H$ is more than $\Delta U$.A.TrueB.FalseC.AmbiguousD.Insufficient data

Last updated date: 10th Aug 2024
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Hint:
An energy change is always accompanied by a physical and chemical reaction. The total amount of energy associated with a fixed amount of a substance under a set of given conditions is also known as the internal energy of that substance.

Step 1
To study the heat changes at constant pressure a thermodynamic function enthalpy is used which is represented by H. The sum of the total internal energy and the pressure-volume energy of a system, under certain conditions is known as enthalpy. So, we can express it mathematically as $H = U + PV$.
Step 2
Enthalpy a thermodynamic parameter is a state function, so its change depends upon the initial and the final stage.
$\Delta H = \Delta U + \Delta (PV)$
$\Delta H = \Delta U + P\Delta V + V\Delta P$
When the chemical change takes place at a constant pressure then the above reaction can be written as: $\Delta H = \Delta U + P\Delta V + V\Delta P$
Hence, $\Delta H = \Delta U + P\Delta V$
In this case, $\Delta H$ is regarded as the measure of heat evolved or absorbed in the process occurring at constant pressure.
Step 3
From the gas equation, we know that,
$P\Delta V = \Delta nRT$
Where $\Delta n$represents the difference between the number of moles of gaseous products and the number of moles of gaseous reactants where we subtract the stoichiometric coefficient of the gaseous reactants from the products.
Substituting the value of $P\Delta V = \Delta nRT$we get,
$\Delta H = \Delta U + \Delta nRT$
Step 4
The combustion of benzene to form carbon dioxide and water the reaction take place as the following:
$2{C_6}{H_6}(l) + 15{O_2}(g)\xrightarrow{{}}12C{O_2}(g) + 6{H_2}O(l)$
$\Delta n = \sum\limits_{}^{} {{n_{gasprd}} - \sum\limits_{}^{} {{n_{gasreac}}} }$
$\Delta n = 12 - 15 = - 3$
So, the final entropy value comes as:
$\Delta H = \Delta U - 3RT$

Hence, the value of $\Delta H$ is less than $\Delta U$. So, we can understand that the option b) is the correct answer for the given question

Note:The conditions under which $\Delta H = \Delta U$ are when $\Delta n = 0$ i.e there is no change in the stoichiometric coefficient values of the reactants and the products,$\Delta V = 0$ the reaction takes place at a constant volume and the reaction does not involve any gaseous reactants or products.