Question
Question: For the combustion of benzene to gaseous carbon dioxide and liquid water, \[\Delta H\] is more than ...
For the combustion of benzene to gaseous carbon dioxide and liquid water, ΔH is more than ΔU.
A.True
B.False
C.Ambiguous
D.Insufficient data
Solution
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.
Complete answer:
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.
ΔH=ΔU+Δ(PV)
ΔH=ΔU+PΔV+VΔP
When the chemical change takes place at a constant pressure then the above reaction can be written as: ΔH=ΔU+PΔV+VΔP
Hence, ΔH=ΔU+PΔV
In this case, ΔH 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,
PΔV=ΔnRT
Where Δnrepresents 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 PΔV=ΔnRTwe get,
ΔH=ΔU+ΔnRT
Step 4
The combustion of benzene to form carbon dioxide and water the reaction take place as the following:
2C6H6(l)+15O2(g)12CO2(g)+6H2O(l)
Δn=∑ngasprd−∑ngasreac
Δn=12−15=−3
So, the final entropy value comes as:
ΔH=ΔU−3RT
Hence, the value of ΔH is less than ΔU. So, we can understand that the option b) is the correct answer for the given question
Note: The conditions under which ΔH=ΔU are when Δn=0 i.e there is no change in the stoichiometric coefficient values of the reactants and the products,ΔV=0 the reaction takes place at a constant volume and the reaction does not involve any gaseous reactants or products.