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Question: The difference between heat of combustion of liquid benzene at constant pressure and at constant vol...

The difference between heat of combustion of liquid benzene at constant pressure and at constant volume at 300 K is (R=8.314 JKml)\left( {R = 8.314\,J\dfrac{K}{{ml}}} \right)
A.-3741 J
B.3741 J
C.37.41 KJ
D.-37.41 KJ

Explanation

Solution

The equation for combustion of liquid benzene is given by,
C6H6(l)+152O2(g)→6CO2(g)+3H2O(l){C_6}{H_6}\left( l \right) + \dfrac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( l \right)
Since, the temperature is given in the question. So further we can calculate the no of moles. We know that (ΔH−ΔU)(\Delta H-\Delta U) is the difference between heat of combustion at constant pressure and at constant volume. So, we can substitute the values in the equation ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT and calculate (ΔH−ΔU)(\Delta H-\Delta U).

Complete step by step answer:
Given in the question is,
Temperature, T=300KT = 300K
Combustion of benzene-
C6H6(l)+152O2(g)→6CO2(g)+3H2O(l){C_6}{H_6}\left( l \right) + \dfrac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( l \right)
Δng=np−nr=6−152=−32\Delta {n_g} = {n_p}-{n_r} = 6 - \dfrac{{15}}{2} = \dfrac{{ - 3}}{2}
We know,
The relation between ΔH\Delta H and ΔE\Delta E is given by,
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
⇒\Rightarrow ΔH−ΔU=ΔngRT\Delta H-\Delta U = \Delta {n_g}RT
⇒\Rightarrow ΔH−ΔU\Delta H-\Delta U is the difference between heat of combustion at constant pressure and at constant volume

Substituting the values in the above equation,
ΔH−ΔU=ΔngRT\Delta H-\Delta U = \Delta {n_g}RT
⇒\Rightarrow ΔH−ΔU=−32×8.314×300\Delta H-\Delta U = \dfrac{{ - 3}}{2} \times 8.314 \times 300 (Since, R=8.314 JKmlR = 8.314\,J\dfrac{K}{{ml}} )
⇒\Rightarrow ΔH−ΔU=−3741 J\Delta H-\Delta U = - 3741\,J

Therefore, the correct answer is option (A).

Note: The change in the internal energy of a system is the sum of the heat transferred and work done. At constant pressure, heat flow and the internal energy are related to the system’s enthalpy. The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU\Delta U.