Relation between ΔH and ΔE for the reaction in gaseous phase at constant temperature and pressure


> 
> Which of the following statements is correct for the reaction at constant temperature and pressure:
> 
> 
> $$\ce{CO(g) + \frac{1}{2}O\_2(g) -> CO2(g)}$$
> 
> 
> 1. $\Delta H = \Delta E$
> 2. $\Delta H > \Delta E$
> 3. $\Delta H < \Delta E$
> 4. None of these.
> 
> 
> 


Since this occurs at constant temperature, $\Delta E = 0$ and $W < 0$. So, $\Delta H$ turns out to be less than $\Delta E$, but this is not the answer. Why?




By definition, $$\Delta H=\Delta E+\Delta(PV)$$For a reacting ideal gas mixture at constant temperature, $$\Delta(PV)=(\Delta n)RT$$where $\Delta n$ is the change in the number of moles between reactants and products. Therefore, $$\Delta H=\Delta E+(\Delta n)RT$$For the reaction under consideration, $$\Delta n=-\frac{1}{2}$$


 2

$$\Delta H=\Delta E + p \cdot \Delta V$$


$\Delta E$ is energy change of a system at constant volume.


$\Delta H$ as is energy change of a system at constant pressure.


As the molar amount of gases is decreased by the reaction, $p \cdot \Delta V \lt 0$ at constant $T,p$.


Therefore, $\Delta H \lt \Delta E$


Neither that $\Delta E<0$ even at constant $T$, as energy released by the combustion reaction is dissipated in the system surrounding.


 1