# 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?

• Constant T says nothing about delta E or delta H., as the system is not isolated. – Poutnik May 6 at 21:54
• Constant temperature does not make the change in energy zero – Charlie Crown May 7 at 2:21
• For a chemically reacting system (or a single component change of phase), the internal energy of the system changes even if the temperature does not. – Chet Miller May 7 at 2:53
• Wanted to confirm if $\Delta E$ represents internal energy. Sorry, I'm not used to this notation (I've always used $\Delta U$ as internal energy). – Eashaan Godbole May 7 at 18:18

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, $$\Delta n=-\frac{1}{2}$$

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

$$\Delta E$$ 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.