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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?

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  • $\begingroup$ Constant T says nothing about delta E or delta H., as the system is not isolated. $\endgroup$ – Poutnik May 6 at 21:54
  • $\begingroup$ Constant temperature does not make the change in energy zero $\endgroup$ – Charlie Crown May 7 at 2:21
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    $\begingroup$ 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. $\endgroup$ – Chet Miller May 7 at 2:53
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    $\begingroup$ 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). $\endgroup$ – Eashaan Godbole May 7 at 18:18
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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, $$\Delta n=-\frac{1}{2}$$

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

$\Delta E$ 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.

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