# Question #fb528

Jan 7, 2017

$\Delta H > \Delta E$ for $\textcolor{red}{\left(2\right)}$, and $\Delta H \approx \Delta E$ for $\textcolor{red}{\left(1\right)}$.

For ideal gases, $\Delta H = \Delta E$ for $\textcolor{red}{\left(1\right)}$.

Recall that $\Delta H$ is the change in enthalpy and $\Delta E$ is the change in internal energy.

In an open container, we should recognize the condition of constant pressure (the phrase "open to the air", or "coffee-cup calorimeter", imply constant pressure as well). As a result, we can use this equation:

$\Delta H = \Delta E + \Delta \left(P V\right)$

$= \Delta E + P \Delta V + \cancel{V \Delta P + \Delta P \Delta V}$

$= \Delta E + P \Delta V$

where the result is something you should recognize (it was given in your textbook). Note that $E$ should not be confused with the total energy.

Your two reactions were:

$\text{H"_2(g) + "Br"_2(g) -> 2"HBr} \left(g\right)$ $\text{ "" "" "" "" "" } \textcolor{red}{\left(1\right)}$

${\text{C"(s) + 2"H"_2"O"(g) -> 2"H"_2(g) + "CO}}_{2} \left(g\right)$ $\text{ "" } \textcolor{red}{\left(2\right)}$

If we assume ideal gases, then in $\textcolor{red}{\left(1\right)}$ we have two $\text{mol}$s of gases going to two $\text{mol}$s of gases, and in $\textcolor{red}{\left(2\right)}$, we have two $\text{mol}$s of gases going to three $\text{mol}$s of gases.

Note that the change in volume due to gas formation is significantly more than due to liquid formation, for instance (gases take up way more space, since they generally have a density over 1000 times as small as liquids or solids).

In this approximation, $\textcolor{red}{\left(2\right)}$ has a nonnegative in volume, since $\Delta {n}_{g a s} > 0$, and $P \Delta V = \Delta {n}_{g a s} R T$ for a process at constant pressure containing ideal gases.

Therefore, $\Delta H > \Delta E$ for $\textcolor{red}{\left(2\right)}$, and $\Delta H \approx \Delta E$ for $\textcolor{red}{\left(1\right)}$.