# The standard enthalpy of reaction, DeltaH^@"_(rxn), for the reaction NH_3(g) + HCl(g) -> NH_4Cl(s) is -39.65 kJ. How do you determine the value of DeltaE^@"_(rxn) for this reaction at 298 K?

Aug 2, 2017

I got $- \text{34.70 kJ/mol}$ at constant pressure for 1 mol of each substance. This is significantly different from the enthalpy of reaction, as the volume of the system significantly decreased.

Well, at constant atmospheric pressure,

$\Delta H = \Delta E + P \Delta V$

Thus, if we know the change in volume, and the pressure, we would be able to get $\Delta {E}_{r x n}^{\circ}$ from $\Delta {H}_{r x n}^{\circ}$... you did not specify the pressure, so it is on you to decide what pressure makes sense.

I will choose $\text{1 bar}$, a completely arbitrary choice.

$\Delta \overline{V} = {\overline{V}}_{N {H}_{4} C l} - \left({\overline{V}}_{N {H}_{3} \left(g\right)} + {\overline{V}}_{H C l \left(g\right)}\right)$

And of course, if we are to know the volumes, we assume 1 mol of each substance... this gives a molar volume $\overline{V}$ based on the densities.

${\rho}_{N {H}_{4} C l \left(s\right)} = {\text{1.53 g/cm"^3 = "1530 g/dm}}^{3}$ at $\text{298.15 K}$

For ideal gases, we assume that $\frac{M}{\rho} = \frac{R T}{P} \approx \overline{V}$. Thus, their molar volumes are:

${\overline{V}}_{N {H}_{4} C l \left(s\right)} = \frac{M}{{\rho}_{N {H}_{4} C l \left(s\right)}} = \left(\text{53.4916 g/mol")/("1530 g/L}\right)$

$\text{ "" "" "" } =$ $\text{0.03496 L/mol}$

barV_(NH_3(g)) = (RT)/P = (("0.083145 L"cdot"bar/mol"cdot"K")("298.15 K"))/("1 bar")

$\text{ "" "" "" } \approx$ $\text{24.79 L/mol}$

barV_(HCl(g)) = (RT)/P = (("0.083145 L"cdot"bar/mol"cdot"K")("298.15 K"))/("1 bar")

$\text{ "" "" "" } \approx$ $\text{24.79 L/mol}$

So, for one mol of each of these substances, the change in molar volume of the system due to the reaction is:

DeltabarV = "0.03496 L/mol" - ("24.79 L/mol" + "24.79 L/mol")

$= - \text{49.54 L/mol}$

As a result, at $\text{1 bar}$ pressure for one mol of each substance...

$\textcolor{b l u e}{\Delta {\overline{E}}_{r x n}^{\circ}} = \Delta {\overline{H}}_{r x n}^{\circ} - P \Delta {\overline{V}}_{r x n}$

= -"39.65 kJ/mol" - (cancel"1 bar" xx -49.54 cancel"L""/mol" xx ("0.008314472 kJ")/(0.083145 cancel"L"cdotcancel"bar"))

$=$ $\textcolor{b l u e}{- \text{34.70 kJ/mol}}$