Question

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?

Answer 1

I got `-"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.

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Well, at constant atmospheric pressure,

`DeltaH = DeltaE + PDeltaV`

Thus, if we know the change in volume, and the pressure, we would be able to get `DeltaE_(rxn)^@` from `DeltaH_(rxn)^@`... you did not specify the pressure, so it is on you to decide what pressure makes sense.

I will choose `"1 bar"`, a completely arbitrary choice.

`DeltabarV = barV_(NH_4Cl) - (barV_(NH_3(g)) + barV_(HCl(g)))`

And of course, if we are to know the volumes, we assume 1 mol of each substance... this gives a molar volume `barV` based on the densities.

`rho_(NH_4Cl(s)) = "1.53 g/cm"^3 = "1530 g/dm"^3` at `"298.15 K"`

For ideal gases, we assume that `M/rho = (RT)/P ~~ barV`. Thus, their molar volumes are:

`barV_(NH_4Cl(s)) = M/(rho_(NH_4Cl(s))) = ("53.4916 g/mol")/("1530 g/L")`

`" "" "" "" "=` `"0.03496 L/mol"`

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

`" "" "" "" "~~` `"24.79 L/mol"`

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

`" "" "" "" "~~` `"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")`

`= -"49.54 L/mol"`

As a result, at `"1 bar"` pressure for one mol of each substance...

`color(blue)(DeltabarE_(rxn)^@) = DeltabarH_(rxn)^@ - PDeltabarV_(rxn)`

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

`=` `color(blue)(-"34.70 kJ/mol")`