# 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?

##### 1 Answer

I got

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

I will choose

#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

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

For ideal gases, we assume that

#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

#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")#