For the aqueous reaction of #"100 mL"# of #"0.50 M"# ammonia with #"300 mL"# of #"0.50 M"# hydrochloric acid to form ammonium chloride, if #"HCl"# is in excess, and the solution heated up by #"1.6 K"#, what is the enthalpy of reaction?

1 Answer
Jul 10, 2017

#DeltaH_(rxn) = -"53.56 kJ/mol"#.


The reaction is an acid-base neutralization:

#"NH"_3(aq) + "HCl"(aq) -> "NH"_4"Cl"(aq)#

(the #"NH"_3(aq)# is introduced as #"NH"_4"OH"(aq)#, which equilibrates to be primarily #"NH"_3#.)

This reaction generates heat, and the heat transferred out into the solution from the reaction at constant pressure is given by

#q_P = mC_PDeltaT#,

where:

  • #q_P# is the heat flow with respect to the solution at constant pressure, i.e. lab bench conditions.
  • #m# is the mass of the solution in #"g"#.
  • #C_P# is the specific heat capacity of water at this #T# and #P#.
  • #DeltaT# is the change in temperature in either #""^@ "C"# or #"K"# (why can we say that?).

We know that the #"HCl"# is in excess because we were told that, so the #"NH"_3(aq)# is clearly the limiting reagent. That means it must be equimolar with #"NH"_4"Cl"# as per the reaction stoichiometry above.

#"0.50 mol NH"_3/cancel"L soln" xx 100 cancel"mL soln" xx cancel"1 L"/(1000 cancel"mL")#

#= "0.050 mols NH"_3 = "0.050 mols NH"_4"Cl"#

The total volume of the solution is

#"100 mL NH"_3 + "300 mL HCl" ~~ "400 mL"#,

if the solution volumes are assumed to be truly additive.

We can also assume the density and specific heat capacity of the solution is the same as for water at #25^@ "C"#. For simplicity, we can use #"1.00 g/mL"# and #"4.184 J/g" cdot "K"#, respectively.

Thus, the approximate mass of the solution is

#400 cancel"mL" xx "1.00 g"/cancel"mL" ~~ "400 g"#,

So, for the heat transferred into the solution (coming out from the reaction!) at constant pressure to increase its temperature is given by:

#q_P = (400 cancel"g")("4.184 J/"cancel"g"cdot cancel"K")(1.6 cancel"K")#

#=# #"2677.76 J"#

But for the REACTION, we have, from conservation of thermal energy...

#q_(rxn) + q_(P) = 0#.

So, at constant pressure (there's a reason why I keep saying this!), where the heat flow is DEFINED to be the change in enthalpy in the same units, we actually have:

#DeltaH_(rxn) = q_(rxn) = -q_P#

And thus, with just #q_P#, we currently have the negative change in enthalpy of reaction in #"J"#. But traditionally we report #DeltaH_(rxn)# in #"kJ/mol"# for the reaction.

Now hopefully you recognize that this is why we calculated the mols of #"NH"_4"Cl"# before:

#color(blue)(DeltaH_(rxn)) -= q_(rxn)/(n_("NH"_4"Cl"))#

#= -q_(P)/(n_("NH"_4"Cl"))#

#= -(2677.76 cancel"J")/("0.050 mols NH"_4"Cl") xx "1 kJ"/(1000 cancel"J")#

#=# #color(blue)(-"53.56 kJ/mol")#