Question

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?

Answer 1

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

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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")`