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

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

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Answer 1 · 2017-07-10T19:31:11

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

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:

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

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

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

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

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

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