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I was looking for some help on understanding this practice question from my text book, I know how to figure out enthalpy, but this is my first temperature change question and I am stumped. Any guidance would be greatly appreciated.

Nitric acid is neutralized with potassium hydroxide in the following reaction:

$$\ce{NHO3(aq) + KOH3(aq) -> KNO3(aq) + H2O(l)}$$

$$\Delta_\mathrm rH= -53.4\ \mathrm{kJ/mol}\ \ce{HNO3}$$.

$55.0\ \mathrm{mL}$ of a $1.30\ \mathrm{mol/L}$ solutions of both reactants, at $21.40\ \mathrm{^\circ C}$, are mixed in a calorimeter. What is the final temperature of the mixture? Assume that the density of both solutions is $1.00\ \mathrm{g/mL}$. Also assume that the specific heat capacity of both solutions is the same as the specific heat capacity of water. No heat is lost to the calorimeter itself.

I know the answer is 29.7 degrees Celsius, I'm just not sure how to get there. I tried juggling around my equation "delta t r H= m c delta t" but since I don't know the final temperature i don't think the delta t works in this case. I've tried omitting the delta t but I get no where near 29.7 degrees Celsius.

So I took my formula Q=mc delta t and rearranged it, I now have delta t= Q / mc. Delta t= -53.4 kJ/mol / (55.0g)(4.19 J/g C) I changed the 55mL to g as the question says to assume the density is 1.00 g/mL. I also used the heat capacity as water given in my course data booklet). And I get -0.2317. I tried doubling the mass as there is two solutions but that doesn't work either

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  • $\begingroup$ How many moles of each reactant are in the reaction mixture? Are you aware that a negative heat of reaction means that heat is given off? $\endgroup$
    – Chet Miller
    Jan 9, 2017 at 22:02
  • $\begingroup$ Thank you for bringing up moles!! Once I figured out the moles I was able to use my equation. I will post my answer right away. I know my editing skills need work but thanks Loong for fixing my question. $\endgroup$
    – user39717
    Jan 10, 2017 at 0:47

1 Answer 1

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n(HNO3)= 0.055L * 1.30 mol\L = 0.0715 mol
Delta H = 0.0715 mol * -53.4 kJ/mol =3.818kJ
Delta H = mc delta t
3818J = (110g)(4.19J/g*C)(delta t)
Delta t = 8.3*C
Ti + Tf = 21.4*C + 8.3*C = 29.7*C

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