1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '17 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. $\endgroup$ – Matt Jan 10 '17 at 0:47
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.