# When 10.0 g of ammonium nitrate is dissolved in 100 cm3 of water, the temperature of the solution decreases from 19.0°C to 10.5°C, how do I solve the following?

## a) Compare the total enthalpy of the reactants with that of the products and state which is greater. b) Classify this reaction as endothermic or exothermic. c) State which chemical bonds were broken when the ammonium nitrate dissolved in the water.

Jan 1, 2017

(a) ΔH_text(products) > ΔH_text(reactants); (b) endothermic; (c) ionic bonds between $\text{NH"_4^+}$ and $\text{NO"_3^"-}$.

#### Explanation:

(a) Enthalpy of formation of reactants and products

We have the reaction

${M}_{r} \text{/g·mol"^"-1} \textcolor{w h i t e}{m m m l l} 80.04$
$\textcolor{w h i t e}{m m m m m m m m} \text{NH"_4"NO"_3"(s)" ⇌ "NH"_4^"+""(aq)" + "NO"_3^"-""(aq)}$
Δ_fH^°"/kJ·mol"^"-1"color(white)(m)"-365.1"color(white)(mmmm)"-132.8"color(white)(mmm)"-206.6"

Δ_fH_text(reactants)^° = "-365.1 kJ.mol"

Δ_fH_text(products)^° = "(-132.8 - 206.6) kJ" = "-339.4 kJ/mol"

$\text{Moles of NH"_4"NO"_3 = 10.0 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × ("1 mol NH"_4"NO"_3)/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3)))) = "0.1249 mol NH"_4"NO"_3 = "0.1249 mol reaction}$

ΔH_text(reactants) = 0.1249 color(red)(cancel(color(black)("mol reaction"))) × "-365.1 kJ"/(1 color(red)(cancel(color(black)("mol reaction")))) = "-45.6 kJ"

ΔH_text(products) = 0.1249 color(red)(cancel(color(black)("mol reaction"))) ×"-339.4 kJ"/(1 color(red)(cancel(color(black)("mol reaction")))) = "-42.4 kJ"

ΔH_text(products) > ΔH_text(reactants)

(b) Exothermic or endothermic?

The temperature decreased when the ammonium nitrate dissolved in the water.

Energy left the system, so the reaction is endothermic.

(c) Bonds broken

The ionic bonds between the $\text{NH"_4^+}$ ions and $\text{NO"_3^"-}$ ions were broken when the ammonium nitrate dissolved in the water.