Question

Ammonium nitrate dissolves in water according to the reaction

${\text{NH}}_4{\text{NO}}_3\left(s\right)\to {\text{NH}}_4^{+}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)$.

(a) Calculate the standard enthalpy change $\Delta H^{\circ }$for this reaction, using data from Appendix D.

(b) Suppose 15.0gNHNH₃ is dissolved in 0.100Lwater at 20.0⁰C. Calculate the temperature reached by the solution, assuming it to be an ideal solution with a heat capacity close to that 100g ofpure water(418JK^-1).

(c) From a comparison with the results of problem 43 and 44, suggest a practical application of this dissolution reaction?

Short answer

a) The answer is .

b) The answer is $T_f=7.5^{\circ }C$.

c) The dissolution of calcium chloride in water releases heat to increase the temperature of water (exothermic reaction) ammonium nitrate absorbs the heat (endothermic reaction).

Step-by-step solution

Given data

${\text{NH}}_4{\text{NO}}_3\left(s\right)\to {\text{NH}}_4^{+}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)$

$\Delta H_{\text{f}}^o$Value of ${\text{NH}}_4^{+}\left(aq\right)$is$-132.51{\text{kJmol}}^{-1}$

$\Delta H_{\text{f}}^o$Value of ${\text{NO}}_3^{-}\left(aq\right)$is $-205.0{\text{kJmolm}}^{-1}$

$\Delta H_{\text{f}}^o$Value of ${\text{NH}}_4{\text{NO}}_3\left(s\right)$is $-365.5.56{\text{kJmol}}^{-1}$

$\Delta H_{\text{f}}^e$Represents the standard enthalpy value.

15.0gNHNH₃ is dissolved in 0.100L water at 20.0⁰C with heat capacity (418JK^-1).

Concept of Enthalpy

Enthalpy change is amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in a chemical reaction.

It is represented by $\Delta H$.

Negative enthalpy change represents the exothermic reaction when energy is released from the reaction and positive enthalpy change represents an endothermic reaction in which energy is taken in from surroundings.

Calculate the enthalpy change ΔH∘

(a)

The given expression is, ${\text{NH}}_4{\text{NO}}_3\left(s\right)\to {\text{NH}}_4^{+}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)$.

The enthalpy change is calculated as:

$$\begin{gathered} \Delta H^{\circ }=\left(1mol\right)(-132.51kJmo{l}^{-1})+\left(1mol\right)(-205.0k{J}^{-1})-\left(1mol\right)(-365.5.56kJmo{l}^{-1} \\ =-132.51kJ-205.0kJ+365.56kJ \\ =+28.1kJ \end{gathered}$$

Calculate the molar mass of NH4NO3

(b)

Molar mass:

$$\begin{gathered} \text{M}\left({\text{NH}}_4{\text{NO}}_3\right)=2\times (\text{atomic mass of N) +}4\times (\text{atomic mass of}\text{H})+3\times (\text{atomic mass of}O) \\ M\left({\text{NH}}_4{\text{NO}}_3\right)=2\times \left(14.01{\text{gmol}}^{-1}\right)+4\times \left(1.008{\text{gmol}}^{-1}\right)+3\times \left(16.00{\text{gmol}}^{-1}\right) \\ =28.02{\text{gmol}}^{-1}+4.032{\text{gmol}}^{-1}+48.00{\text{gmol}}^{-1} \\ =80.05{\text{gmol}}^{-1} \end{gathered}$$

Calculate the number of moles of NH4NO3

Molar mass ofNH₄NO₃ is 80.05hmol^-1.

Calculating moles of 15.0gNH₄NO₃.

$$\begin{gathered} n=\frac{15.0\text{g}}{80.05\text{gmol}} \\ =0.187\text{mol} \end{gathered}$$

Calculate the value of ΔT

The change in temperature is calculated as:
$$\begin{gathered} q=C\Delta T \\ \Delta T=\frac{q}{C} \end{gathered}$$ .......... (1)

Where, q= Heat capacity, T=Change in temperature in K, C= Heat capacity of water.

Heat capacity of water = 418JK^-1.

$\Delta H^{\circ }=28.1kJ$ Is associated with 1.00mol of NH₄NO₃, heat absorbed associated with 0.187mol of NH₄NO₃ is $0.187\text{mol}\times \frac{28.1\text{kJ}}{1.00\text{mol}}=5.25\text{kJ}$.

Substitute these values in equation (1).

$$\begin{gathered} \Delta T=\frac{5.25\times {10}^3\text{J}}{418\text{JK}} \\ =-12.5\text{K} \end{gathered}$$

Calculate the final temperature of NH4NO3

The final temperature has been calculated by the formula.

$\Delta T=T_{\text{f}}-T_{\text{i}}$ ...... (2)

Where, T_f is the final temperature and T_iis the initial temperature.

$0^{\circ }C=273K$

$$\begin{gathered} T_i=20.0^{\circ }C \\ ={(20.0+273)}^{\circ }C\times \frac{K}{\circ C} \\ =293K \end{gathered}$$

$\Delta T=-12.5\text{K}$

Substitute these values in equation (2)

$$\begin{gathered} T_{\text{f}}-293\text{K}=-12.5\text{K} \\ {T}_{\text{f}}=-12.5\text{K}+293\text{K} \\ {T}_{\text{f}}=280.5\text{K} \end{gathered}$$

Convert Kelvin to degree Celsius;

$$\begin{gathered} T_f=280.5K \\ =(280.5-273)K\times {}^{\circ }\frac{C}{K} \\ =7.5^{\circ }C \end{gathered}$$

Compare the results

(c)

$N{H}_{4}N{O}_3\left(s\right)\to N{H}_4*\left(aq\right)+N{O}_3^{-}\left(aq\right)\Delta H^{\circ }=+28.1kJmo{l}^{-1}$ (Endothermic reaction)

role="math" localid="1663745873510" $CaC{l}_2\left(s\right)\to C{a}^2\left(aq\right)+2C{l}^{-}\left(aq\right)\Delta H^{\circ }=-81.4kJmo{l}^{-1}$ (Exothermic reaction)

The dissolution of calcium chloride in water releases heat to increase the temperature of water (exothermic reaction) ammonium nitrate absorbs the heat (endothermic reaction).