Question

The dissolution of calcium chloride in water${\text{CaCl}}_2\left(s\right)\to {\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$is used in first-aid hot packs. In these packs, an inner pouch containing the salt is broken, allowing the salt to dissolve in the surrounding water.

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

(b) Suppose 20.0g CaCl2is dissolved in 0.100L water at role="math" localid="1663746929702" $20.0^{\circ }C$. Calculate the temperature reached by the solution, assuming it to be an ideal solution with a heat capacity close to that of 100gpure water (418JK^-1 ).

Short answer

(a) The enthalpy change, $\Delta H^{\circ }=-81.4kJ$.

(b) The final temperature of calcium chloride is 55.1⁰C.

Step-by-step solution

Given data

Dissolution of calcium chloride ${\text{CaCl}}_2\left(s\right)\to {\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$.

role="math" localid="1663747379938" $∆H_f^{\circ }$value of ${\text{Ca}}^{2+}\left(aq\right)$is $-542.83{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$value of ${\text{Cl}}^{-}\left(aq\right)$is $-167.18{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$value of ${\text{CaCl}}_2\left(s\right)$is $-795.8{\text{kJmol}}^{-1}$.

$∆H_f^{\circ }$Represents the standard enthalpy value.

$20.0gCaCl2$is dissolved in 0.100L water at 20.0⁰C.

Concept of Enthalpy

Enthalpy change is the amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in the 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, ${\text{CaCl}}_2\left(s\right)\to {\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$.

The enthalpy change is calculated as follows:

$\Delta H^{\circ }=n\Delta H_f^{\circ }\left(Products\right)-n{H}_f^{\circ }\left(Reac\tan ts\right)$

$\Delta H_f^{\circ }$Represents the standard enthalpy value.

localid="1663748757571" $$\begin{gathered} ∆H^{\circ }=∆H_f^{\circ }\left[C{a}^{2+}\right(aq\left)\right]+2∆H_f^{\circ }\left[C{l}^{-}\right(aq\left)\right]-∆H_f^{\circ }\left[CaC{l}_2\right(s\left)\right] \\ =\left(1mol\right)(-542.83kJmo{l}^{-1})+\left(2mol\right)(-167.18kJmo{l}^{-1})-\left(1mol\right)(-795.8kJmo{l}^{-1}) \\ =-542.83kJ-334.36kJ+795.8kJ \\ =-81.4kJ \end{gathered}$$

Calculate the molar mass of CaCl2

$\text{M}\left({\text{CaCl}}_2\right)=(\text{atomic mass of Ca) + 2}\times (\text{atomic mass of Cl})$

$$\begin{gathered} M\left({\text{CaCl}}_2\right)=\left(40.08{\text{gmol}}^{-1}\right)+2\times \left(35.45{\text{gmol}}^{-1}\right) \\ =40.08{\text{gmol}}^{-1}+70.90{\text{gmol}}^{-1} \\ =111.0{\text{gmol}}^{-1} \end{gathered}$$

Calculate the number of moles of CaCl2

Molar mass of CaCl₂ is 111.0gmol^-1.

Calculating moles of 20.0g CaCl₂.

$$\begin{gathered} n=\frac{20.0\text{g}}{111.0{\text{gmol}}^{-1}} \\ =0.180\text{mol} \end{gathered}$$

Calculate the value of ∆T

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.

$∆H^{\circ }=81.4kJ$ is associated with 1.00mol of CaCl₂, heat evolved associated with 0.180mol of CaCl₂is $0.180\text{mol}\times \frac{-81.4\text{kJ}}{1.00\text{mol}}=14.7\text{kJ}$.

Substitute these values in equation (1).

$$\begin{gathered} \Delta T=\frac{14.7\times {10}^3\text{J}}{418.{\text{JK}}^{-1}} \\ =35.1\text{K} \end{gathered}$$

Calculate the final temperature of CaCl2

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_i is the initial temperature.

$$\begin{gathered} 0^{\circ }C=273K \\ {T}_i=20.0^{\circ }C \\ ={(20.0+273)}^{\circ }C\times \frac{K}{\circ C} \\ =293K \end{gathered}$$

So, $\Delta T=35.1\text{K}$.

Substitute these values in equation (2).

$$\begin{gathered} T_{\text{f}}-293\text{K}=35.1\text{K} \\ {T}_{\text{f}}=35.1\text{K}+293\text{K}{T}_{\text{f}} \\ =328.1\text{K} \end{gathered}$$

Convert Kelvin to degree Celsius.

$$\begin{gathered} T_f=328.1K \\ =(328.1-273)K\times \frac{{}^{\circ }C}{K} \\ =55.1^{\circ }C \end{gathered}$$