1.24 g of LiCl(s) is dissolved in 24.0 mL of water that is initially 21.2°C.

1.24 g of LiCl(s) is dissolved in 24.0 mL of water that is initially 21.2°C.
a) Write the chemical equation for the dissolution of LiCl(s) into aqueous lithium and chloride ions.
b) Using the thermodynamic tables in Appendix II of Tro, calculate the molar enthalpy of dissolving LiCl(s) in water. Is this process endothermic or exothermic?
c) Calculate the amount of heat absorbed/released when 1.24 g of LiCl(s) dissolves.
d) Calculate the final temperature of the solution. You can assume the heat capacity of solution is equal to the heat capacity of water. (dwater = 1g/mL; Cp(water) = 4.184 J/g°C)

The Correct Answer and Explanation is:

Let’s work through each part of this problem in detail.

Part a) Chemical Equation for Dissolution

When lithium chloride (LiCl) dissolves in water, it dissociates into lithium (Li⁺) and chloride (Cl⁻) ions:
[
\text{LiCl(s)} \rightarrow \text{Li}^+(\text{aq}) + \text{Cl}^-(\text{aq})
]
This equation shows that solid LiCl separates into its ions when placed in water, a typical process for ionic compounds.

Part b) Molar Enthalpy of Dissolution (∆H_dissolution)

To find the molar enthalpy of dissolution, we refer to Appendix II in Tro’s thermodynamic tables, which lists the enthalpy of solution (ΔH_sol) for LiCl(s) as approximately -37.0 kJ/mol. This negative value indicates that the process is exothermic (releases heat).

Part c) Heat Absorbed or Released (q)

To calculate the amount of heat (q) released when 1.24 g of LiCl dissolves in water, we need to convert the mass of LiCl into moles and then use the molar enthalpy of solution.

1. Calculate moles of LiCl: First, find the molar mass of LiCl:
   [
   \text{Molar mass of LiCl} = 6.94 \, (\text{Li}) + 35.45 \, (\text{Cl}) = 42.39 \, \text{g/mol}
   ] Next, calculate moles of LiCl:
   [
   \text{Moles of LiCl} = \frac{1.24 \, \text{g}}{42.39 \, \text{g/mol}} \approx 0.0293 \, \text{mol}
   ]
2. Calculate the heat (q) released: Using the enthalpy of solution:
   [
   q = \text{moles of LiCl} \times \Delta H_{\text{sol}}
   ]
   [
   q = 0.0293 \, \text{mol} \times (-37.0 \, \text{kJ/mol})
   ]
   [
   q \approx -1.0841 \, \text{kJ} = -1084.1 \, \text{J}
   ]

Since ( q ) is negative, heat is released, confirming the process is exothermic.

Part d) Final Temperature of the Solution

To determine the final temperature of the solution, we use the relationship:
[
q = m \cdot C_p \cdot \Delta T
]
where:

• ( m ) = mass of the solution (since the density of water is 1 g/mL, 24.0 mL water ≈ 24.0 g)
• ( C_p ) = specific heat capacity of water = 4.184 J/g°C
• ( \Delta T ) = ( T_{\text{final}} – T_{\text{initial}} )

1. Rearrange to solve for ( T_{\text{final}} ): [
   T_{\text{final}} = T_{\text{initial}} + \frac{q}{m \cdot C_p}
   ]
2. Calculate ( T_{\text{final}} ): Substituting the values:
   [
   T_{\text{final}} = 21.2^\circ \text{C} + \frac{-1084.1 \, \text{J}}{24.0 \, \text{g} \times 4.184 \, \text{J/g°C}}
   ] [
   T_{\text{final}} = 21.2^\circ \text{C} – 10.8^\circ \text{C}
   ] [
   T_{\text{final}} \approx 10.4^\circ \text{C}
   ]

Summary Explanation

When 1.24 g of LiCl dissolves in water, the exothermic reaction releases approximately 1084.1 J of heat. This energy release causes the temperature of the solution to drop as the system adjusts to release this thermal energy. The final temperature of the solution is approximately 10.4°C, which is significantly lower than the initial temperature of 21.2°C due to the heat released during dissolution. This exercise illustrates the effect of exothermic processes on solution temperatures and how thermodynamic properties can predict temperature changes during dissolution.