AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) 21. In an experiment a student mixes a 50.0 mL sample of — Thermodynamics Chemistry Question

STEPS:

1. Identify the temperature change ($\Delta T$) of the mixture:
The temperature of the reaction mixture increased from $20.0^\circ\text{C}$ to $21.0^\circ\text{C}$. The change in temperature is calculated as:
$$\Delta T = T_f - T_i = 21.0^\circ\text{C} - 20.0^\circ\text{C} = 1.0^\circ\text{C} \quad$$

2. Calculate the heat gained by the aqueous mixture ($q_{\text{solution}}$):
Use the calorimetry formula relating heat transfer to temperature change:
$$q = mc\Delta T \quad$$
Substitute the total mass of the mixture ($100.\text{ g}$), the specific heat capacity ($4.2\text{ J/(g }^\circ\text{C)}$), and the temperature change ($1.0^\circ\text{C}$) into the formula:
$$q_{\text{solution}} = (100.\text{ g})\left(4.2\text{ J/(g }^\circ\text{C)}\right)(1.0^\circ\text{C}) = 420\text{ J} = 0.42\text{ kJ} \quad$$

3. Relate the heat of the solution to the heat of the reaction ($q_{\text{rxn}}$):
Because the calorimeter is assumed to be perfectly insulated, the heat absorbed by the solution must equal the heat released by the chemical reaction. Since the temperature of the surroundings (the water) increased, the reaction is exothermic, meaning it released heat and has a negative enthalpy sign:
$$q_{\text{rxn}} = -q_{\text{solution}} = -0.42\text{ kJ}$$

4. Determine the moles of reaction ($n_{\text{rxn}}$) that occurred:
The moles of both $\text{AgNO}_3$ and $\text{NaCl}$ can be computed using their volumes and molarity ($\text{moles} = \text{Volume in L} \times \text{Molarity}$):
* $\text{Moles of AgNO}_3 = 0.0500\text{ L} \times 0.100\text{ M} = 0.00500\text{ mol} \quad$
* $\text{Moles of NaCl} = 0.0500\text{ L} \times 0.100\text{ M} = 0.00500\text{ mol} \quad$
Because they react in a $1:1$ stoichiometric ratio according to the balanced equation, they react completely without any excess reactant, meaning exactly $0.00500\text{ mol}$ of reaction occurred.

5. Calculate the molar enthalpy of the precipitation reaction ($\Delta H_{\text{rxn}}$):
To find the enthalpy change per mole of reaction, divide the heat change of the reaction by the moles of reaction that occurred:
$$\Delta H_{\text{rxn}} = \frac{q_{\text{rxn}}}{n_{\text{rxn}}} = \frac{-0.42\text{ kJ}}{0.00500\text{ mol}} = \mathbf{-84\text{ kJ/mol}_{\text{rxn}}} \quad$$
This confirms that Option A is the correct answer.

*

WHY_OTHERS_WRONG:

• Option B is incorrect: The value of $-0.42\text{ kJ/mol}_{\text{rxn}}$ represents only the total heat energy released by the physical sample mixture inside the coffee cup ($q_{\text{rxn}}$). It fails to divide by the moles of reaction to obtain a standard molar enthalpy value.
• Option C is incorrect: The value of $0.42\text{ kJ/mol}_{\text{rxn}}$ represents the unscaled heat absorbed by the calorimeter water ($q_{\text{solution}}$). It incorrectly assigns a positive sign (which implies an endothermic process that would cool the mixture) and fails to divide by the moles of reaction.
• Option D is incorrect: The value of $84\text{ kJ/mol}_{\text{rxn}}$ is the correct numerical magnitude but has a positive sign, representing an endothermic reaction. In an endothermic process, the reaction would absorb heat from the surroundings, causing the temperature of the mixture to *decrease* rather than increase.