Let's break down the problem step by step: ### **1. Write the Neutralization Reaction** \[ \text{NaOH (aq) + HCl (aq) → NaCl (aq) + H}_2\text{O (l)} \] ### **2. Find the Number of Moles Reacted** Both are 1.0 M and 100.0 mL: \[ \text{Moles of NaOH} = 1.0\,\text{mol/L} \times 0.100\,\text{L} = 0.100\,\text{mol} \] \[ \text{Moles of HCl} = 1.0\,\text{mol/L} \times 0.100\,\text{L} = 0.100\,\text{mol} \] The reaction is 1:1, so all react. ### **3. Calculate the Heat (q) Absorbed by the Solution** #### **a. Total Volume and Mass** \[ \text{Total volume} = 100.0\,\text{mL} + 100.0\,\text{mL} = 200.0\,\text{mL} \] \[ \text{Density} = 1.00\,\text{g/mL} \] \[ \text{Total mass} = 200.0\,\text{mL} \times 1.00\,\text{g/mL} = 200.0\,\text{g} \] #### **b. Temperature Change (\(\Delta T\))** \[ T_{\text{initial}} = 1.850 \times 10^1 = 18.50\,^\circ\text{C} \] \[ T_{\text{final}} = 2.96 \times 10^1 = 29.6\,^\circ\text{C} \] \[ \Delta T = 29.6 - 18.5 = 11.1\,^\circ\text{C} \] #### **c. Heat Absorbed** \[ q = \text{mass} \times \text{specific heat} \times \Delta T \] \[ q = 200.0\,\text{g} \times 4.18\,\text{J/g}^\circ\text{C} \times 11.1^\circ\text{C} \] \[ q = 200.0 \times 4.18 \times 11.1 \] \[ q = 200.0 \times 46.398 \] \[ q = 9279.6\,\text{J} \] ### **4. Enthalpy Change Per Mole** This is the heat released for **0.100 mol** of HCl neutralized. \[ \Delta H = \frac{q}{\text{moles reacted}} \] \[ \Delta H = \frac{9279.6\,\text{J}}{0.100\,\text{mol}} \] \[ \Delta H = 92796\,\text{J/mol} \] But since this is an **exothermic reaction**, the enthalpy change is **negative**: \[ \Delta H = -92796\,\text{J/mol} \] Convert to kJ: \[ \Delta H = -92.796\,\text{kJ/mol} \] ### **5. Final Answer (rounded appropriately)** \[ \boxed{-9.28 \times 10^1\,\text{kJ}} \] #### **Summary Table** | Step | Value | |----------------------------------|--------------------------| | Initial temp (\(T_i\)) | \(1.850 \times 10^1\) °C = 18.50 °C | | Final temp (\(T_f\)) | \(2.96 \times 10^1\) °C = 29.6 °C | | ΔT | 11.1 °C | | Mass of solution | 200.0 g | | Specific heat | 4.18 J/g°C | | q (heat released) | 9279.6 J | | Moles reacted | 0.100 mol | | ΔH per mole | -92796 J/mol = -92.8 kJ/mol | --- ## **Final Answer** \[ \boxed{-9.28\times10^1\,\text{kJ}} \] This is the enthalpy change (ΔH) for the neutralization of HCl by NaOH under the given conditions.