6.111 In a calorimetric experiment, 6.48 g of lithium hy- droxide, LİOH, was dissolved in water. The temperature of the calorimeter rose from 25.00°C to 36.66°C. What is AH for the solution process? LIOH(s) → Li*(aq) + OH(aq) The heat capacity of the calorimeter and its contents is 547 J/°C.

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**Calorimetry Experiment: Determining ΔH for LiOH Dissolution**

In a calorimetric experiment, 6.48 grams of lithium hydroxide (LiOH) was dissolved in water. This caused the temperature of the calorimeter to increase from 25.00°C to 36.66°C. The task is to determine the enthalpy change (ΔH) for the solution process.

**Chemical Reaction:**

\[ \text{LiOH}_{(s)} \rightarrow \text{Li}^{+}_{(aq)} + \text{OH}^{-}_{(aq)} \]

**Heat Capacity:**

The combined heat capacity of the calorimeter and its contents is 547 J/°C.

---

**Calculating ΔH:**

To find ΔH, we need to calculate the heat absorbed or released using the formula:

\[ q = C \times \Delta T \]

Where:
- \( q \) is the heat absorbed or released,
- \( C \) is the heat capacity,
- \( \Delta T \) is the change in temperature.

\[ \Delta T = 36.66°C - 25.00°C = 11.66°C \]

\[ q = 547 \, \text{J/°C} \times 11.66 \, \text{°C} \] 

Calculate \( q \) to find the enthalpy change, considering the mass of the solute and stoichiometry of the reaction.
Transcribed Image Text:**Calorimetry Experiment: Determining ΔH for LiOH Dissolution** In a calorimetric experiment, 6.48 grams of lithium hydroxide (LiOH) was dissolved in water. This caused the temperature of the calorimeter to increase from 25.00°C to 36.66°C. The task is to determine the enthalpy change (ΔH) for the solution process. **Chemical Reaction:** \[ \text{LiOH}_{(s)} \rightarrow \text{Li}^{+}_{(aq)} + \text{OH}^{-}_{(aq)} \] **Heat Capacity:** The combined heat capacity of the calorimeter and its contents is 547 J/°C. --- **Calculating ΔH:** To find ΔH, we need to calculate the heat absorbed or released using the formula: \[ q = C \times \Delta T \] Where: - \( q \) is the heat absorbed or released, - \( C \) is the heat capacity, - \( \Delta T \) is the change in temperature. \[ \Delta T = 36.66°C - 25.00°C = 11.66°C \] \[ q = 547 \, \text{J/°C} \times 11.66 \, \text{°C} \] Calculate \( q \) to find the enthalpy change, considering the mass of the solute and stoichiometry of the reaction.
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