Solid KBr is dissolved in H20. The AH, = -690 kJ/mol for the solute dissociation and AH2 = +19.9 kJ/mol for making soluțe-solvent interactions. The AHspluition is kJ/mol and the solution is described as Enter your AHsoluition to the ones place, i.e. 45 kJ.

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### Dissolution of KBr in Water

**Concept: Enthalpy of Solution**

When solid KBr (Potassium Bromide) is dissolved in H₂O (Water), the process involves two main enthalpy changes:
- **ΔH₁: Enthalpy of Solute Dissociation**
  \[ ΔH₁ = -690 \, \text{kJ/mol} \]
- **ΔH₂: Enthalpy of Solute-Solvent Interaction**
  \[ ΔH₂ = +19.9 \, \text{kJ/mol} \]

The overall enthalpy change of the solution, noted as **ΔH_solution**, is the sum of these two enthalpies:
\[ \Delta H_{\text{solution}} = \Delta H_{1} + \Delta H_{2} \]

**Formula:**
\[ ΔH_{\text{solution}} = (-690 \, \text{kJ/mol}) + (19.9 \, \text{kJ/mol}) \]

**Instructions for Calculation:**
1. Substitute the given values into the formula.
2. Calculate the sum.
3. Enter the result in kilojoules per mole (kJ/mol) to the nearest ones place. For example, 45 kJ.

**Input Fields:**
- The **ΔH_solution** is \( \_\_\_\_\_\_ \text{kJ/mol} \)
- The solution is described as: **\[ \text{Exothermic}/\text{Endothermic} \]**

**Example Calculation:**
\[ \Delta H_{\text{solution}} = -690 + 19.9 = -670.1 \]
So, ΔH_solution is -670 kJ/mol.

**Type of Process:**
- If **ΔH_solution** is negative: The solution is **Exothermic**.
- If **ΔH_solution** is positive: The solution is **Endothermic**.

**Interactive Tool:**
Enter your calculated **ΔH_solution** to the ones place, e.g., 45 kJ, and select the correct description of the solution.
Transcribed Image Text:### Dissolution of KBr in Water **Concept: Enthalpy of Solution** When solid KBr (Potassium Bromide) is dissolved in H₂O (Water), the process involves two main enthalpy changes: - **ΔH₁: Enthalpy of Solute Dissociation** \[ ΔH₁ = -690 \, \text{kJ/mol} \] - **ΔH₂: Enthalpy of Solute-Solvent Interaction** \[ ΔH₂ = +19.9 \, \text{kJ/mol} \] The overall enthalpy change of the solution, noted as **ΔH_solution**, is the sum of these two enthalpies: \[ \Delta H_{\text{solution}} = \Delta H_{1} + \Delta H_{2} \] **Formula:** \[ ΔH_{\text{solution}} = (-690 \, \text{kJ/mol}) + (19.9 \, \text{kJ/mol}) \] **Instructions for Calculation:** 1. Substitute the given values into the formula. 2. Calculate the sum. 3. Enter the result in kilojoules per mole (kJ/mol) to the nearest ones place. For example, 45 kJ. **Input Fields:** - The **ΔH_solution** is \( \_\_\_\_\_\_ \text{kJ/mol} \) - The solution is described as: **\[ \text{Exothermic}/\text{Endothermic} \]** **Example Calculation:** \[ \Delta H_{\text{solution}} = -690 + 19.9 = -670.1 \] So, ΔH_solution is -670 kJ/mol. **Type of Process:** - If **ΔH_solution** is negative: The solution is **Exothermic**. - If **ΔH_solution** is positive: The solution is **Endothermic**. **Interactive Tool:** Enter your calculated **ΔH_solution** to the ones place, e.g., 45 kJ, and select the correct description of the solution.
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