To determine the heat of the reaction for the dissociation of [tex]\( 7.5 \, \text{g} \)[/tex] of [tex]\( \text{KNO}_3 \)[/tex] in [tex]\( 49.0 \, \text{g} \)[/tex] of water in a coffee cup calorimeter, we will follow these steps:
1. Determine the total mass of the solution
2. Calculate the temperature change ([tex]\(\Delta T\)[/tex])
3. Calculate the heat gained or lost by the solution ([tex]\( q_{\text{solution}} \)[/tex])
4. Calculate the heat gained or lost by the calorimeter ([tex]\( q_{\text{calorimeter}} \)[/tex])
5. Determine the total heat of the reaction ([tex]\( q_{\text{reaction}} \)[/tex])
Let's walk through each step with the details provided.
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### Step 1: Determine the total mass of the solution
The total mass of the solution is the sum of the mass of [tex]\( \text{KNO}_3 \)[/tex] and the mass of water.
[tex]\[ \text{Total mass} = \text{mass}_{\text{water}} + \text{mass}_{\text{KNO}_3} \][/tex]
Given:
- [tex]\(\text{mass}_{\text{water}} = 49.0 \, \text{g}\)[/tex]
- [tex]\(\text{mass}_{\text{KNO}_3} = 7.5 \, \text{g}\)[/tex]
[tex]\[ \text{Total mass} = 49.0 \, \text{g} + 7.5 \, \text{g} = 56.5 \, \text{g} \][/tex]
---
### Step 2: Calculate the temperature change ([tex]\(\Delta T\)[/tex])
The temperature change is the difference between the final temperature and the initial temperature.
[tex]\[ \Delta T = \text{final temperature} - \text{initial temperature} \][/tex]
Given:
- Initial temperature = [tex]\(20.4^\circ \text{C}\)[/tex]
- Final temperature = [tex]\(9.7^\circ \text{C}\)[/tex]
[tex]\[ \Delta T = 9.7^\circ \text{C} - 20.4^\circ \text{C} = -10.7^\circ \text{C} \][/tex]
The negative sign indicates a decrease in temperature.
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### Step 3: Calculate the heat gained or lost by the solution ([tex]\( q_{\text{solution}} \)[/tex])
The heat change in the solution can be calculated using the formula:
[tex]\[ q_{\text{solution}} = \text{total mass} \times c_{\text{solution}} \times \Delta T \][/tex]
Given:
- [tex]\( \text{Total mass} = 56.5 \, \text{g} \)[/tex]
- [tex]\( c_{\text{solution}} = 4.18 \, \text{J/g} \, ^\circ \text{C} \)[/tex]
[tex]\[ q_{\text{solution}} = 56.5 \, \text{g} \times 4.18 \, \text{J/g} \, ^\circ \text{C} \times (-10.7^\circ \text{C}) = -2527.019 \, \text{J} \][/tex]
---
### Step 4: Calculate the heat gained or lost by the calorimeter ([tex]\( q_{\text{calorimeter}} \)[/tex])
The heat change in the calorimeter can be calculated using the formula:
[tex]\[ q_{\text{calorimeter}} = C_{\text{calorimeter}} \times \Delta T \][/tex]
Given:
- [tex]\( C_{\text{calorimeter}} = 6.5 \, \text{J/}^\circ \text{C} \)[/tex]
[tex]\[ q_{\text{calorimeter}} = 6.5 \, \text{J/}^\circ \text{C} \times (-10.7^\circ \text{C}) = -69.55 \, \text{J} \][/tex]
---
### Step 5: Determine the total heat of the reaction ([tex]\( q_{\text{reaction}} \)[/tex])
The total heat of the reaction is the sum of the heat changes in both the solution and the calorimeter.
[tex]\[ q_{\text{reaction}} = q_{\text{solution}} + q_{\text{calorimeter}} \][/tex]
[tex]\[ q_{\text{reaction}} = -2527.019 \, \text{J} + (-69.55 \, \text{J}) = -2596.569 \, \text{J} \][/tex]
---
### Conclusion:
The heat of the reaction for the dissociation of [tex]\( 7.5 \, \text{g} \)[/tex] of [tex]\( \text{KNO}_3 \)[/tex] in [tex]\( 49.0 \, \text{g} \)[/tex] of water is [tex]\( -2596.569 \, \text{J} \)[/tex] (or approximately -2596.57 J). The negative sign indicates that the reaction is endothermic, meaning it absorbs heat from the surroundings.
