madey21
Answered

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[tex]$7.5 \, \text{g} \, \text{KNO}_3$[/tex] dissociated in [tex]$49.0 \, \text{g}$[/tex] of water in a coffee cup calorimeter. The thermometer reading changed from [tex]$20.4^{\circ} \text{C}$[/tex] to [tex]$9.7^{\circ} \text{C}$[/tex]. What is the heat of the reaction?

[tex]\[
\begin{array}{c}
\text{KNO}_3 \rightarrow \text{K}^{+} + \text{NO}_3^{-} \\
C_{\text{soln}} = 4.18 \, \text{J/g}^{\circ} \text{C} \quad C_{\text{cal}} = 6.5 \, \text{J/}^{\circ} \text{C} \\
q_{\text{r}} = \, [?] \, \text{J}
\end{array}
\][/tex]

Hint: Make sure to account for all parts of the solution and calorimeter.

Enter either a '+' or '-' sign AND the magnitude.

Sagot :

To determine the heat of the reaction for the dissociation of [tex]\( KNO_3 \)[/tex] in water in a coffee cup calorimeter, we need to follow several steps involving the specific heat capacities of the solution and the calorimeter, as well as the temperature change.

### Step-by-Step Solution:

1. Given Data:

- Mass of [tex]\( KNO_3 \)[/tex]: [tex]\( 7.5 \, \text{g} \)[/tex]
- Mass of water: [tex]\( 49.0 \, \text{g} \)[/tex]
- Initial temperature: [tex]\( 20.4^\circ \mathrm{C} \)[/tex]
- Final temperature: [tex]\( 9.7^\circ \mathrm{C} \)[/tex]
- Specific heat capacity of solution ([tex]\( C_{\text{soln}} \)[/tex]): [tex]\( 4.18 \, \text{J/g}^\circ \mathrm{C} \)[/tex]
- Heat capacity of the calorimeter ([tex]\( C_{\text{cal}} \)[/tex]): [tex]\( 6.5 \, \text{J/}^\circ \mathrm{C} \)[/tex]

2. Calculate the change in temperature ([tex]\( \Delta T \)[/tex]):
[tex]\[ \Delta T = \text{Final Temperature} - \text{Initial Temperature} = 9.7^\circ \mathrm{C} - 20.4^\circ \mathrm{C} = -10.7^\circ \mathrm{C} \][/tex]

3. Calculate the heat absorbed by the solution ([tex]\( q_{\text{soln}} \)[/tex]):
The total mass of the solution is the sum of the mass of [tex]\( KNO_3 \)[/tex] and the mass of water:
[tex]\[ \text{Total mass of solution} = 7.5 \, \text{g} + 49.0 \, \text{g} = 56.5 \, \text{g} \][/tex]

Using the specific heat capacity of the solution and the change in temperature:
[tex]\[ q_{\text{soln}} = \text{Total mass of solution} \times C_{\text{soln}} \times \Delta T \][/tex]
[tex]\[ q_{\text{soln}} = 56.5 \, \text{g} \times 4.18 \, \text{J/g}^\circ \mathrm{C} \times (-10.7^\circ \mathrm{C}) = -2527.019 \, \text{J} \][/tex]

4. Calculate the heat absorbed by the calorimeter ([tex]\( q_{\text{cal}} \)[/tex]):
[tex]\[ q_{\text{cal}} = C_{\text{cal}} \times \Delta T \][/tex]
[tex]\[ q_{\text{cal}} = 6.5 \, \text{J/}^\circ \mathrm{C} \times (-10.7^\circ \mathrm{C}) = -69.55 \, \text{J} \][/tex]

5. Calculate the total heat of reaction ([tex]\( q_{\text{rxn}} \)[/tex]):
The heat of the reaction is the negative sum of the heat absorbed by the solution and the calorimeter. Since the temperature decreased, the system released heat, so [tex]\( q_{\text{rxn}} \)[/tex] should be positive:
[tex]\[ q_{\text{rxn}} = -(q_{\text{soln}} + q_{\text{cal}}) \][/tex]
[tex]\[ q_{\text{rxn}} = -(-2527.019 \, \text{J} - 69.55 \, \text{J}) = 2596.569 \, \text{J} \][/tex]

### Conclusion:
The heat of the reaction, [tex]\( q_{\text{rxn}} \)[/tex], is [tex]\( +2596.569 \, \text{J} \)[/tex].