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To solve this problem, we will follow a systematic approach involving several steps. Each step will involve calculating a specific component of the entire process, ultimately leading to the calculation of the enthalpy change, [tex]\( \Delta H \)[/tex], for the reaction. Let's start the step-by-step solution:
1. Determine the Mass of the Solution:
Given:
- Volume of [tex]\( \text{CuSO}_4 \)[/tex] solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
- Volume of [tex]\( \text{KOH} \)[/tex] solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
- Density of the solution (same as water): [tex]\( 1.0 \, \text{g/mL} \)[/tex]
The total volume of the solution is:
[tex]\[ \text{Total volume} = 50.0 \, \text{mL} (\text{CuSO}_4) + 50.0 \, \text{mL} (\text{KOH}) = 100.0 \, \text{mL} \][/tex]
Given the density of the solution, the mass of the solution is:
[tex]\[ \text{Mass of solution} = \text{Total volume} \times \text{Density} = 100.0 \, \text{mL} \times 1.0 \, \text{g/mL} = 100.0 \, \text{g} \][/tex]
2. Calculate the Temperature Change:
Given:
- Initial temperature: [tex]\( 20.2^\circ \text{C} \)[/tex]
- Final temperature: [tex]\( 26.3^\circ \text{C} \)[/tex]
The temperature change is:
[tex]\[ \Delta T = 26.3^\circ \text{C} - 20.2^\circ \text{C} = 6.1^\circ \text{C} \][/tex]
3. Calculate the Heat Absorbed by the Solution:
Given:
- Specific heat capacity of water (same as the solution): [tex]\( 4.18 \, \text{J/g}^\circ \text{C} \)[/tex]
- Mass of the solution: [tex]\( 100.0 \, \text{g} \)[/tex]
- Temperature change: [tex]\( 6.1^\circ \text{C} \)[/tex]
The heat absorbed by the solution is:
[tex]\[ q_{\text{solution}} = \text{mass} \times \text{specific heat capacity} \times \Delta T = 100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ \text{C} \times 6.1^\circ \text{C} = 2549.8 \, \text{J} \][/tex]
4. Calculate the Heat Absorbed by the Calorimeter:
Given:
- Heat capacity of the calorimeter: [tex]\( 12.1 \, \text{J/K} \)[/tex]
- Temperature change: [tex]\( 6.1^\circ \text{C} \)[/tex]
The heat absorbed by the calorimeter is:
[tex]\[ q_{\text{calorimeter}} = \text{heat capacity} \times \Delta T = 12.1 \, \text{J/K} \times 6.1 \, \text{K} = 73.81 \, \text{J} \][/tex]
5. Calculate the Total Heat Absorbed by the System:
The total heat absorbed by the system is the sum of the heat absorbed by the solution and the calorimeter:
[tex]\[ q_{\text{total}} = q_{\text{solution}} + q_{\text{calorimeter}} = 2549.8 \, \text{J} + 73.81 \, \text{J} = 2623.61 \, \text{J} \][/tex]
6. Calculate the Number of Moles of [tex]\( \text{CuSO}_4 \)[/tex]:
Given:
- Molarity of [tex]\( \text{CuSO}_4 \)[/tex] solution: [tex]\( 1.00 \, \text{M} \)[/tex]
- Volume of [tex]\( \text{CuSO}_4 \)[/tex] solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
The number of moles of [tex]\( \text{CuSO}_4 \)[/tex] is:
[tex]\[ \text{moles of CuSO}_4 = \text{Volume in L} \times \text{Molarity} = \frac{50.0 \, \text{mL}}{1000} \, \text{L/mL} \times 1.00 \, \text{M} = 0.050 \, \text{mol} \][/tex]
7. Calculate the Enthalpy Change ([tex]\( \Delta H \)[/tex]):
The enthalpy change per mole of [tex]\( \text{CuSO}_4 \)[/tex] is:
[tex]\[ \Delta H = \frac{\text{Total heat absorbed}}{\text{moles of CuSO}_4} = \frac{2623.61 \, \text{J}}{0.050 \, \text{mol}} = 52472.2 \, \text{J/mol} \][/tex]
Converting this to kJ/mol:
[tex]\[ \Delta H = \frac{52472.2 \, \text{J/mol}}{1000} = 52.4722 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change [tex]\( \Delta H \)[/tex] for the process is [tex]\( 52.4722 \, \text{kJ/mol} \)[/tex].
1. Determine the Mass of the Solution:
Given:
- Volume of [tex]\( \text{CuSO}_4 \)[/tex] solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
- Volume of [tex]\( \text{KOH} \)[/tex] solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
- Density of the solution (same as water): [tex]\( 1.0 \, \text{g/mL} \)[/tex]
The total volume of the solution is:
[tex]\[ \text{Total volume} = 50.0 \, \text{mL} (\text{CuSO}_4) + 50.0 \, \text{mL} (\text{KOH}) = 100.0 \, \text{mL} \][/tex]
Given the density of the solution, the mass of the solution is:
[tex]\[ \text{Mass of solution} = \text{Total volume} \times \text{Density} = 100.0 \, \text{mL} \times 1.0 \, \text{g/mL} = 100.0 \, \text{g} \][/tex]
2. Calculate the Temperature Change:
Given:
- Initial temperature: [tex]\( 20.2^\circ \text{C} \)[/tex]
- Final temperature: [tex]\( 26.3^\circ \text{C} \)[/tex]
The temperature change is:
[tex]\[ \Delta T = 26.3^\circ \text{C} - 20.2^\circ \text{C} = 6.1^\circ \text{C} \][/tex]
3. Calculate the Heat Absorbed by the Solution:
Given:
- Specific heat capacity of water (same as the solution): [tex]\( 4.18 \, \text{J/g}^\circ \text{C} \)[/tex]
- Mass of the solution: [tex]\( 100.0 \, \text{g} \)[/tex]
- Temperature change: [tex]\( 6.1^\circ \text{C} \)[/tex]
The heat absorbed by the solution is:
[tex]\[ q_{\text{solution}} = \text{mass} \times \text{specific heat capacity} \times \Delta T = 100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ \text{C} \times 6.1^\circ \text{C} = 2549.8 \, \text{J} \][/tex]
4. Calculate the Heat Absorbed by the Calorimeter:
Given:
- Heat capacity of the calorimeter: [tex]\( 12.1 \, \text{J/K} \)[/tex]
- Temperature change: [tex]\( 6.1^\circ \text{C} \)[/tex]
The heat absorbed by the calorimeter is:
[tex]\[ q_{\text{calorimeter}} = \text{heat capacity} \times \Delta T = 12.1 \, \text{J/K} \times 6.1 \, \text{K} = 73.81 \, \text{J} \][/tex]
5. Calculate the Total Heat Absorbed by the System:
The total heat absorbed by the system is the sum of the heat absorbed by the solution and the calorimeter:
[tex]\[ q_{\text{total}} = q_{\text{solution}} + q_{\text{calorimeter}} = 2549.8 \, \text{J} + 73.81 \, \text{J} = 2623.61 \, \text{J} \][/tex]
6. Calculate the Number of Moles of [tex]\( \text{CuSO}_4 \)[/tex]:
Given:
- Molarity of [tex]\( \text{CuSO}_4 \)[/tex] solution: [tex]\( 1.00 \, \text{M} \)[/tex]
- Volume of [tex]\( \text{CuSO}_4 \)[/tex] solution: [tex]\( 50.0 \, \text{mL} \)[/tex]
The number of moles of [tex]\( \text{CuSO}_4 \)[/tex] is:
[tex]\[ \text{moles of CuSO}_4 = \text{Volume in L} \times \text{Molarity} = \frac{50.0 \, \text{mL}}{1000} \, \text{L/mL} \times 1.00 \, \text{M} = 0.050 \, \text{mol} \][/tex]
7. Calculate the Enthalpy Change ([tex]\( \Delta H \)[/tex]):
The enthalpy change per mole of [tex]\( \text{CuSO}_4 \)[/tex] is:
[tex]\[ \Delta H = \frac{\text{Total heat absorbed}}{\text{moles of CuSO}_4} = \frac{2623.61 \, \text{J}}{0.050 \, \text{mol}} = 52472.2 \, \text{J/mol} \][/tex]
Converting this to kJ/mol:
[tex]\[ \Delta H = \frac{52472.2 \, \text{J/mol}}{1000} = 52.4722 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change [tex]\( \Delta H \)[/tex] for the process is [tex]\( 52.4722 \, \text{kJ/mol} \)[/tex].
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