Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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].
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
