The molar enthalpy of dissolution of sodium iodide is 7.61 kJ/mol NaI as shown below: NaI(s) → Na+ (aq) + I- (aq) ∆H = +7.61 kJ/mol A 35.7 g sample of NaI is dissolved in 50.0 g of water in a calorimeter at 21.5°C. The final temperature of the solution is __
The Correct Answer and Explanation is:
To solve this problem, we will need to use the concept of heat transfer during the dissolution process and apply the principle of conservation of energy. The heat absorbed by the water will equal the heat released by the dissolution of sodium iodide. Here’s how we can break it down:
Given Data:
- Molar enthalpy of dissolution (ΔHdissolution\Delta H_{\text{dissolution}}ΔHdissolution) = 7.61 kJ/mol
- Mass of NaI = 35.7 g
- Mass of water = 50.0 g
- Initial temperature of the solution = 21.5°C
- Final temperature of the solution = unknown
- Specific heat capacity of water (cwaterc_{\text{water}}cwater) = 4.18 J/g°C
Step 1: Convert the mass of NaI to moles
To find how many moles of NaI are dissolving, we need the molar mass of NaI:
- Molar mass of NaI = Na (22.99 g/mol) + I (126.90 g/mol) = 149.89 g/mol
- Moles of NaI = Mass of NaI / Molar mass of NaI = 35.7 g / 149.89 g/mol = 0.238 mol
Step 2: Calculate the heat released during dissolution
The heat released during the dissolution of NaI is given by: qdissolution=moles of NaI×ΔHdissolutionq_{\text{dissolution}} = \text{moles of NaI} \times \Delta H_{\text{dissolution}}qdissolution=moles of NaI×ΔHdissolution qdissolution=0.238 mol×7.61 kJ/mol=1.81 kJ=1810 Jq_{\text{dissolution}} = 0.238 \, \text{mol} \times 7.61 \, \text{kJ/mol} = 1.81 \, \text{kJ} = 1810 \, \text{J}qdissolution=0.238mol×7.61kJ/mol=1.81kJ=1810J
Step 3: Calculate the temperature change of the water
Now, we will calculate the temperature change of the water based on the heat absorbed. The heat absorbed by the water is: qwater=mwater×cwater×ΔTq_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta Tqwater=mwater×cwater×ΔT
where:
- mwater=50.0 gm_{\text{water}} = 50.0 \, \text{g}mwater=50.0g
- cwater=4.18 J/g°Cc_{\text{water}} = 4.18 \, \text{J/g°C}cwater=4.18J/g°C
- ΔT\Delta TΔT is the change in temperature (final temperature minus initial temperature)
Rearranging the equation to solve for ΔT\Delta TΔT: ΔT=qwatermwater×cwater\Delta T = \frac{q_{\text{water}}}{m_{\text{water}} \times c_{\text{water}}}ΔT=mwater×cwaterqwater ΔT=1810 J50.0 g×4.18 J/g°C=1810209≈8.65°C\Delta T = \frac{1810 \, \text{J}}{50.0 \, \text{g} \times 4.18 \, \text{J/g°C}} = \frac{1810}{209} \approx 8.65°CΔT=50.0g×4.18J/g°C1810J=2091810≈8.65°C
Step 4: Calculate the final temperature of the solution
The initial temperature of the solution is 21.5°C. The final temperature will be: Tfinal=Tinitial+ΔTT_{\text{final}} = T_{\text{initial}} + \Delta TTfinal=Tinitial+ΔT Tfinal=21.5°C+8.65°C=30.15°CT_{\text{final}} = 21.5°C + 8.65°C = 30.15°CTfinal=21.5°C+8.65°C=30.15°C
Final Answer:
The final temperature of the solution is approximately 30.2°C.
