∆Hrxn = -58.3 kj/mol
Cs of NaCl = 3.91 J g-1 oC-1
Density of 1 M NaCl = 1.037 g/mL
∆T = 13.6 oC
50 mL of Naoh reacted with 50 mL of HCl
How to calculate the calorimeter's heat capacity from the given?
2017-11-07 9:15 am
回答 (2)
2017-11-07 9:40 am
NaOH + HCl → NaCl + H₂O …… ΔHrxn = -58.3 kJ/mol
Volume of final solution = (50 + 50) mL = 100 mL = 0.1 L
No. of moles of H₂O formed = No. of moles of NaCl formed = (1 mol/L) × (0.1 L) = 0.1 mol
Heat released in the reaction = (58.3 kJ/mol) × (0.1 mol) = 5.83 kJ = 5830 J
Mass of NaCl solution = (100 mL) × (1.037 g/mL) = 103.7 g
Heat absorbed by NaCl solution = (103.7 g) × [3.91 J/(g °C)] × (13.6 °C) = 5514 J
Heat absorbed by the calorimeter = (5830 - 5514) J = 316 J
Heat capacity of the calorimeter = (316 J) / (13.6 °C) = 22.5 J/°C
Volume of final solution = (50 + 50) mL = 100 mL = 0.1 L
No. of moles of H₂O formed = No. of moles of NaCl formed = (1 mol/L) × (0.1 L) = 0.1 mol
Heat released in the reaction = (58.3 kJ/mol) × (0.1 mol) = 5.83 kJ = 5830 J
Mass of NaCl solution = (100 mL) × (1.037 g/mL) = 103.7 g
Heat absorbed by NaCl solution = (103.7 g) × [3.91 J/(g °C)] × (13.6 °C) = 5514 J
Heat absorbed by the calorimeter = (5830 - 5514) J = 316 J
Heat capacity of the calorimeter = (316 J) / (13.6 °C) = 22.5 J/°C
2017-11-09 4:19 am
is this right?
收錄日期: 2021-04-18 17:56:02
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