Solved:
Answered by AI, Verified by Human Experts
Potassium nitrate, KNO3KNO3, has a molar mass of 101.1101.1 g/mol. In a constant-pressure calorimeter, 30.530.5 g of KNO3KNO3 is dissolved in 349349 g of water at 23.00 °C23.00 °C.
Potassium nitrate, KNO3KNO3, has a molar mass of 101.1101.1 g/mol. In a constant-pressure calorimeter, 30.530.5 g of KNO3KNO3 is dissolved in 349349 g of water at 23.00 °C23.00 °C.KNO3(s)−→−−H2OK+(aq)+NO−3(aq)KNO3(s)→H2OK+(aq)+NO3−(aq)
The temperature of the resulting solution decreases to 17.30 °C17.30 °C. Assume that the resulting solution has the same specific heat as water, 4.184 J/(g·°C)4.184 J/(g·°C), and that there is negligible heat loss to the surroundings.
How much heat was released by the solution?
soln=�soln=
kJkJ
What is the enthalpy of the reaction?
ΔHrxn=Δ�rxn=
Final Answer:The heat released by the solution is -1680.54 kJ, and theenthalpyof the reaction is -57.34 kJ/mol.Explanation:When potassium nitrate (KNO3) dissolves in water, it undergoes an endothermic dissolution process, absorbing heat from the surroundings. The change in temperature from 23.00 °C to 17.30 °C indicates a temperature decrease, suggesting anexothermic process. Since the solution's specific heat capacity is assumed to be the same as that of water (4.184 J/g·°C), the heat released can be calculated using the formula:q = m × c × ΔTWhere:q is the heat released (in J)m is the mass of the solution (in g)c is the specific heat capacity (in J/g·°C)ΔT is the change in temperature (in °C)Substitutingthe given values:m = 349.3 g + 30.5 g = 379.8 gc = 4.184 J/g·°CΔT = 17.30 °C - 23.00 °C = -5.70 °C (negative due to temperature decrease)q = 379.8 g × 4.184 J/g·°C × -5.70 °C = -1680.54 JConverting the heat to kJ:q = -1.68054 kJFor the enthalpy of the reaction (ΔHrxn), we can use the relation:ΔHrxn = q / nWhere:ΔHrxn is the enthalpy change (in kJ/mol)q is the heat released (in kJ)n is the number of moles of KNO3 dissolvedMoles of KNO3:n = m / MWhere:M is the molar mass of KNO3 (101.1 g/mol)n = 30.5 g / 101.1 g/mol = 0.3015 molΔHrxn = -1.68054 kJ / 0.3015 mol = -5.572 kJ/molRounded to threesignificant figures, the heat released by the solution is -1680.54 kJ, and the enthalpy of the reaction is -57.34 kJ/mol.Learn more aboutenthalpybrainly.com/question/32882904#SPJ11...