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Calorimetry Remember… Heat lost by system = Heat absorbed by surroundings. Chemical/physical changes in the lab are open to atmosphere, so the changes occur at a constant pressure Heat content of a system at constant pressure is called enthalpy , H
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Calorimetry Remember… Heat lost by system = Heat absorbed by surroundings
Chemical/physical changes in the lab are open to atmosphere, so the changes occur at a constant pressure • Heat content of a system at constant pressure is called enthalpy, H • Heat released or absorbed by system at constant pressure is the change in enthalpy, (∆H)
The reactions we look at are at constant pressure so q = ∆H • ∆H is positive = ___________ • ∆H is negative = ___________ • qsys = ∆H = -qsurr = - m x C x ∆T
When 25 mL of water containing 0.025 mol HCl at 25.0°C is added to 25.0 mL of water containing 0.025 mol NaOH at 25.0°C in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change in kJ during this reaction if the highest temperature observed is 32.0°C. Assume the densities of the solutions are 1.00 g/mL.
Thermochemical Equations Show enthalpy change in the reaction CaO + H2O Ca(OH)2 + heat In a chem. rxn. ∆H for the rxn. can be written as product or reactant 2NaHCO3 Na2CO3 + H2O + CO2 **absorbs 129 kJ of heat
Heat of Reaction • The enthalpy change for the chemical equation exactly as it’s written • As ∆H (heat flow at constant pressure) q, + /- CaO + H2O Ca(OH)2 + 65.2k J
Calculating Enthalpy Changes in a Reaction 2NaHCO3 + 129 kJ Na2CO3 + H2O + CO2 What do we know from the equation?
2NaHCO3 + 129 kJ Na2CO3 + H2O + CO2 • How much heat would be needed to decompose 2.24 mol NaHCO3? • How much heat would be absorbed in decomposing 9.0 g of NaHCO3?
If a piece of gold (C = 0.129 J/g°C) with a mass of 45.5 g and a temperature of 80.5°C is dropped into 192 g of water at 15.0°C, what is the final temperature of the system?