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

The enthalpy change for the chemical equation exactly as it’s written 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 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?