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Ch 17 Thermochemistry.

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Presentation on theme: "Ch 17 Thermochemistry."— Presentation transcript:

1 Ch 17 Thermochemistry

2 What is the difference between heat and temperature?
HEAT is energy that transfers from one object/substance to another TEMPERATURE is a measure of the amount of energy an object/substance has

3 A. Energy Transformations
Energy that is stored in the chemical bonds of a substance is called CHEMICAL POTENTIAL ENERGY Heat ALWAYS flows from hot to cold

4 B. Endothermic and Exothermic Processes
System=the reaction Surroundings=everything around the reaction Surroundings Law of Conservation of Energy- energy can be neither created nor destroyed System

5 B. Endothermic and Exothermic Processes
Endothermic Process- heat is absorbed from the surroundings Endo = Into Endothermic processes are represented by a positive “q” HEAT

6 B. Endothermic and Exothermic Processes
Exothermic process- heat is released into the surroundings Exo = Exit Exothermic processes are represented by a negative “q” HEAT

7 C. Measuring Heat Flow Two Common Units 1J = 0.2390 cal
Joule calorie 1J = cal 4.184 J = 1 cal 1Calorie = 1 kilocal = 1000 cal

8 D. Heat Capacity and Specific Heat
Heat Capacity depends on: The mass of the object The chemical composition of the object “the amount of heat needed to increase the temperature of an object by 1 oC Specific heat capacity- amount of heat needed to raise the temperature of 1g by 1 oC

9 D. Heat Capacity and Specific Heat
C = q / (m X ΔT) C =Specific Heat q = heat (joules or calories) m = mass (grams) ΔT = change in temperature The change in temperature can be measure in Kelvin or degrees Celsius

10 C= 849J/ (95.4g)(23.0 oC) = 0.387 J/(g· oC)
The temperature of a 95.4g piece of copper increases from 25.0 oC to 48.0 oC when it absorbs 849 J of heat. What is the specific heat of copper? Known: m= 95.4g q=849 J ΔT= =23.0 oC Work: C= 849J/ (95.4g)(23.0 oC) = J/(g· oC)

11 E. Calorimetry Measures the heat flow into or out of a system
Heat released by the system is equal to heat absorbed by the surroundings ENTHALPY: (H) the heat constant of a system at constant pressure

12 E. Calorimetry The terms heat and enthalpy change are interchangeable
q = ΔH qsys = ΔH = -m x C x ΔT Negative enthalpy = exothermic Positive enthalpy = endothermic

13 When 25. 0mL of water containing HCl at 25. 0 oC is added to 25
When 25.0mL of water containing HCl at 25.0 oC is added to 25.0mL of water containing NaOH at 25.0 oC in a calorimeter a rxn occurs. Calculate the enthalpy change (in kJ) during the rxn if the highest temperature observed was 32.0 oC. Assume all densities =1g/mL KNOWN: Cwater=4.18J/g oC V=25.0mL+25.0mL ΔT=7 oC Density= 1g/mL ΔH = ?

14 m= (50mL) x (1. 00g/mL) = 50g ΔT= TF – Ti = 32. 0 – 25. 0 =7
m= (50mL) x (1.00g/mL) = 50g ΔT= TF – Ti = 32.0 – 25.0 =7.0 ΔH= -mCΔT= -(50.0g)(4.18J/goC)(7.0oC) ΔH= J = -1500J = -1.5kJ

15 F. Thermochemical Equations
In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product Endothermic (positive ΔH) 2NaHCO kJ Na2CO3 + H2O + CO2 Exothermic (negative ΔH) CaO + H2O Ca(OH) kJ

16 G. Heat of Combustion The heat of reaction for the complete burning of one mole of a substance Written the same was as change in enthalpy

17 Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= kJ Fe(s) + O2(g)→ Fe2O3(s) How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? 2 4 3 10.00g Fe 1 mol = mol Fe 55.85g Fe mol Fe 1652 kJ =73.97 kJ of heat 4 mol Fe

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