Chemistry I Honors Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 1 Energy is the capacity for doing work or supplying heat. Every substance has a certain amount of energy stored inside it. During a chemical reaction, a substance with a different amount of chemical potential energy is formed. The energy change that occurs during this transformation can be used to do work. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 2 (page 1) When energy changes are studied, a system is the part of the universe that is studied. The surroundings include everything else in the universe. When energy leaves the system and moves into the surroundings, the reaction is said to be exothermic Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 2 (page 2) . When energy flows into the system from the surroundings, the reaction is said to be endothermic. During an exothermic reaction, the system gains energy (heat) while the surroundings cool down. During an endothermic reaction, the system loses energy (heat) to the surroundings. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 3 calorie joule Chemistry I H - Chapter 17 Notes

Objective 4 Heat capacity Specific heat Depends on mass Is the heat capacity divided by mass (g) Depends on composition Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 5 A calorimeter is an insulated device used to measure the absorption or release of heat in chemical or physical processes. In Calorimetry, the heat released by the system is equal to the heat absorbed by the surroundings. Conversely, the heat absorbed by a system is equal to the heat released by the surroundings. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 6 A thermochemical equation is the chemical equation that includes the enthalpy change. CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 7 Using this equation, 2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g), calculate the amount of heat (in kJ) required to decompose 2.24 mol NaHCO3(s). Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 8 (page 1) The molar heat of fusion is the heat absorbed by one mole of a solid substance as it melts to a liquid at constant temperature. The molar heat of solidification is the heat lost when one mole of a liquid solidifies at constant temperature. The quantity of heat absorbed by a melting solid is the same as the quantity of heat released as the liquid solidifies. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 8 (page 2) The molar heat of vaporization is the heat absorbed by one mole of a liquid substance as it vaporizes to a gas at constant temperature. The molar heat of condensation is the heat lost when one mole of a gas condenses at constant temperature. The quantity of heat absorbed by a vaporizing liquid is the same as the quantity of heat released as the gas condenses. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 8 (page 3) The molar heat of solution is the enthalpy change causes by the dissolution of one mole of a substance. Chemistry I H - Chapter 17 Notes

NaOH(s)  Na+1 (aq) + OH-1 (aq) Objective 9 How much heat (in kJ) is released when 2.500 mol NaOH(s) is dissolved in water? NaOH(s)  Na+1 (aq) + OH-1 (aq) Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 10 Hess’s law of heat summation states that if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 11 (page 1) When it is not practical or convenient to calculate enthalpy changes directly, Hess’s law of heat summation allows the calculation to be done indirectly. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 11 (page 2) If we know some of the possible steps involved in the reaction AND the thermodynamic properties of each of the steps, we can use these steps (or the reverse of these steps) to give the final equation of the reaction we want to calculate. By adding the heats of reaction for each of the steps (or their reverses) we can determine the final heat of reaction. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 12 Use Hess’s Law to calculate the enthalpy change for the conversion of diamond to graphite. Equations needed: C(graphite) + O2(g)  CO2(g) ΔH = -393.5 kJ C(diamond) + O2(g)  CO2(g) ΔH = -395.4 kJ Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 12 Write equation a in reverse to give: c. CO2(g)  C(graphite) + O2(g) ΔH = 393.5 kJ When you write a reverse reaction, the sign on the ΔH is changed. Chemistry I H - Chapter 17 Notes

Chemistry I H - Chapter 17 Notes Objective 12 If you add the two equations, you get the equation for the conversion of diamond to graphite. C(diamond) + O2(g)  CO2(g) ΔH = -395.4 kJ + CO2(g)  C(graphite) + O2(g) ΔH = 393.5 kJ = C(diamond)  C (graphite) ΔH = -1.9 kJ Chemistry I H - Chapter 17 Notes