1. Thermochemistry Chapter 10 thermo #3.ppt

2. Today, you will learn… How to solve problems that include stoichiometry and thermochemistry The definition of specific heat How to solve problems using Q=mC  T How heat is transferred with and without a phase change

3. Reaction Enthalpies All chemical reactions either release or absorb heat Exothermic reactions: heat is a product Reactants products + energy as heat (  H = negative) Endothermic reactions: heat is a reactant. Reactants + energy as heat products (  H = positive)

4. Reaction Enthalpies Example: __ SO 2 (g) + __ O 2 (g)  __ SO 3 (g)  H = -198.2 kJ How much heat is released when 454 g of SO 2 are burned in air? First, balance the equation: 2 SO 2 (g) + 1 O 2 (g)  2 SO 3 (g)  H = -198.2 kJ 454 g ? kJ Use stoichiometry (treat enthalpy like a coefficient, ignore sign):

5. Specific Heat (C), (called S in your textbook, but usually C. S stands for entropy… a coming attraction!) Things heat up or cool down (change temperature) at different rates. Land heats up and cools down faster than water, and aren’t we lucky for that!?

6. Specific heat (also called specific heat capacity) is the amount of heat required to raise the temperature of 1 g a material by one degree (C or K, they’re the same size). C water = 4.184 J / g C (“holds” its heat) C sand = 0.664 J / g C (less E to change) This is why land heats up quickly during the day and cools quickly at night and why water takes longer to heat/cool.

7. Why does water have such a high specific heat? Water molecules form strong attractions with each other water molecule (including H-bonds!); so it takes more heat energy to break the attractive forces. Metals have weak bonds (remember the “sea of e-) and do not need as much energy to break them. water metal

8. Q = m x C x  T Q = change in thermal energy m = mass of substance  T = change in temperature (T f – T i ) C = specific heat of substance Memorize this!

9. Specific Heat Capacity If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are involved? (C Al ) = 0.897 J/gK heat gain / loss = Q = (m)(C)(∆T) where ∆T = T final - T initial Q = (25.0 g)(0.897 J/gK)(37 - 310)K Q = - 6120 J Notice that the negative sign on Q signals heat “lost by” or transferred OUT of Al.

10. Transferred heat can cause a change in temperature q transferred = (c)(mass)(∆T)

11. Or… Heat Transfer can cause a Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water Use the units to help you solve!

12. Heat Transfer and Changes of State Liquid (l)  Vapor (g) Requires energy (heat). Why do you… cool down after swimming ? use water to put out a fire?

13. Remember this – it’s the Heating/Cooling Curve for Water! Note that T is constant as phase changes (Evaporate water)

14. WHY DO I NEED  H vap or  H fus WHEN I HAVE Q = (m)(C)(∆T)? Well, when a phase changes THERE IS NO change in temperature (  T = 0)… but there is definitely a change in energy!

15. A separate step is required for EACH SEGMENT on the graph. Slanted segments = temperature changes, use Slanted segments = temperature changes, use Q = (m)(C)(∆T) Flat segments = phase changes Use Q = (mass)(  H fus ) or Q = (mass)(  H vap ) WATCH YOUR UNITS!!!! J and kJ are not the same thing.

16. For ice at 0˚C to steam at 100˚C, there are 3 steps. Q = (mass)(  H fus ) Q = (mass)(  H vap ) Q = (mass)(c)(  T)

17. Example: What quantity of heat is required to melt 500.0 g of ice and heat the water to steam at 100 o C? Heat of fusion of ice = 333 J/g Specific heat of water = 4.184 J/gK Heat of vaporization = 2,260 J/g +333 J/g +2260 J/g

18. How much heat is required to melt 500. g of ice and heat the water to steam at 100 o C? 1. To melt ice: Q 1 = (500.0 g)(333 J/g) = 1.665 x 10 5 J 2.To raise water from 0  C to 100  C: Q 2 = (500.0 g)(4.184 J/gK)(100 - 0)K = 2.092 x 10 5 J 3.To evaporate water at 100  C: Q 3 = (500.0 g)(2,260 J/g) = 1.130 x 10 6 J 4. Total heat energy = Q 1 + Q 2 + Q 3 Q total = 1.665 x 10 5 J + 2.092 x 10 5 J + 1.130 x 10 6 J Q total = 1.506 x 10 6 J = 1,506 kJ

19. Molar Heat of Combustion The amount of energy released in completely burning one mole of substance. Molar heat of solidification (freezing) and molar heat of condensation – related to molar heat of fusion and molar heat of vaporization. Same value, opposite sign!  H fus = -  H solid  H vap = -  H cond Some more terms!