2. Active Ingredients KCl NaCl NH 2 CONH 2 (urea) C 7 O 6 H 14 (methyl-D-glucopyranoside; a surfactant) Ace Ice Melter RAPIDLY MELTS ICE AND SNOW AND PREVENTS RE ‑ FREEZING Ace Ice Melter features a special custom blend of superior ice melting ingredients. Together they melt even the most stubborn ice and snow and work to prevent re-freezing. Melts ice down to 0°F (-18 °C).

3. What have we learned? Sometimes heat is given off during a chemical reaction. This makes it feel hotter. Sometimes heat is absorbed during a chemical reaction. This makes it feel colder. What causes it to be different?

4. Chemical bonds contain energy. Add the energy of all of the bonds in the reactants together to find their total energy. Add the energy of all of the bonds in the products together to find their total energy. If the two numbers aren’t the same (and they almost never are), then there will be heat energy given off or taken in.

5. 2 H 2 + O 2 2 H 2 O Energy 2 H 2 + O 2 2 H 2 O If the products contain less energy, energy must have been given off during the reaction. Energy barrier

6. 2 H 2 + O 2 2 H 2 O Energy 2 H 2 + O 2 2 H 2 O If the products contain more energy, energy must have been absorbed during the reaction. Energy barrier

7. If heat energy is given off during a reaction, it is called an EXERGONIC or an EXOTHERMIC REACTION. Heat exits = exergonic = exothermic Exergonic reactions can be recognized by a temperature INCREASE.

8. If heat energy is absorbed during a reaction, it is called an ENDERGONIC or an ENDOTHERMIC REACTION. Heat enters = endergonic = endothermic Endergonic reactions can be recognized by a temperature DECREASE.

10. 1.0 gram of solid sodium metal is added to 100 g of water. The reaction produces sodium hydroxide and hydrogen gas. Calculate the molar heat of reaction if the water’s temperature increased by 2  C. Step 1: Write out the chemical equation and balance it. Na (s)+ H 2 O (l)NaOH +1 + H 2 (g)(aq)222 Step 2: Determine if there’s a limiting reagent.* Na is the limiting reagent. Only 1.26 mol of H 2 will be formed. *Choose a product that has a coefficient of 1 for best results.

11. Step 3: Determine the amount of heat involved in the reaction. q = mC p  T q = ?m = 100 g C p = 4.184 g/J  C  T = 2  C Step 4: Calculate your molar heat of reaction. If a reaction that produced 1.26 moles of H 2 also released 837 J of heat, then the molar enthalpy (heat) change for this reaction would be:

12. A simpler problem: How much heat is given off when 1.6 g of CH 4 are burned in an excess of oxygen if  H comb = -802 kJ/mol? Step 1: Write the reaction equation. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) Step 2: Calculate molar amount involved Step 3: Calculate amount of heat given off  H rxn = (-802 kJ/mol)(0.100 mol CH 4 ) = -80.2 kJ Q: Is this an exothermic or endothermic reaction?

13. EX 3: What is the molar heat of combustion of propene (C 3 H 6 ) if burning 3.2 g releases 156 kJ of heat? Step 1: Write the reaction equation. 2 C 3 H 6 (g) + 9 O 2 (g)  6 CO 2 (g) + 6 H 2 O (g) This reaction equation involves the combustion of 2 moles of C 3 H 6 and we want to find out what it is for one mole. Save yourself a headache and simplify future calculations by dividing the reaction equation through by 2. C 3 H 6 (g) + 4.5 O 2 (g)  3 CO 2 (g) + 3 H 2 O (g) Step 2: Convert grams of propene to moles. Step 3: Divide the heat released by moles of propene.

14. EX 4: How many grams of potassium chlorate were originally present if its decomposition into oxygen gas and potassium chloride had a  H rxn = -111 kJ? (  H decomp = -44.44 kJ/mol) Step 1: Write out the reaction equation. 2 KClO 3 (s)  3 O 2 (g) + 2 KCl (s) Then divide through by 2 so that there’s a 1 in front of the KClO 3 KClO 3 (s)  1.5 O 2 (g) + KCl (s)  Hrxn = -111 kJ Step 2: Solve for the number of moles of KClO 3 x = 2.50 mol KClO 3 Step 3: Convert moles to grams. (2.50 mol KClO 3 )(122.55 g/mol) = 306.38 g KClO 3 decomposed