Active IngredientsKClNaClNH2CONH2 (urea)C7O6H14 (methyl-D-glucopyranoside; a surfactant)

Ace Ice MelterRAPIDLY 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).

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?

• 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.

2 H2 + O2 2 H2O

Energy

2 H2 + O2

2 H2O

•If the products contain less energy, energy must have been given off during the reaction.

Energy barrier

2 H2 + O22 H2O

Energy

2 H2 + O2

2 H2O

•If the products contain more energy, energy must have been absorbed during the reaction.

Energy barrier

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.

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.

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 2C.

Step 1: Write out the chemical equation and balance it. Na (s)+ H2O (l) NaOH-1+1 + H2 (g)(aq)22 2

Step 2: Determine if there’s a limiting reagent.*

2H mol 1.26Na 22H 1

x g/mol 22.98977Na g 1.0

2H mol 2.77O2H 22H 1

x g/mol 18.015

O2H g 100.0

Na is the limiting reagent. Only 1.26 mol of H2 will be formed.

Na is the limiting reagent. Only 1.26 mol of H2 will be formed.

*Choose a product that has a coefficient of 1 for best results.

Step 3: Determine the amount of heat involved in the reaction.

q = mCpT

q = ? m = 100 g Cp = 4.184 g/JC T = 2 C

J 837C)C)(2J/g O)(4.1842H g (100 q

Step 4: Calculate your molar heat of reaction.

If a reaction that produced 1.26 moles of H2 also released 837 J of heat, then the molar enthalpy (heat) change for this reaction would be:

J/mol 664mol 1.26J 837

rxnΔH

A simpler problem:

How much heat is given off when 1.6 g of CH4 are burned in an excess of oxygen if Hcomb = -802 kJ/mol?

Step 1: Write the reaction equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Step 2: Calculate molar amount involved

combusted 4CH moles 0.100g/mol 16.042

4CH g 1.6

Step 3: Calculate amount of heat given off

Hrxn = (-802 kJ/mol)(0.100 mol CH4) = -80.2 kJ

Q: Is this an exothermic or endothermic reaction?

EX 3: What is the molar heat of combustion of propene (C3H6) if burning 3.2 g releases 156 kJ of heat?

Step 1: Write the reaction equation.

2 C3H6 (g) + 9 O2 (g) 6 CO2 (g) + 6 H2O (g)

• This reaction equation involves the combustion of 2 moles of C3H6 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.

C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g)

Step 2: Convert grams of propene to moles.

6363 HC mol 0.076

g/mol 42.081HC g 3.2

Step 3: Divide the heat released by moles of propene.

kJ/mol 2053HC mol 0.076

kJ 156combHΔ

63

EX 4: How many grams of potassium chlorate were originally present if its decomposition into oxygen gas and potassium chloride had a Hrxn = -111 kJ? (Hdecomp = -44.44 kJ/mol)

Step 1: Write out the reaction equation.

2 KClO3 (s) 3 O2 (g) + 2 KCl (s)

Then divide through by 2 so that there’s a 1 in front of the KClO3

KClO3 (s) 1.5 O2 (g) + KCl (s) Hrxn = -111 kJ

Step 2: Solve for the number of moles of KClO3

kJ/mol 44.443KClO mol

kJ 111 x

x = 2.50 mol KClO3

Step 3: Convert moles to grams.

(2.50 mol KClO3)(122.55 g/mol) = 306.38 g KClO3 decomposed