Chapter 6 Chemical Reactions: an Introduction

Honors Chemistry Mr. Pendolino

Chapter 6 Chemical Reactions: An Introduction

Chemical Reactions • Reactions involve chemical changes in matter resulting in new substances • Reactions involve rearrangement and exchange of atoms to produce new molecules – Elements are not transmuted during a reaction Reactants  Products

Evidence of Chemical Reactions • a chemical change occurs when new substances are made • visual clues (permanent) – color change, precipitate formation, gas bubbles, flames, heat release, cooling, light • other clues – new odor, permanent new state

Chemical Equations • Shorthand way of describing a reaction • Provides information about the reaction – Formulas of reactants and products – States of reactants and products – Relative numbers of reactant and product molecules that are required – Can be used to determine weights of reactants used and of products that can be made

Conservation of Mass • Matter cannot be created or destroyed • In a chemical reaction, all the atoms present at the beginning are still present at the end • Therefore the total mass cannot change • Therefore the total mass of the reactants will be the same as the total mass of the products

Combustion of Methane • methane gas burns to produce carbon dioxide gas and liquid water – whenever something burns it combines with O2(g)

CH4 (g) + O2 (g)  CO2(g) + H2O(l)

O H H O C + O O C + H H H H O

of 6 Honors Chemistry Mr. Pendolino

1 C + 4 H + 2 O - 1 C + 2 O + 2 H + O 1 C + 2 H + 3 O

Combustion of Methane Balanced

• to show the reaction obeys the Law of Conservation of Mass it must be balanced

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

O O O O H H H H C + + C + + H H O O O O H H 1 C + 4 H + 4 O  1 C + 4 H + 4 O

Writing Equations • Use proper formulas for each reactant and product • proper equation should be balanced – obey Law of Conservation of Mass – all elements on reactants side also on product side – equal numbers of atoms of each element on reactant side as on product side • balanced equation shows the relationship between the relative numbers of molecules of reactants and products – can be used to determine mass relationships

Symbols Used in Equations • symbols used after chemical formula to indicate state – (g) = gas; (l) = liquid; (s) = solid – (aq) = aqueous, dissolved in water

Sample – Recognizing Reactants and Products • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 – Metals are solids, except for Hg which is liquid ¬ write the equation in words – identify the state of each chemical Magnesium + oxygen (g) magnesium oxide(s) write the equation in formulas – identify diatomic elements – identify polyatomic ions – determine formulas

Mg(s) + O2 (g)  MgO(s)

Balancing by Inspection ¬ Count atoms of each element a polyatomic ions may be counted as one “element” if it does not change in the reaction

Al + FeSO4  Al2(SO4)3 + Fe 1 SO4 3 b if an element appears in more than one compound on the same side, count each separately and add

CO + O2  CO2 1 + 2 O 2

Balancing by Inspection Pick an element to balance a avoid elements from 1b ® Find Least Common Multiple and factors needed to make both sides equal ¯ Use factors as coefficients in equation a if already a coefficient then multiply by new factor ° Recount and Repeat until balanced

Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 ¬ write the equation in words – identify the state of each chemical Magnesium + oxygen (g)  magnesium oxide(s) write the equation in formulas – identify diatomic elements – identify polyatomic ions – determine formulas

Mg(s) + O2 (g)  MgO(s)

Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 ® count the number of atoms of on each side – count polyatomic groups as one “element” if on both sides – split count of element if in more than one compound on one side

Mg(s) + O2 (g)  MgO(s) 1  Mg 1 2  O  1

Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 ¯ pick an element to balance – avoid element in multiple compounds ° find least common multiple of both sides & multiply each side by factor so it equals LCM

Mg(s) + O2 (g)  MgO(s) 1  Mg 1 1 x 2  O  1 x 2

Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 ± use factors as coefficients in front of compound containing the element 3 if coefficient already there, multiply them together

Mg(s) + O2 (g)  2 MgO(s) 1  Mg 1 1 x 2  O  1 x 2

Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 ² Recount

Mg(s) + O2 (g)  2 MgO(s) 1  Mg 2 2  O  2 ³ Repeat

2 Mg(s) + O2 (g)  2 MgO(s) 2 x 1  Mg 2 2  O  2

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water • write the equation in words – identify the state of each chemical Ammonia (g) + oxygen (g)  nitrogen monoxide (g) + water (g) write the equation in formulas – identify diatomic elements – identify polyatomic ions – determine formulas

NH3 (g) + O2 (g)  NO (g) + H2O (g)

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ® count the number of atoms of on each side – count polyatomic groups as one “element” if on both sides – split count of element if in more than one compound on one side

NH3 (g) + O2 (g)  NO (g) + H2O (g) 1  N 1 3  H  2 2  O  1 + 1

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ¯ pick an element to balance – avoid element in multiple compounds ° find least common multiple of both sides & multiply each side by factor so it equals LCM

NH3 (g) + O2 (g)  NO (g) + H2O (g) 1  N 1 2 x 3  H  2 x 3 2  O  1 + 1

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ± use factors as coefficients in front of compound containing the element

2 NH3 (g) + O2 (g)  NO (g) + 3 H2O (g) 1  N 1 2 x 3  H  2 x 3 2  O  1 + 1

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ² Recount

2 NH3 (g) + O2 (g)  NO (g) + 3 H2O (g) 2  N 1 6  H  6 2  O  1 + 3 ³ Repeat

2 NH3 (g) + O2 (g) 2 NO (g) + 3 H2O (g) 2  N 1 x 2 6  H  6 2  O  1 + 3

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ´ Recount

2 NH3 (g) + O2 (g)  2 NO (g) + 3 H2O (g) 2  N 2 6  H  6 2  O  2 + 3

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water µ Repeat – A trick of the trade, when you are forced to attack an element that is in 3 or more compounds – find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction! – We want to make the O on the left equal 5, therefore we will multiply it by 2.5

2 NH3 (g) + 2.5 O2 (g) 2 NO (g) + 3 H2O (g) 2  N 2 6  H  6 2.5 x 2  O  2 + 3

Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ³ Multiply all the coefficients by a number to eliminate fractions – x.5  2, x.33  3, x.25  4, x.67  3

2 x [2 NH3 (g) + 2.5 O2 (g) 2 NO (g) + 3 H2O (g)]

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g) 4  N 4 12  H  12 10  O  10