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Chapter 6 Chemical Reactions: An Introduction Chemical Reactions • Reactions involve chemical changes in matter that result in new substances. • Reactions involve rearrangement and exchange of atoms to produce new molecules. • 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 Evidence of Chemical Reactions (cont.) 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. Conservation of Mass (cont.) 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 C H H H + O O C + O H H O 1C+4H + 2O 1C+2O +2H +O 1C+2H+3O Combustion of Methane Balanced • To show a reaction obeys the Law of Conservation of Mass, it must be balanced. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) H H C H H O + O C + O 1C + 4H + 4O O O O O + H H + O H H 1C + 4H + 4O 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 equations show 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 – e. g. NH3(aq) indicates ammonia 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. Recognizing Reactants and Products (cont.) • Write the equation in words – Identify the state of each chemical magnesium(s) + oxygen(g) → magnesium oxide(s) • Write the equation in formulas – Identify diatomic elements – Identify polyatomic ions – Determine formulas Mg(s) + O2(g) → MgO(s) (unbalanced) Balancing by Inspection • Count atoms of each element – Polyatomic ions may be counted as one “element” if they do not change in the reaction. Al + FeSO4 → Al2(SO4)3 + Fe 1 SO4 3 – If an element appears in more than one compound on the same side, count each element separately and add. CO + O2 → CO2 1 + 2 O 2 Balancing by Inspection (cont.) • Pick an element to balance. • Find least common multiple (LCM) and factors needed to make both sides equal. • Use factors as coefficients in equation. – If already a coefficient, then multiply by new factor • Recount and repeat until balanced. Example #1 • When magnesium metal burns in air it produces a white, powdery compound, magnesium oxide. – Burning in air means reacting with O2 Example #1 (cont.) • Write the equation in words. – Identify the state of each chemical magnesium(s) + oxygen(g) → magnesium oxide(s) • Write the equation in formulas. – Identify diatomic elements – Identify polyatomic ions – Determine formulas Mg(s) + O2(g) → MgO(s) (unbalanced) Example #1 (cont.) • 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 Example #1 (cont.) • 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 1x2←O→1x2 Example #1 (cont.) • Use factors as coefficients in front of compound containing the element – If coefficient is already there, multiply them together Mg(s) + O2(g) → 2 MgO(s) 1 ← Mg → 1 x 2 1x2←O→1x2 Example #1 (cont.) • Recount Mg(s) + O2(g) → 2 MgO(s) 1 ← Mg → 2 2←O→2 • Repeat 2 Mg(s) + O2(g) → 2 MgO(s) 2 x 1 ← Mg → 2 2←O→2 Example #2 • Under appropriate conditions, at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water. Example #2 (cont.) • 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) Example #2 (cont.) • 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 Example #2 (cont.) • Pick an element to balance – Avoid elements 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 2x3←H→2x3 2←O→1+1 Example #2 (cont.) • Use factors as coefficients in front of compound containing the element. 2 NH3(g) + O2(g) → NO(g) + 3 H2O(g) 1←N→1 2x3←H→2x3 2←O→1+1 Example #2 (cont.) • Recount 2 NH3(g) + O2(g) → NO(g) + 3 H2O(g) 2←N→1 6←H→6 2←O→1+3 • Repeat 2 NH3(g) + O2(g) → 2 NO(g) + 3 H2O(g) 2←N→1x2 6←H→6 2←O→1+3 Example #2 (cont.) • Recount 2 NH3(g) + O2(g) → 2 NO(g) + 3 H2O(g) 2←N→2 6←H→6 2←O→2+3 Example #2 (cont.) • Repeat – 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 Example #2 (cont.) • 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