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SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 5 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university.

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Presentation on theme: "SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 5 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university."— Presentation transcript:

1 SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 5 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

2 CHAPTER 5 CHEMICAL REACTIONS

3 STOICHIOMETRY - The area of study involved with calculation of the quantities of substances consumed or produced in a chemical reaction - Chemical reactions are represented by chemical equations - Reactants are substances that are consumed - Products are substances that are formed

4 CHEMICAL EQUATIONS - Reactants are written on the left side of a chemical equation and products on the right side - An arrow pointing towards the products, is used to separate the reactants from the products - The plus sign (+) is used to separate different reactants or different products

5 - Chemical equations must be consistent with experimental facts (reactants and products in a reaction that actually takes place) - Chemical equations must be balanced (equal numbers of atoms of each kind on both sides) (Daltons atomic theory) CHEMICAL EQUATIONS

6 C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) States of reactants and products Physical states of reactants and products are represented by: (g): gas (l): liquid (s): solid (aq): aqueous or water solution CHEMICAL EQUATIONS

7 - Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged) - The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation C 2 H 5 OH(l) + O 2 (g) → 2CO 2 (g) + H 2 O(g) 2 C atoms Place the coefficient 2 in front of CO 2 to balance C atoms BALANCING CHEMICAL EQUATIONS

8 C 2 H 5 OH(l) + O 2 (g) → 2CO 2 (g) + 3H 2 O(g) (5+1)=6 H atoms 3(1x2)=6 H atoms Place 3 in front of H 2 O to balance H atoms - Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged) - The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation BALANCING CHEMICAL EQUATIONS

9 C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) 1+(3x2)=7 O atoms(2x2)+3=7 O atoms Place 3 in front of O 2 to balance O atoms - Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged) - The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation BALANCING CHEMICAL EQUATIONS

10 C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) 2 C atoms (5+1)=6 H atoms 1+(3x2)=7 O atoms 2 C atoms (3x2)=6 H atoms (2x2)+3=7 O atoms - Check to make sure equation is balanced - When the coefficient is 1, it is not written - Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged) - The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation BALANCING CHEMICAL EQUATIONS

11 Balance the following chemical equations Fe(s) + O 2 (g) → Fe 2 O 3 (s) C 12 H 22 O 11 (s) + O 2 (g) → CO 2 (g) + H 2 O(g) (NH 4 ) 2 Cr 2 O 7 (s) → Cr 2 O 3 (s) + N 2 (g) + H 2 O(g) BALANCING CHEMICAL EQUATIONS

12 TYPES OF CHEMICAL REACTIONS Five Types of Chemical Reactions - Combination reaction - Decomposition reaction - Single-replacement reaction - Double-replacement reaction - Combustion reaction

13 COMBINATION REACTION - Addition or synthesis reaction - Two or more reactants produce a single product X + Y → XY N 2 (g) + 3H 2 (g) → 2NH 3 (g) 2Mg(s) + O 2 (g) → 2MgO(s) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq)

14 DECOMPOSITION REACTION - Two or more products are formed from a single reactant XY → X + Y 2H 2 O(l) → 2H 2 (g) + O 2 (g) BaCO 3 (s) → BaO(s) + CO 2 (g) 2NaN 3 (s) → 2Na(s) + 3N 2 (g)

15 SINGLE-REPLACEMENT REACTION - Substitution or Displacement reaction - An atom or molecule replaces another atom or molecule A + BY → B + AY Fe(s) + CuSO 4 (aq) → Cu(s) + FeSO 4 (aq) Zn(s) + 2HCl(aq) → ZnCl 2 (aq) + H 2 (g) Cl 2 (g) + 2NaBr(aq) → 2NaCl(aq) + Br 2 (g) - Metal replaces metal and nonmetal replaces nonmetal - Cation replaces cation and anion replaces anion

16 DOUPLE-REPLACEMENT REACTION - Exchange or metathesis (transpose) reaction - Parts of two compounds switch places to form two new compounds AX + BY → AY + BX AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq) BaCl 2 (aq) + Na 2 SO 4 (aq) → BaSO 4 (s) + 2NaCl(aq) CaCl 2 (aq) + Na 2 CO 3 (aq) → CaCO 3 (s) + 2NaCl(aq)

17 COMBUSTION REACTION - Reaction between a substance and oxygen (air) accompanied by the production of heat and light - A common synonym for combustion is burn CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) + heat 2CH 3 OH + 3O 2 (g) → 2CO 2 (g) + 4H 2 O(g) + heat 2Mg(s) + O 2 (g) → 2MgO(s) + heat Hydrocarbons are the most common type of compounds that undergo combustion producing CO 2 and H 2 O

18 - Also called redox reactions - Involve transfer of electrons Oxidation - loss of electrons Reduction - gain of electrons Ionic solid sodium chloride (Na + and Cl - ions) formed from solid sodium and chlorine gas 2Na(s) + Cl 2 (g) → 2NaCl(s) The oxidation (rusting) of iron by reaction with moist air 4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s) OXIDATION-REDUCTION REACTIONS

19 - There is no transfer of electrons from one reactant to another reactant BaCO 3 (s) → BaO(s) + CO 2 (g) - Double replacement reactions - Most reactions we have already come across NONREDOX REACTIONS

20 OXIDATION NUMBER (STATE) The concept of oxidation number - provides a way to keep track of electrons in redox reactions - not necessarily ionic charges Conventionally - actual charges on ions are written as n+ or n- - oxidation numbers are written as +n or -n Oxidation - increase in oxidation number (loss of electrons) Reduction - decrease in oxidation number (gain of electrons)

21 OXIDATION NUMBERS 1. Oxidation number of uncombined elements = 0 Na(s), O 2 (g), H 2 (g), Hg(l) 2. Oxidation number of a monatomic ion = charge Na + = +1, Cl - = -1, Ca 2+ = +2, Al 3+ = +3 3. Oxygen is usually assigned -2 H 2 O, CO 2, SO 2, SO 3 Exceptions: H 2 O 2 (oxygen = -1) OF 2 (oxygen = +2)

22 4. Hydrogen is usually assigned +1 (-1 when bonded to metals) +1: HCl, NH 3, H 2 O -1: CaH 2, NaH 5. Halogens are usually assigned -1 (F, Cl, Br, I) Exceptions: when Cl, Br, andI are bonded to oxygen Cl 2 O: Cl O Cl 6. The sum of oxidation numbers for - neutral compound = 0 - polyatomic ion = charge H 2 O = 0, CO 3 2- = -2, NH 4 + = +1 +1-2+1 OXIDATION NUMBERS

23 CO 2 The oxidation state of oxygen is -2 CO 2 has no charge The sum of oxidation states of carbon and oxygen = 0 1 carbon atom and 2 oxygen atoms 1(x) + 2(-2) = 0 x = +4 CO 2 x-2 for each oxygen OXIDATION NUMBERS

24 CH4CH4 x+1 1(x) + 4(+1) = 0 x = -4 OXIDATION NUMBERS

25 NO3-NO3- x-2 1(x) + 3(-2) = -1 x = +5 OXIDATION NUMBERS

26 CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) +1-4+4+10-2 CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g) -4 +1 +4 8e - loss -2 0 8e - gain OXIDATION NUMBERS

27 Oxidation Loss of electrons Increase in oxidation number Reducing Agent (electron donor) Reduction Gain of electrons Decrease in oxidation number Oxidizing Agent (electron acceptor) Mnemonic OIL RIG Oxidation Involves Loss; Reduction Involves Gain Redox reactions are characterized by transfer of electrons OXIDATION NUMBERS

28 COLLISION THEORY - Conditions necessary for a chemical reaction to occur Molecular Collisions - Reactant particles must collide (interact) with one another Activation Energy - The colliding particles must possess a certain minimum amount of total energy, known as the activation energy Collision Orientation - The particles must collide in a proper orientation (exceptions: single atoms, small and symmetrical molecules)

29 - When all soluble strong electrolytes are shown as ions - Chemical equation is balanced - Soluble compounds (aq) are separated into ions (only strong electrolytes) - Insoluble compounds (s), liquids (l), and gases (g) are NOT separated into ions IONIC EQUATIONS

30 Complete ionic equation - When all ions in both reactants and products are shown AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq) Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq) → AgCl(s) + K + (aq) + NO 3 - (aq) IONIC EQUATIONS

31 Net Ionic Equation - When spectator ions are cancelled from the complete ionic equation - Net charge on reactant side must equal net charge on product side Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq) → AgCl(s) + K + (aq) + NO 3 - (aq) Ag + (aq) + Cl - (aq) → AgCl(s) - Some ions appear on both reactant and product sides - These ions play no direct role in the reaction - These ions are called spectator ions IONIC EQUATIONS

32 Neutralization Reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) Complete Ionic Equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) → Na + (aq) + Cl - (aq) + H 2 O(l) Net Ionic Equation H + (aq) + OH - (aq) → H 2 O(l) IONIC EQUATIONS

33 EXOTHERMIC REACTION - Reaction in which heat energy is released (exo- means ‘out of’) - Heat energy is one of the products of the reaction - Energy required to break bonds in reactants is less than energy released by bond formation in products - Combustion of gasoline CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) + heat

34 ENDOTHERMIC REACTION - Reaction in which heat energy is absorbed (endo- means ‘into’) - Heat energy is one of the reactants of the reaction - Energy required to break bonds in reactants is more than energy released by bond formation in products - Melting of ice (reason why it feels cold) - Photosynthesis in plants N 2 (g) + O 2 (g) + heat → 2NO(g)

35 CHEMICAL EQUATIONS (STOICHIOMETRIC CALCULATIONS) Given: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) - 1 molecule of C 3 H 8 reacts with 5 molecules of O 2 to produce 3 molecules of CO 2 and 4 molecules of H 2 O - 1 mole of C 3 H 8 reacts with 5 moles of O 2 to produce 3 moles of CO 2 and 4 moles of H 2 O

36 - make sure the equation is balanced - calculate moles of propane from given mass and molar mass - determine moles of oxygen from mole ratio (stoichiometry) - calculate mass of oxygen = 349 g O 2 CHEMICAL EQUATIONS (STOICHIOMETRIC CALCULATIONS) Given: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) What mass of oxygen will react with 96.1 g of propane?

37 - make sure the equation is balanced - calculate moles of propane from given mass and molar mass - determine moles of CO 2 from mole ratio (stoichiometry) - calculate mass of CO 2 = 288 g CO 2 CHEMICAL EQUATIONS (STOICHIOMETRIC CALCULATIONS) Given: C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) What mass of CO 2 will be produced from 96.1 g of propane?

38 - Also called limiting reagent - The reactant that is completely consumed in a reaction - The reactant(s) with leftovers is (are) known as the excess reactant(s) or excess reagent(s) To determine the limiting reactant: - Write and balance the equation for the reaction - Use given amount of each reactant to determine amount of desired product - The reactant that gives the smallest amount of product is the limiting LIMITING REACTANT

39 Consider the following reaction for producing nitrogen gas from gaseous ammonia and solid copper(II) oxide: LIMITING REACTANT 2NH 3 (g) + 3CuO(s) → N 2 (g) + 3Cu(s) + 3H 2 O(g) If a sample containing 18.1 g of NH 3 is reacted with 90.4 g of CuO, which is the limiting reactant?

40 - Make sure the equation is balanced - Calculate moles of desired product from each reactant LIMITING REACTANT 2NH 3 (g) + 3CuO(s) → N 2 (g) + 3Cu(s) + 3H 2 O(g) CuO is limiting since it produces smaller amount of N 2

41 PERCENT YIELD Theoretical Yield The calculated quantity of product formed, assuming all of the limiting reactant is used up Actual Yield The amount of product actually obtained in a reaction (always less than the theoretical yield)

42 PERCENT YIELD Given actual yield: - Determine the limiting reactant - Calculate the theoretical yield from the limiting reactant - Calculate percent yield

43 PERCENT YIELD Calculate the percent yield of N 2 from the previous example if 9.04 g of N 2 is actually produced - CuO is the limiting reactant - Calculate the theoretical yield 2NH 3 (g) + 3CuO(s) → N 2 (g) + 3Cu(s) + 3H 2 O(g)


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