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Chapter 4

# CHM110 Chapter Four

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Department
Chemistry
Course
CHM110H5
Professor
Heather Miller
Semester
Summer

Description
Chemistry Chapter Four Chemical Reactions and Chemical Equations A chemical reaction is a process in which one set of substances (reactants) is converted to a new set of substances (products). The main evidences that prove that a chemical reaction has occurred are: a) Colour change b) Formation of a solid (precipitate) within a clear solution c) Evolution of a gas d) Evolution or absorption of heat To write the shorthand representation of a chemical equation certain steps must be followed. Consider the chemical reaction: nitrogen monoxide + oxygen  nitrogen dioxide 1. Write the reaction using chemical symbols NO + O 2NO 2 2. Balance the chemical equation 2NO + 1O2  2 NO2 *** Balancing equations*** Remember these rules when balancing an equation 1. An equation can be balanced only by adjusting the coefficients of formulas. 2. Never introduce extraneous atoms to balance. NO + O 2NO + O2 3. Never change a formula for the purpose of balancing an equation. NO + O 2NO 3 When balancing equations, using these strategies will result in efficient balancing of chemical equations.  Balance elements that occur in only one compound of each side first.  Balance free elements last.  Balance unchanged polyatomics (or other groups of atoms) as groups.  Fractional coefficients are acceptable and can be cleared at the end by multiplication. Problem: Balance C H O 6 O14 4O + H2O 2 2 Solution: Balance C: C H 6 +14 4 6 CO2+ H O 2 2 Balance H: C H O + O  6 CO + 7H O 6 14 4 2 2 2 At this point, the right side contains 19 O atoms whereas the left only contains 4. To get 15 more O atoms we need to add 15/2 more O atoms by multiplying it with O . 2 Balance O: C H 6 14(14/2) O  6 CO 2 7H O 2 2 Now remove the fraction by multiplying the whole equation by 2. 2C 6 O14 14 O  12 2O + 14 H O 2 2 Remember to check both sides to see that both sides contain equal amounts of each element present. Problem: Write a balanced equation for the complete combustion of C H , a 8 18 gasoline component. Solution: 2C H8 18+ 25O  26CO + 18H 2 2 States of Matter The states of matter in a chemical equation are represented by subscript symbols next to the compound. (g) gas (l) liquid (s) solid Example: 2C H + 25O  16CO + 18H O 8 18(l) 2(g) 2(g) 2 (l) Another common symbol which is used for reactants or products dissolved in water is (aq): aqueous solution. Reaction Conditions A simple chemical equation cannot convey the information of how to carry out the reaction in an experiment. However there are symbols that tell us what must be done for the reaction to occur. Example: Δ which means to heat up as in the decomposition reaction of silver oxide: 2 Ag O2 (s) 4 Ag (s)+ O 2 (g) General equations and Stoichiometry Stoichiometry includes all the quantitative relationships involving atomic and formula masses, chemical formulas, and chemical equations. Consider: 2H 2 (g) O 2(g) 2H O 2 (l) This means that : 2x molecules H 2 (g) x molecules O 2(g) 2x molecules H O 2 (l) And since x = Avogadro’s number which represents 1 mole, we can also write: 2 mol H 2 (g) 1 mol O 2(g) 2 mol H O 2 (l) The coefficients in the equation allow us to make statements such as: • Two moles of H 2 are produced for every two moles of H consumed. 2 • Two moles of H O are2produced for every one mole of O consumed. 2 • Two moles of H are c2nsumed for every one mole of O consumed. 2 These statements can be turned into conversion factors called stoichiometric factors. A stoichiometric factor relates the amounts, on a mole basis, of any two substances involved in a chemical reaction. Thus a stoichiometric factor is also a mole ratio. Relating the Numbers of Moles of Reactant and Products Problem: How many moles of CO are prod2ced in the combustion of 2.72 mol of triethylene glycol, C H6O 14i4 an excess of O ? 2 An excess of oxygen gas means that triethylene glycol is the limiting reactant. Balance the equation: 2 C 6 14+ 45O  122O +14H O 2 2 ? mol CO =2[2.72 mol C H O 6 (14mo4 CO / 2 mol C 2 O )] = 1663 14l4CO 2 Chemical Reactions in Solutions Most chemical reactions take place in solution partly because mixing the reactants helps to achieve the close contact between atoms, ions, or molecules necessary for the reaction to occur. Solvent determines whether the solution exists as a solid, liquid, or a gas. However, for learning purposes, the aqueous (aq) solution will be used as a solvent. Solute is the material being dissolved by the solvent. Aqueous reactions can be grouped into three general categories: Precipitation reactions, acid-base neutralization reactions and oxidation-reduction (red ox) reactions. Molarity Molarity is a solu
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