1A4PPT - Chemistry 1A Chapter 4 Chemical Reaction • A...

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Unformatted text preview: Chemistry 1A Chapter 4 Chemical Reaction • A chemical change or chemical reaction is a process in which one or more pure substances are converted into one or more different pure substances. Chemical Reactions - Example Chemical Equations (1) • Chemical equations show the formulas for the substances that take part in the reaction. – The formulas on the left side of the arrow represent the reactants, the substances that change in the reaction. The formulas on the right side of the arrow represent the products, the substances that are formed in the reaction. If there are more than one reactant or more than one product, they are separated by plus signs. The arrow separating the reactants from the products can be read as “goes to” or “yields” or “produces.” Chemical Equations (2) • The physical states of the reactants and products are provided in the equation. – A (g) following a formula tells us the substance is a gas. Solids are described with (s). Liquids are described with (l). When a substance is dissolved in water, it is described with (aq) for “aqueous,” which means “mixed with water.” Chemical Equations (3) • The relative numbers of particles of each reactant and product are indicated by numbers placed in front of the formulas. – These numbers are called coefficients. An equation containing correct coefficients is called a balanced equation. – If a formula in a balanced equation has no stated coefficient, its coefficient is understood to be 1. Chemical Equations (4) • If special conditions are necessary for a reaction to take place, they are often specified above the arrow. – Some examples of special conditions are electric current, high temperature, high pressure, or light. Chemical Equation Example Special Conditions Balancing Chemical Equations • Consider the first element listed in the first formula in the equation. – If this element is mentioned in two or more formulas on the same side of the arrow, skip it until after the other elements are balanced. – If this element is mentioned in one formula on each side of the arrow, balance it by placing coefficients in front of one or both of these formulas. • Moving from left to right, repeat the process for each element. • When you place a number in front of a formula that contains an element you tried to balance previously, recheck that element and put its atoms back in balance. Balancing Equations – Strategies (1) • Strategy 1: Often, an element can be balanced by using the subscript for this element on the left side of the arrow as the coefficient in front of the formula containing this element on the right side of the arrow and vice versa (using the subscript of this element on the right side of the arrow as the coefficient in front of the formula containing this element on the left side). Balancing Equations – Strategies (2) • Strategy 2: The pure nonmetallic elements (H2, O2, N2, F2, Cl2, Br2, I2, S8, Se8, and P4 ) can be temporarily balanced with a fractional coefficient (1/2, 3/2, 5/2, etc.). If you do use a fraction during the balancing process, you can eliminate it later by multiplying each coefficient in the equation by the fraction’s denominator. Balancing Equations – Strategies (3) • Strategy 3: If polyatomic ions do not change in the reaction, and therefore appear in the same form on both sides of the chemical equation, they can be balanced as if they were single atoms. • Strategy 4: If you find an element difficult to balance, leave it for later. Equation Stoichiometry (1) • Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction. • General Steps 1. If you are not given it, write and balance the chemical equation for the reaction. Equation Stoichiometry (2) 2. Start your dimensional analysis in the usual way. 3. If you are not given grams of substance 1, convert from the unit that you are given to grams. This may take one or more conversion factors. 4. Convert from grams of substance 1 to moles of substance 1. Equation Stoichiometry (3) 5. Convert from moles of substance 1 to moles of substance 2 using the coefficients from the balanced equation to create the molar ratio used as a conversion factor. 6. Convert from moles of substance 2 to grams of substance 2, using its molar mass. 7. If necessary, convert from grams of 2 to the desired unit for 2. This may take one or more conversion factors. Equation Stoichiometry Steps Equation Stoichiometry Shortcut Questions to Ask When Designing a Process for Making a Substance • How much of each reactant should be added to the reaction vessel? • What level of purity is desired for the final product? If the product is mixed with other substances (such as excess reactants), how will this purity be achieved? Limiting Component Limiting Component (2) Limiting Reactant • The reactant that runs out first in a chemical reaction limits the amount of product that can form. This reactant is called the limiting reactant. Why substance limiting? (1) • To ensure that one or more reactants are converted to products most completely. – Expense P4(s) + 5O2(g) → P4O10(s) + excess O2(g) – Importance SiO2(s) + 2C(s) → Si(l) + 2CO(g) + excess O2(g) Why substance limiting? (2) • Concern for excess reactant that remains – danger P4(s) + 5O2(g) → P4O10(s) + excess O2(g) – ease of separation SiO2(s) + 2C(s) → Si(l) + 2CO(g) + excess O2(g) Limiting Reactant Problems • Tip-off - You are given two amounts of reactants in a chemical reaction, and you are asked to calculate the maximum amount of a product that can form from the combination of the reactants. • General Steps 1. Do two separate calculations of the maximum amount of product that can form from each reactant. 2. The smaller of the two values calculated in the step above is your answer. It is the maximum amount of product that can be formed from the given amounts of reactants. Percent Yield Actual Yield Percent Yield = x 100% Theoretical Yield • Actual yield is measured. It is given in the problem. • Theoretical yield is the maximum yield that you calculate. Why not 100% Yield? • • • • Reversible reactions Side Reactions Slow Reactions Loss during separation/purification More Types of Chemical Reactions • • • • Combination Decomposition Combustion Single Displacement Combination Reactions • In combination reactions, two or more elements or compounds combine to form one compound. 2Na(s) + Cl2(g) → 2NaCl(s) C(s) + O2(g) → CO2(g) MgO(s) + H2O(l) → Mg(OH)2(s) Decomposition Reactions • In decomposition reactions, one compound is converted into two or more simpler substances. Electric current 2H2O(l) → 2H2(g) + O2(g) Electric current 2NaCl(l) → 2Na(l) + Cl2(g) Combustion Reactions • A combustion reaction is a redox reaction in which oxidation is very rapid and is accompanied by heat and usually light. The combustion reactions that you will be expected to recognize have oxygen, O2, as one of the reactants. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) Combustion Products (1) • When any substance that contains carbon is combusted (or burned) completely, the carbon forms carbon dioxide. • When a substance that contains hydrogen is burned completely, the hydrogen forms water. 2C6H14(g) + 19O2(g) → 12CO2(g) + 14H2O(l) Combustion Products (2) • The complete combustion of a substance, like ethanol, C2H5OH, that contains carbon, hydrogen, and oxygen also yields carbon dioxide and water. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) Combustion Products (3) • When any substance that contains sulfur burns completely, the sulfur forms sulfur dioxide. CH3SH(g) + 3O2(g) → CO2(g) + 2H2O(l) + SO2(g) Single Displacement Single Displacement Reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) oxidation: Zn(s) → Zn2+(aq) + 2e− reduction: Cu2+(aq) + 2e− → Cu(s) Single Displacement Reaction Example ...
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