TECHNICAL SCIENCE DAS12703 ROZAINITA BT. ROSLEY PUSAT PENGAJIAN DIPLOMA UNVERSITI TUN HUSSEIN ONN MALAYSIA

Transcription

1 TECHNICAL SCIENCE DAS12703 ROZAINITA BT. ROSLEY PUSAT PENGAJIAN DIPLOMA UNVERSITI TUN HUSSEIN ONN MALAYSIA

2 ii TABLE OF CONTENTS TABLE OF CONTENTS... i LIST OF FIGURES... iii Chapter SOLUTIONS OBJECTIVES INTRODUCTION CHEMICAL EQUATION STOICHIOMETRY OF SOLUTION REACTIONS LIMITING REAGENT THEORITICAL YIELD, ACTUAL YIELD AND PERCENTAGE YIELD SOLUTION CONCENTRATION VOLUMETRIC ANALYSIS CHEMICAL KINETIC OBJECTIVES INTRODUCTION RATE OF REACTION AND ITS MEASUREMENT RATE LAW... 21

3 iii LIST OF FIGURES FIGURE 1 SIMPLE REACTION OF A AND B FIGURE 2 THE RATE OF REACTION A B, REPRESENTED AS THE DECREASE OF A MOLECULES WITH TIME AND AS INCREASE OF B MOLECULES WITH TIME FIGURE 3 REACTION RATE VS. CONCENTRATION OF REACTANT... 23

4 4 CHAPTER 1 SOLUTIONS 1.1 OBJECTIVES Upon completion of this chapter, students should have: 1. To understand the process of dissolving. 2. To learn why certain components dissolve in water. 1.2 INTRODUCTION Most of the important chemistry that keeps plants, animals and humans functioning occurs in aqueous solution. Even the water that comes out of a tap is not pure water but a solution of various materials in water. For example, tap water may contain dissolve chlorine to disinfect it, dissolved materials that make it hard and traces of many other substances that result from natural and human-initiated pollution. A solution is a homogenous mixture, a mixture in which the components are uniformly intermingled. This means that a sample from one part is the same as a sample from any other part. For example, the first sip of coffee is the same as the last sip. The atmosphere that surrounds us is a gaseous solution containing O2 (g), N2 (g), and other gases randomly dispersed. Solutions can also be solids. For example, brass is a homogenous mixture a solution of copper and zinc. The substance present in largest amount is called the solvent, and the other substance/substances are called solutes. Aqueous solutions are solutions with water as the solvents.

5 5 1.3 CHEMICAL EQUATION A chemical equation is a description of a chemical reaction that uses symbols and formula to represent the elements and compounds involved in the reaction. Numerical coefficients preceding each symbol or formula and indicating molar proportions may be needed to balance a chemical equation. A chemical equation is a shorthand way to describe chemical change using symbols and formulas to represent the elements and compounds involved in the change. We can describe a chemical reaction in words. For example, Carbon reacts with oxygen to form carbon dioxide We can also describe the same reaction with chemical symbols and formulas. C + O2 CO2 Substances on the left of the arrow (C + O2 in this case) ae reactants or starting material. Those on the right (here, O2) are the products of the reaction. Reactants and products need not be written in any particular order in a chemical equation, except that all the reactants must be to the left of the arrow and all products to the right. In other words, we could also write the preceding equation as O2 + C CO2 At the submicroscopic (atomic or molecular) level, the chemical equation means that one atom of carbon (C) reacts with one molecule of oxygen (O2) to produce one molecule of carbon dioxide (CO2).

6 6 Sometimes we indicate the physical states of the reactants and products by writing the initial letter of the state immediately following the formula. Thus, (g) indicates gaseous substance, (l) a liquid, and (s) a solid. The label (aq) indicates an aqueous solution that is, a water solution. Using these labels, our equation becomes C(s) + O2(g) CO2(g) Balancing Chemical Equations. In general, we can balance a chemical equation by using the following steps: 1. Identify all reactants and products and write their correct formula on the left side and right side of the equation. 2. Begin balancing the equation by trying suitable coefficients that will give the same number of each element on both sides of the equation. Only the coeffiecient (the number before the formula), and not the subscripts (the numbers within the formula) can be changed. 3. Look for elements that appear only once on each side of the equation and with equal numbers of atoms on each side the formula containing these elements must have the same coefficient. 4. Look for elements that appear only once on each side of the equation but in unequal numbers of atoms. Balance these elements. Next, balance elements that appear in two or more formula on the same side of the equation. If a reactant or products exists as a free a free elements, try balancing that elements last. 5. Check your balance equation to be sure that you have the same total number of each type of atom on both sides of the equation arrow.

7 7 Example Balance the equation below: a) Na + H2O NaOH +H2 Solution a) Step 1: Balance each atom in the equation. The number of Na and O atoms is balanced. The number of H atoms is not balanced. It is balanced by putting the fraction ½ before H2. Na + H2O NaOH + ½ H2 (balanced) b) Step 2: Multiply each coefficient by 2 to obtain whole number of coefficients. 2Na + 2H2O 2NaOH + H2 (balanced) Exercise: 1. Balance the following representation of the chemical reaction that occurs when an airbag deploys. NaN3 Na + N2 2. The reaction between hydrogen and nitrogen to give ammonia, called the Haber process, is typically the first step in the industrial production of nitrogen fertilizers, represented as Balance the equation. H2 + N2 NH3 3. Iron ores like Fe2O3 are smelted by reaction with carbon to produce metallic iron and carbon dioxide

8 8 1.4 STOICHIOMETRY OF SOLUTION REACTIONS Because so many important reactions occur in solution, it is important to be able to do stoichiometric calculations for solution reactions. Steps for Solving Stoichiometric Problems Involving Solutions Step 1 Step 2 Step 3 Step 4 Step 5 Write the balanced equation for the reaction. For reactions involving ions, it is best to write the net ionic equation. Calculate the moles of reactants. Determine which reactant is limiting. Calculates the moles of other reactants or products, as required. Convert to grams or other units, if required. Molarity A concentration units that chemists often use is molarity. For reactions involving solutions, the amount of solute is usually measured in moles and the quantity of solution in liters or millimeters. The molarity (M) is the amount of solutes, in moles, per liter of solution. moles of solutes Molarity ( M ) liters of solution

9 9 Example: Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. Solution: Mass of solute = 11.5 g NaOH Volume of solution: 1.50 L Molarity ( M ) moles of solutes liters of solution Mass of solutes Use molar mass Moles of solutes Molarity Moar mass, NaOH = 40.0 g = = mass molar mass 11.5 g 40.0 g = mol NaOH moles = volume = 0.288mol NaOH 1.50L of solution = M NaOH

10 10 Exercise: 1. Calculate the molarity of a solution prepared by dissolving 1.00 g of ethanol, C2H5OH, in enough water to give a final volume of 101 ml. Solution Stoichiometry (Reactions Involving Only One Solution) Apply the mole-mole relationship in a balanced chemical equation to solve problem involving only one solution. Exercise: 11.9 g of NH3 is produced when x grams of (NH4)2SO4 reacted completely in v cm 3 of 2.5 M NaOH according to the equation below: (NH4)2SO4 + 2 NaOH Na2SO4 2H2O + 2NH3 Calculate the value x and v. Titration (Reactions Involving Two Solutions) Titration is a procedure in which two reactants in solutions are made to react in their stoichiometric proportions as indicated in a balanced chemical equation. Example: VA cm3 of solution A of unknown concentration is transferred to a flask. A solution B of known concentration, MB, is added carefully from a burette until a reaction of A with B is just complete. This is called the equivalence point of the titration. The volume, VB, of solution B required for a complete reaction with A is recorded. The equation for the reaction is aa + bb cc + dd The molarity of solution A, MA, is calculated from the relationship below: MAV M V B A B a b

11 11 Exercise: Calculate the molarity of phosphoric acid, H3PO4 if 40.0 cm 3 of it requires 25.0 cm 3 of M potassium hydroxide, KOH for its neutralization. Balance the equation below for the neutralization reaction described above. H3PO4 + KOH K3PO4 + H2O 1.5 LIMITING REAGENT When two or more reactants are combined in non-stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess (limiting reactant). Limiting reactant problems always involve 2 steps: 1. Identify the limiting reactant - Convert all masses to moles - Compare actual mole ratio to mole ratio given by the balanced chemical equation. OR - Calculate the number of moles obtained from each reactant in turn. - The reactant that gives the smaller amount of product is the limiting reactant. 2. Calculate the number amount of product obtained from the limiting reactant.

12 12 Example: Sodium hydrogen carbonate is prepared from NaCl and ammonium hydrogen carbonate, according to the equation: NH4HCO3 (aq) + NaCl (aq) NaHCO3 (aq) + NH4Cl (aq) If moles of NH4HCO3 are reacted with moles of NaCl, how many grams of NaHCO3 are obtained? Solution: 1 NH4HCO3 (aq) + 1 NaCl (aq) 1 NaHCO3 (aq) + 1 NH4Cl (aq) 0300 mol mol? g 3 = mol x 1mol NaHCO 1mol NaCl g NaHCO 1mole NaHCO 3 3 = 21.5 g NaHCO3 Exercise: Suppose 25.0 kg (2.50 x 10 4 g) of nitrogen gas and 5.00 kg (5.00 x 10 3 g) of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass ammonia produced when this reaction is run to completion. 1.6 THEORITICAL YIELD, ACTUAL YIELD AND PERCENTAGE YIELD. In previous section we learned how to calculate the amounts of products formed when specified amounts of reactant are mixed together. In doing these calculations, we used the fact that amount of product is controlled by the limiting reagent. Products stop forming when one reactant runs out.

13 13 The amount of product calculated in this way is called the theoretical yield of the product. It is the amount of product predicted from the amounts of reactants used. The actual yield of product, which is the amount of product actually obtained, is often compared to the theoretical yield. This comparison, usually expressed as a percentage yield. Actual yield 100% Theoretical yield Exercise Methanol can be produced by the reaction between carbon monoxide and hydrogen. Suppose 68.5 kg of CO (g) is reacted with 8.60 kg of H (g). 2 a) Calculate the theoretical yield of methanol. b) If 3.57 x 10 4 g of CH OH is actually produced, what is the % yield of methanol SOLUTION CONCENTRATION Units for Concentration The concentration of a solution is the amount of solute present in a given amount of solvent, or a given amount of solution. For this discussion, we will assume the solute is a liquid, or a solid and the solvent is a liquid. The concentration of a solution can be expressed in many different ways. There are four most common units of concentration: percent by mass, mole fraction, molarity and molality.

14 14 Percent by Mass The percent by mas (also called percent by weight or weight percent) is the ratio of the mass of a solute to the mass of the solution, multiplied by 100%) Percent by mass mass of mass of solute 100% solute mass of solvent mass of solute 100% mass of solution The percent by mass is unit less number because it is a ratio of two similar quantities. Mole Fraction (x) The mole fraction of a component of a solution component is defined as: Mole fraction of component moles of component sum of moles of all components The mole fraction is also unit less, because it too is a ratio of two similar quantities. Molarity (M) Molarity is defined as the number of moles of solute in 1 L of solution. Molarity moles of solutes liters of solution Thus, the units of molarity are mol/l

15 15 Molality (m) Molality is the number of moles of solute dissolved in 1 kg of solvent. Molality moles of solute mass of solvent (kg) Comparison of Concentration Units The choice of concentration unit is based on the purpose of the experiment. For instance, the mole fraction is not used to express the concentration of solutions for titrations and gravimetric analyses, but it is appropriate for calculating partial pressure for gases and for dealing with vapor pressure of solutions. The advantage of molarity is that it is generally easier to measure the volume of a solution, using precisely calibrated volumetric flask, than to weigh the solvent. For this reason, molarity is often preferred over molality. On the other hand, molality is independent of temperature, because the concentration is expressed in number of moles of solute and mass of solvent. Percent by mass is similar to molality in that it is independent of temperature. Furthermore, because it is defined in term of ratio of mass of solute to mass of solution, we do not need to know the molar mass of the solute in order to calculate the percent by mass Preparation of Solutions from Solids A standard solution is a solution whose concentration is accurately known. When the appropriate solute is available in pure form, a standard solution can be prepared by

16 16 weighing out a sample of solute, transferring it completely to a volumetric flask and adding enough solvent to bring the volume up to the mark on the neck of the flask Preparation of Solutions by Dilutions To save time and space in laboratory, solutions that are routinely used are often purchased or prepared in concentrated form (called stock solutions). Water (or another solvent) is then added to achieve the molarity desired for a particular solution. The process of adding more solvent to a solution is called dilution. A typical dilution calculation involves determining how much water must be added to an amount of stock solution to achieve a solution of the desired concentration. The key to doing these calculations is to remember that only water is added in the dilution. The amount of solute in the final, more dilute solution is the same as the amount of solute in the original, concentrated stock solution. That is, Moles of solute after dilution = Moles of solute before dilution The number of moles of solutes stays the same but more water is added, increasing the volume, so the molarity decreases. Decerase M Remains constant moles of solute volume (L) Increase (water added) Initial Conditions Final Conditions M1 V1 = moles of solutes = M2 V2

17 VOLUMETRIC ANALYSIS Volumetric analysis is a widely used quantitative analytical method. As the name implies, this method involved the measurement of volume of a solution of known concentration, which is used to determine the concentration of the analyte. Volumetric analysis is a practical technique whereby one uses reacting volume to analyse and calculate a variety of unknown values. Volumetric analysis can be used to find the: a) Concentration of a solution. b) Molecular mass of a substance. c) Percentage purity of a substance. d) Formula of a substance. e) Percentage composition of an element present. f) Stoichiometry of an equation.

18 18 CHAPTER 2 CHEMICAL KINETIC 2.1 OBJECTIVES Upon completion of this chapter, student should have to: 1. Identify a rate law of reaction for a given chemical reaction. 2. Describe the factors that influence the reaction rate including the size of matter, concentration of reactant, temperature and catalyst. 3. Calculate the rate constant for a given chemical reaction by using rate law equation. 2.2 INTRODUCTION Chemical kinetics is the study of the rate of reactions, or how fast they occur under different conditions. It usually includes a study of the mechanism of reactions, which is look at how the reacting molecules break apart and then form the new molecules. Chemical kinetics is the area of chemistry concerned with the speed or rates at which a chemical reaction occurs. The word kinetics suggests movement or change. Kinetics refers to the rate of reaction or reaction rate which is the change in the concentration of a reactant or product with time (M/s). Any reaction can be represented by the general equation: Reactant Products This equation tells us that during the course of a reaction, reactants are consumed while products are form. As a result, we can follow the progress of a reaction by monitoring either the decrease in concentration of the reactants or the increase in concentration of the products.

19 19 Figure 1 shows the progress of a simple reaction in which A molecules are converted to B molecules. A B FIGURE 1 Simple Reaction of A and B The decrease in the number of A (reactant) molecules and the increase in the number of B (product) molecules with the time are shown in Figure 2. Figure 2 The rate of reaction A B, represented as the decrease of A molecules with time and as increase of B molecules with time.

20 20 In general, it is more convenient to express the reaction rate in terms of the change in concentration with time. Thus, for the reaction A B, the rate can be expressed as:- - [ A] rate t or [ B] rate t Where [A] and [B] are the changes in concentration (molarity) over a time period, t. Because the concentration of A decreases during the time interval, [A] is a negative quantity. The rate of reaction is a positive quantity, so a minus sign is needed in the rate expression to make the rate positive. On the other hand, the rate of product formation does not require a minus sign because [B] is positive quantity (the concentration of B increases with time). These rates are average rates because they are averaged over a certain time period, t. 2.3 RATE OF REACTION AND ITS MEASUREMENT Any rate process represents some sort of output or occurrence per unit time. Examples include the flow of water (gallons per minute), the output of heater (BTU s per hour) or the velocity of vehicle (mile traveled per hour). In the case of a chemical reaction, the output is the product of the reaction, and the reaction rates represent the amount of product form per unit time. Therefore, a reaction rate is typically expressed as moles of product formed per liter of volume per unit time. A mole per liter is familiar concentration unit, so that a reaction rate is equivalent to an increase in concentration of product per unit time. Time can be expressed in seconds, minutes or hours. The use of seconds is the most common. We define the rate law for a reaction in terms of the time rate of change in concentration of one of the reactants or products. In general, the rate of change of the chosen species will be a function of the concentration of the reactants and products species as well as of external parameters such as the temperature. For example, in Figure 2, the rate of change for a species at any time is proportional to the slope of its concentration curve. The slope

21 21 varies with the time and generally approaches zero as the reaction approaches equilibrium. The stoichiometry of the reaction determines the proportionality constant. Consider the general reaction: aa + bb cc + dd We find that 1 a [ A] 1 t b [ B] 1 t c [ C] 1 t d [ D] t By convention, since we would like the rate to be positive if the reaction 2.4 RATE LAW The rate law / rate equation for a chemical reaction relates the reaction rates of the concentrations of reactants. For a reaction: aa + bb cc + dd The rate law would be: Rate of reaction = k [A] x [B] y where k = rate constant A and B = reactants The exponents in a rate law; x and y are called the rate order. Rate order is defined as the exponent to which the concentration of each reactant is raised in the rate equation. They are determined by experiment and not from the stoichiometric coefficient in an overall chemical equation. o The order of reaction with respect to reactant A = x o The order of reaction with respect to reactant B = y o The overall order of a reaction = x + y

22 22 o Notice that the exponents (rate order) of the rate law are NO derived from the coefficients of the chemical equation. The value of the rate constant k, also known as the proportionality constant depends on the reaction, temperature and the presence or absence of a catalyst. The units of k depend on the values of x, y and so on. For a reaction, the relationship between the concentration of its reactants and time can be derived from its rate law using calculus. However, mathematical expressions the relate concentrations and time in complex reactions can be complicated. As such, we will consider three simple cases of reactions that involve only one reactant. o Zero Order Reaction o First Order Reaction o Second Order Reaction Zero Order Reaction Zero Order Reaction is one of which the reaction does not rely upon the concentration of its reactants. The rate law for zero order reaction involves only one reactant The rate would be A Products Rate = k [A] 0 = k.(1) Therefore the reaction rate vs. concentration of reactant graph would be yield a horizontal line. The rate of reaction also can be expressed mathematically as: [A] Rate.(2) t

23 23 Figure 3 Reaction rate vs. concentration of reactant Equation (1) and (2) would give: [A] k t k. t [ A] In order to get an equation that shows the relationship between the concentrations of A at t seconds, the equation is integrated over time. t 0 [ A] k t [ A] t [ A] 0 kt ([ A] [ A] ) t 0 [ A] t kt [ A] 0 (3) Where [A]t = concentration of A at t seconds [A]0 = initial concentration of A When rearranged: [A]0 - [A]t = kt.(4)

24 First Order Reaction A first order reaction is a reaction whose rate depends on the reactant concentration raised to the first power. In a first order reaction of the type The rate is A product From the rate law we also know that, [A] rate t Rate k[ A] To obtain the units of k for this rate law, we write k rate [A] M / s M 1/ s or s 1 Combining the first two equations for the rate we get [A] kt t Using calculus, we can show from above equation that [A] t ln [ A] 0 kt Where [A]0 and [A]t are the concentrations of A at times, t = 0 and t = t, respectively. Above equation can be arranged as: ln[ A t ] kt ln[ A ] 0