Chapter-2- Chemical reactions and the mole concept التفاعلات الكيميائية ومفهوم المول

1. General Chemistry I (Chem 1010) 1437/1438 (Semester 2) Chapter-2-Chemical reactions and the mole concept التفاعلات الكيميائية ومفهوم المول Dr/ El Hassane ANOUAR Chemistry Department, College of Sciences and Humanities, Prince Sattam bin Abdulaziz University, P.O. Box 83, Al-Kharij 11942, Saudi Arabia. Faculty of Applied Sciences, UiTM

2. 2.1. Chemical reactions and chemicalequations • Chemical reaction: A chemical process in which a substance (or substances) is changed into one or more new substances. Chemical reactions are represented by chemical equations. • Chemical equation: A symbolic representation of a chemical reaction in of chemical . Products or substances that results from a reaction Example React to form or yield Reactants or starting substances 1 Coefficients in front of the formula give the relative number of molecules or formula units involved in the reaction.

3. 1.1. Chemical reactions and chemicalequations (l)=liquid (s) = solid (aq) = aqueous (solution In waterr States or phases of the substances 1 Example (∆)=Heating temperature the conditions under which a reaction takes place.

4. 2.1. Chemical reactions and chemicalequations • Balancing Chemical Equations: • In general, to balance a chemical equation we need: • Step 1: • Identify all reactants and products and write their correct formulas on the left side and right side of the equation, respectively. • Step 2 • Begin balancing the equation by trying different coefficients to make the number of atoms of each element the same on both sides of the equation.

5. 2.1. Chemical reactions and chemicalequations • Balancing Chemical Equations: Step 3: First, look for elements that appear only once on each side of the equation with the same number of atoms on each side. The formulas containing these elements must have the same coefficient. Next, look for elements that appear only once on each side of the equation but in unequal numbers of atoms. Balance these elements. Finally, balance elements that appear in two or more formulas on the same side of the equation. Step 4: Check your balanced equation to be sure that you have the same total number of each type of atoms on both sides of the equation arrow.

6. 2.1. Chemical reactions and chemicalequations • Balancing Chemical Equations: Example

7. 2.1. Chemical reactions and chemicalequations • Balancing Chemical Equations: Example: Combustion of propane in O2 Unbalance equation Balance equation Propane (C3H8) reacts with oxygen, O2to yield CO2, and H2O.

8. 2.1. Chemical reactions and chemicalequations • Balancing Chemical Equations: Example 2.1 (P.52) Write and balance a chemical reaction for the complete combustion of benzene C6H6 in oxygen. Solution

9. 2.2-2.3. Stoichiometry: Quantitative relations in chemical reactions • To interpret a reaction quantitatively, we need to apply our knowledge of molar masses and the mole concept. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. • The above approach is called the mole method: The stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance.

10. 2.2. Stoichiometry: Quantitative relations in chemical reactions Example • In stoichiometric calculations, we say that 3 moles of H2are equivalent to 2 moles of NH3, that is,

11. 2.2-2.3. Stoichiometry: Quantitative relations in chemical reactions Example 2.2 (P.56) How many moles of oxygen are produced from the thermal decomposition of 50.0 g of potassium chlorate according to the following reaction? Solution

12. 2.2-2.3. Stoichiometry: Quantitative relations in chemical reactions Example 2.3 (P.57) Given the following balanced equation: Calculate the number of grams of C3H8 which could be burned by 213 moles of O2 ? Solution + 4H2O(l)

13. 2.4-Limiting reactant • Often in a chemical reaction, only one of the reactants may be completely consumed at the end of the reaction, whereas some amounts of other reactants will remain unreacted. The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when a reaction goes to completion. • The reactant that is not completely consumed is often referred to as an excess reactant. • Once one of the reactants is used up (consumed), the reaction stops. This means that: The moles of product are always determined by the starting moles of limiting reactant.

14. 2.4-Limiting reactant • Determining the Limiting Reactant. 2H2O(g) Consider the burning of hydrogen in oxygen. Suppose: 1 mol H2+1 mol O2 into a reaction vessel. How many moles of H2O will be produced? Note that from the reaction: 2 mol H2 produces 2 mol H2O 1 mol O2 produces 2 mol H2O. Now you calculate the moles of H2O that you could produce from each quantity of reactant, assuming that there is sufficient other reactant. Comparing these results, you see that hydrogen, H2, yields the least amount of product, so it must be the limiting reactant.

15. 2.4-Limiting reactant • Determining the Limiting Reactant. 4H2O(g) Suppose: 1 mol H2 +1 mol O2 into a reaction vessel.

16. 2.4-Limiting reactant • Rule of determination of Limiting Reactant. In three steps: 1st step: Convert the given amount of reactants to moles by using molecular mass: that from the reaction: 2ndstep: Divide the moles calculated in the first step by coefficient of reactants (each mol by its coefficient reactant). 3rdstep: Compare the results of 2. The smaller number is correspond to Limiting Reactant.

17. 2.4-Limiting reactant Example 2.3 (P.60) 5.00 g of iron and 5.00 gram of Sulphur are heated together to form iron (II) Sulphide. Which reactant present in excess, and how much of the product is formed ? Solution FeS (s)

18. 2.4-Limiting reactant Example 2.4 (P.61) If 4.00 g of NH3 (g) and 14.00 g of F2(g) are allowed to react according to the following equation: Calculate the mass of N2F4 (g) and HF (g) produced ? Solution N2F4 (g) + 6HF (g)

19. 2.5-Theoretical yield and percentage yield • The amount of limiting reagent present at the start of a reaction determines the theoretical yield of the reaction: The amount of product that would result if all the limiting reagent reacted => Is the maximum obtainable yield, predicted by the balanced equation. • The percent yield, which describes the proportion of the actual yield to the theoretical yield. It is calculated as follows: The actual yield: is the amount of product actually obtained from a reaction, is almost always less than the theoretical yield.

20. 2.5-Theoretical yield and percentage yield Example 2.5 (P.63) How many grams of N2F4 can theoretically be prepared from 4.0 g of NH3 and 14.0 of F2? If 4.80 g of N2F4 is obtained from the experiment, what is the percentage yield? The chemical equation for the reaction is: Solution N2F4 (g) + 6HF (g)

21. 2.5-Theoretical yied and percentage yield Example 2.5 (P.63) Diborane, B2H6, is manufactured by the following reaction: What is the mass of B2H6 obtained from the above reaction, if 26.5g of NaBH4(s) and 45.6g of BF3(g) were used, knowing that the percentage yield of the B2H6 is 80%? Solution 3NaBH43NaBF4(s) + 2B2H6(g)

22. 2.6-Reaction in solution • A solution is a homogeneous mixture of two or more substances. • The substance present in a smaller amount is called the solute. • The substance present in a larger amount is called the solvent. • A solution may be • Gaseous (such as air) • Solid (such as an alloy) • Liquid (such as seawater). • In this section we will discuss only aqueous solutions, in which • The soluteinitially is a liquid or a solid • The solvent is water.

23. 2.6-Reaction in solution • 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. • Here we will consider one of the most commonly used units in chemistry, molarity (M), or molar concentration, which is the number of moles of solute per liter of solution. Molarity is defined as Mol Mol/L L

24. 2.6-Reaction in solution Example 2.7 (P.67) Calculate the mass in grams of NaOH required to prepare 4.0 × 102 mL of 0.6 M NaOH solution? Solution

25. 2.7-Preparing solution by dilution • Dilute Solution is a solution containing a relatively small quantity of solute as compared with the amount of solvent. • Concentrated solution is a solution that has a relatively large amount of solute dissolved in it. • Dilution is the procedure for preparing a less concentrated solution from a more concentrated one. • Dilution does not change the total number of solute particles

26. 2.7-Preparing solution by dilution • Dilution does not change the total number of solute particles Before dilution After dilution MiVi • If water is used for dilution, then • Vf = Vi + Vwater Vwater= Vf - Vi

27. 2.7-Preparing solution by dilution Example 2.8 (P.69) How many milliliters of water must be added to 85.0 mL of 1.00 M H2SO4 to produce 0.650M H2SO4? Solution

28. 2.8-The stoichiometry of reactions in solutions Example 2.8 (P.69) A 42.0 mL of a solution (KOH) requires 15.0 mL of 0.350 M (H2SO4) for complete reaction: 2 KOH (aq) + H2SO4 (aq)  K2SO4 (aq) + 2H2O (l) What is the molarity of the KOH solution ? Solution (see page 71)