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REPRESENTING CHEMICAL CHANGES presented by Mphiriseni Khwanda

REPRESENTING CHEMICAL CHANGES presented by Mphiriseni Khwanda. 25 August 2011. Chemical Changes: Quantitative Aspect. We shall focus on the following: Chemical changes Stoichiometric calculations Limiting reactant Percentage Yield

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REPRESENTING CHEMICAL CHANGES presented by Mphiriseni Khwanda

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  1. REPRESENTING CHEMICAL CHANGESpresented byMphiriseni Khwanda 25 August 2011

  2. Chemical Changes: Quantitative Aspect We shall focus on the following: Chemical changes Stoichiometric calculations Limiting reactant Percentage Yield Energy changes in reactions related to bond energy and Activation energy

  3. The mole concept A mole is an indication of the amount in a substance and its SI unit is mol. The mass of one mole is called molar mass and has a unit g per mol or g/mol In other words,

  4. How many particles does a mole denote? One mole of any substance contains Avogadros’ number ((NA )= 6,02 x 1023 elementary units of a substance. For example, 1 mol of H2 O contains 6,02 x 1023 of water molecules 1 mol of Mg contains 6,02 x 1023 of water molecules

  5. Summary of Number particles, moles, mass and Avogadros’ number

  6. Examples 1. What is the number of moles in 64g of Sulphur? Answer: From mass to moles we use 2. Calculate the number HCl molecules in 1,5 mol HCl. Answer: number of molecules = n . NA = 9.03 x 1023

  7. Stoichiometry Quantitative relationships between reactants and products The balanced chemical equation gives us the relationships in moles. Multiple mol-mol conversion factors or mole ratios can be derived from balanced equations Consider: N2 + 3H2 2NH3

  8. The Mole Ratio • One mol of N2 produces 2 mol of NH3or • One mol of N2 reacts with 3 mol of H2

  9. The Mole Ratio • 3 mol of H2 produce 2 mol of NH3 • Inverse can also be used as a conversion factor

  10. Conversion Factor Rule • Multiplying anything by 1 does not change any thing Example: If a = b then we can say Hence, 1 can be written as any ratio

  11. Example: mass –mol calculations • Consider: N2 + 3H2 2NH3 • How many moles of NH3 can be produced from 33.6 g of N2? • Convert grams of N2 to mol of N2, then convert mol of N2 to mol of NH3

  12. Example The first step in recycling of the element zinc from its ore ZnS, involves the heating of ZnS in air to obtain ZnO and SO2 . (a) Calculate the mass of oxygen that is needed to react with 7 g ZnS. (b) What mass of SO2 is formed? Answer: first balance the equation (a) (b)

  13. Limiting reactant Given the masses of two different reactants, determine the limiting reactant and calculate the yield of product. Procedure Convert amount of each reactant to the number of moles of product using mole ratios. The limiting reactant is the one that produces the smallest amount of product.

  14. Limiting reactant considerations When reactants are mixed in exactly the mass ratio determined from the balanced equation, the mixture is said to be a stoichiometric mixture. Example: 4.0 g H2 + 32.0 g O2 36.0 g H2O Other mass ratios require calculations to determine the limiting reactant.

  15. Limiting reactant example 2 CH3OH(l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (g) Mix 40.0 g of methanol with 40.0 g of O2 - what is the mass of CO2 produced? Methanol, CH3OH, has a molar mass of 40.0 g O2has a molar mass of 32.00 g

  16. Process

  17. Yield Calculations Calculate the percent yield of a reaction from the measured actual yield and the calculated theoretical yield. The actual yield is the amount of product obtained when the reaction is run. The theoretical yield is the calculated amount of product that would be obtained if all of the limiting reactant was converted to a given product. The percent yield is the actual yield in grams or moles divided by the theoretical yield in grams or moles times 100%. In some cases, a reverse reaction occurs whereby reactants are reformed from products. This limits the percent of reactants that are converted to products. The reaction between N2 and H2 to produce NH3 is a reaction which is reversible. This has severe consequences for the commercial production of ammonia

  18. Example calculation • For the conversion of N2 and H2 to NH3 • 4.70 g H2 react with N2 • 12.5 g of NH3 is formed • The theoretical yield is 26.5 g of NH3 in this reaction. The percent yield is 47.5% (commercially the yield is only ca 28%).

  19. Energy Relationships in Chemical Reactions You will extend your knowledge of stoichiometry to include amounts of heat involved in chemical reactions. Given the change in enthalpy for a reaction, calculate the amount of heat gained or released by a given mass of reactant. Heat can be a reactant or a product in a reaction. Endothermic reactions are those that require heat to occur; hence heat is a reactant. Exothermic reactions produce heat; hence heat is a product. A balanced chemical reaction that includes heat energy is referred to as a thermochemical equation.

  20. Heat in chemical reactions A thermochemical reaction can be expressed in one of two ways 2 H2 (g) + O2 (g) 2 H2O (l) H = -572 kJ note that the heat evolved is per 2 mol of H2 or 2 mol of H2O produced H is termed the heat of reaction or enthalpy H is negative for exothermic reactions and positive for endothermic reactions

  21. Heat in chemical reactions The second way to indicate heat in a chemical reaction is as a reactant or product. 2 H2 (g) + O2 (g) 2 H2O (l)+ 572 kJ(an exothermic reaction) N2 (g) + O2 (g) + 181 kJ  2 NO (g)(an endothermic reaction) Heat energy can be treated quantitatively in a manner similar to the amount of reactant or product.

  22. Example • 2 C2H2 (g) + 5O2 (g) 4 CO2 (g) + 2 H2O (l)H = -2602 kJ • If 550 kJ is evolved in the combustion of C2H2, what is the mass of CO2 formed?

  23. Bonding Energy and Activation Energy • Many chemical bonds are broken and made during chemical reaction • When the bond is broken in a compound additional energy called bonding or dissociation energy is needed, • It is the amount of energy that is needed to break atoms and is usually determined experimentally . It is calculated as follows: Consider 3 H2 +N2 → 2 NH2, • Show all bonds in a molecule by drawing a Couper structures • Add all bonding energies of the reactants • Add all bonding energies of the products • The heat of reaction = reactants bonding energies – products bonding energies • If positive, then endothermic, negative hence exothermic Activation energy is the energy required to start a reaction

  24. Ndolivhuwa

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