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Chemical Reactions

Chemical Reactions . Chemical reactions are changes which produce new substances.

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Chemical Reactions

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  1. Chemical Reactions • Chemical reactions are changes which produce newsubstances. • Bonds between atoms are broken (covalent bonds, metallic bonds, and/or ionic bonds), atoms scramble around, and the atoms then re-connect to form different substances. Saying that a “chemical reaction” has occurred, is identical to saying that a “chemical change” has occurred. • Sometimes the new substances produced are in a different state (and sometimes not).

  2. A change of state (by itself) does not produce a new substance. • A change of state is just melting/freezing, boiling/evaporation/condensation, or sublimation/deposition. The substance itself doesn’t change during a change of state. • Atoms are notcreated or destroyed during chemical reactions. • Since atoms have mass: mass is notcreated or destroyed during chemical reactions. • Since mass is not gained or lost during a chemical reaction, we say that mass is conserved. • The name of this fact is called the “law of ­­­­­­­­­­­­­conservationof mass”.

  3. When asked to write a “chemical equation”, we use formulas and symbols (to represent the compounds and elements) which take part in the chemical reaction. • For example, instead of saying, “hydrogen peroxide” changes into “water” and “oxygen”, we would use the formulas, H2O2 and H2O, and the symbol O2. • An arrow is placed within a chemical equation, to separate the substances present before the reaction begins, from those present after the reaction occurs. The arrow means “yields” (or “produces”).

  4. For example, H2O2 H2O + O2 would be the first step toward writing the final chemical equation for this reaction; and it would be read aloud as “H2O2produceswater andoxygen,” or, “H2O2yieldswater andoxygen”

  5. The substances to the left of the arrow are referred to as the reactants(because they are allowed to “react” together, while the substances to the right of the arrow are referred to as the products(because they are what are “produced” during the reaction).

  6. The example, H2O2 H2O + O2 is not the final complete chemical equation; in fact, it is not yet even yet a complete skeletonequation. • Skeleton equations show the reactants to the right, and the products to the left, of the arrow; and, additional information that is known about the substances should also be included within the skeleton equation. The additional information will be shown by using accepted symbols.

  7. Designates a substance as being in the solid state: (s). Appears after the substance. • Designates a substance as being in the liquid state: (l). Appears after the substance. • Designates a substance as being in the gaseous state: (g). Appears after the substance.

  8. Designates the substance as being dissolved (or remains dissolved) in an aqueous solution (a solution containing water): (aq). Appears after the substance. • A double-ended arrow replaces  when a reaction is reversible • This occurs when the new substances just formed (the products), react together and regenerate the original substances (the original reactants)

  9. An arrow with a triangle over it (or an arrow with the word heat over it) indicates that heat is supplied to the reaction.  • An arrow with a chemical (a formula or a symbol) written above or below it indicates that the chemical is being used to speed up the chemical reaction. Specific example: Ni

  10. The general term for a chemical which speeds up a reaction is: catalyst. • A catalyst does NOT “enter into” the reaction; it just quickens it. • A catalyst is present at the start of, during, and at the end of a reaction; but, since it doesn’t change into a different substance, it is NOT a reactant or a product.

  11. What is not written in a skeleton equation? • The relative amountsof each substance in the reaction. • How do you determine the products of a chemical reaction? • Only by actually doingthe reaction! • This is because several chemicals may react together, under slightly different conditions, to produce different products.

  12. You should never see the same substance both to the left side and to the right side of the arrow. • During this chapter, we assume the reaction is complete. This means, each substance that is present before the reaction begins will be completely changed into something different; unless the chemical reaction is said to be “reversible”. In that case, the reaction never “goes to completion”; instead the reactants react to make the products, and simultaneously the products react to make the reactants…. Which react to make the products….. which simultaneously react to make the reactants….. .

  13. Balancing Chemical equations. • For each skeleton equation to become a complete (final form) chemical equation, coefficientsmust be inserted into the skeleton equation. • Coefficients are whole numbers which indicate the relative amounts of each substance. • A chemical equation is said to be “balanced” when, the number of atoms of each element found on the reactant side of the equation is equal to the number of atoms of that same element on the product side of the equation.

  14. In a balanced equation, each side of the equation will have the same numbers of atoms of each element. • To balance an equation, a coefficient is placed beforeeach substance in the equation. The coefficient is a “multiplier”. It doubles, triples, quadruples (etc.) the amount of the substance it is placed in front of. • The coefficients used for an equation should be the lowestwhole numbers possible to achieve balancing of the atoms.

  15. Strategies or tips for balancing chemical equations. • Balance oneelement at a time; don’t try to balance the whole equation at once. • First pick an element that appears in only onereactant and oneproduct.

  16. Determine the LCM for the element you have picked. Ex (below) – let’s pick fluorine. • The subscripts are 2 and 3; the least common multiple of 2 and 3 is 6. • “count by 2” and “count by 3”… when you find the first # that is the same, you have identified the LCM

  17. Example: • Place a coefficient of 3on the left and 2on the right, so as to now have 6 atoms of F both on the left and on the right. • _____Al + ___ F2 ___ AlF3Think momentarily only about the LCM of “F” . • 3F22 F3 (LCM of 2, 3 = 6) • _____Al + _ 3 F2  __2_ AlF3(Remember, coefficients are multipliers

  18. After balancing one element, pick a second element to work with that appears in the samecompound as the already-balanced element. Again, determine the LCM • Think momentarily only about the LCM of “Al”, in __ Al  __2 AlF3 • You need, at the moment, the LCM of 1, 2 which is 2

  19. After balancing one element, pick a second element to work with that appears in the ___________ compound as the already-balanced element. Again, determine the LCM. • 2 Al + 3F2 2AlF3 is correct because 2 Al + 6 F  2 Al + 6 F

  20. Start an atom inventory. The atom inventory for the above example is this: Only when the atom inventory is equal on both sides, is the balancing correctly done. 2 Al 6 F 2 Al 6 F

  21. Balance these two atoms last: hydrogenand oxygen • The reason is, many times H and O occur in 2 substances to the left of the arrow and/or 2 substances to the right of the arrow. • When this occurs, you must addas well as multiply to solve the problem. The example shown below will illustrate this point.

  22. __ CH4 + __ O2__ CO2 + __H2O • Balance the “C” first: 1CH4 + __ O2_1_ CO2 + __H2O atom inventory: 1C 4H 2 Ox 1C 2 Ox 2H 1 Ox 1C 2H 3 Ox

  23. Balance the “H” second LCM of 4, 2 = 4 this means, aim for 4 H’s. ? X 2 = 4 ? = 2 Insert “2” here _1_ CH4 + __ O2_1_ CO2 + _2_H2O

  24. _1_ CH4 + __ O2_1_ CO2 + _2_H2O atom inventory Now work with oxygen Notice, to balance the “O”, you must now add the number of oxygen atoms from two compounds (CO2, H2O): CO2has 2 oxygen atoms; 2 H2O also has 2 oxygen atoms. 2 +2 = 4 1C 4H 2 Ox 1C 2 Ox 4H 2 Ox 1C 4H 4 Ox

  25. Balance the ___________________ atoms second. Save the _______________ atoms till last because you will have to ___________ together the oxygen atoms from CO2 and H2O. • Practice: Predict the products and balance: C3H8 + O2 _1_ CH4 + __ O2_1_ CO2 + _2_H2O Continue working with oxygen- LCM 2,4 is 4 aim for 4 oxy’s on the left; ? X 2 = 4 ? = 2 _1_ CH4 + _2_O2_1_ CO2 + _2_H2O

  26. _1_ CH4 + _2_ O2 _1_ CO2 + _2_H2O This equation is correctly balanced as shown by the atom inventory. 1C 4H 4 Ox 1C 4H 4 Ox

  27. If you are having difficulty with a combustion problem (or any problem), double a coefficient and re-work the problem based on the assumption that the new double coefficient is correct. This corrects problems with needing a “half” coefficient, which might occur during combustion reactions in which there are an even number of carbon atoms. • Ex: 2 ½ can’t be used as a coefficient; but, 5 can be used as a coefficient.

  28. As long as you have a polyatomic ion both to the left and to the right of the arrow, work with it as a whole unit (don’t work with it as individual atoms). • Ex: in Al(NO3)3, there are 3 (NO3) • Re-write H2O as HOH whenever (OH) is present on the opposite side of the arrow from H2O.

  29. The Five General Types of chemical reactions. • When you are shown a series of chemical equations, by distinguishing bothbetween single elements or diatomic elementsand compounds, you will notice that there are 5 different types of common chemical reactions. • It is easiest to identify the 5 types of chemical reactions (or equations) by ignoring the coefficients-- The coefficients identify the relative amount of each substance; but, they do not identify the type of chemical reaction!

  30. Once you know what to look for, you can learn to classify the reactions (or equations). • Unfortunately however, NOT every chemical reaction can be “classified” (placed into) onlyone of the five general types! • Some chemical reactions will fit within more than one of the 5 types. For this reason, the examples that you will be given are hand-chosen for beginning chemistry students. Below, when you see that I have written, “don’t worry about….,” it means that the information is given as enrichment only, not required learning.

  31. Synthesis reaction: A reaction in which separate elements combine to make a compound. A + B AB (element A combines with element B to make compound AB) • Be careful to remember the 7 diatomic elements: N2, O2, F2, Cl2, Br2, I2, H2 • Ex: Al + F2 AlF3

  32. [Do not worry about “photosynthesis”; but, photosynthesis is a synthesis reaction in which small compounds combine to form a larger compound: • CO2 + H2O  C6H12O6 + O2

  33. Decomposition reaction: A reaction in which a single compound breaks apart into its “constituent” elements. AB  A + B • In other words, “a single reactant is the identifying characteristic of a decomposition reaction,” and, during “the decomposition of a simple binary compound”, the products are “predictable”, as they are the elements found in the compound.

  34. The starting compound might be either a molecular compound or an ioniccompound. • Do not worry about “respiration”; but, during respiration (which begins with an intake of oxygen) a decomposition reaction occurs, during which blood sugar (glucose, C6H12O6) is broken down into carbon dioxide and water vapor: • C6H12O6 + O2  CO2 + H2O

  35. Combustion reaction: A reaction in which there are 2 reactants. One reactant is usually a hydrocarbon (a substance containing both hydrogenand carbon); the second reactant must be oxygen. Remember, the symbol for oxygen is O2. • We will assume that the combustion is complete; therefore, there will only ever be 2 specific products: CO2and H2O, both present in the gaseous state.

  36. When complete combustion occurs, all of the carbon found within the reactant will be totally converted into the product, carbon dioxide (CO2). • Ex: _1_ CH4 + _2_O2_1_ CO2 + _2_H2O

  37. Notice how it is easier to classify the example after removing the coefficients: CH4 + O2 CO2 + H2O

  38. Single-Replacement reaction: A reaction in which the reactants are 1 element + 1 ionic compound; and the products are 1 different element+ 1 different ionic compound.

  39. In other words, “one element reacts with a compound to form a new compound and a different element”. A + XY X + AY Na + KCl K + NaCl or A + XY  Y + XA F2+ NaCl Cl2 + NaF

  40. Double-Replacement reaction: A reaction in which the reactants are 2 ioniccompounds, and the products are both compounds; however, at least one of the following conditions must occur.

  41. At least 1 product is a binary molecular compound (this means that the compound has 2 different non-metal elements). Ex: H2O • A least 1 product is in the gaseous state. Ex: CO2 • Or, at least 1 product is in the solid state (or said to “precipitate” or be a “precipitate”). Ex: PbCl2 (s)

  42. Learning how to “predict the products”: • You will be asked to predict the products for (and then balance) a given combustion reaction. • Don’t worry about it! The products of any combustion reaction are CO2and H2O. • Balance the carbon atoms first. If there are an even number of carbon atoms, doublethe coefficients that you think you should use!

  43. Balance the hydrogen atoms second. Save the oxygenatoms till last because you will have to addtogether the oxygen atoms from CO2 and H2O. • Practice: Predict the products and balance: C3H8 + O2

  44. You will be asked to predict the products for (and then balance) a given single-replacement equation, which is given in the form of: ionic compound + metallic element  ? Change places (Remember: XY + A X + AY ionic compound + metallic element  single atom + new ionic compound

  45. Step 1: Take the cation out of the ionic compound, and place the cation (as a single atom) to the right of the arrow. Position the “+” sign after it. • Step 2: Take the symbol of the metallic element and position it after the “+” sign. The metallic element will become the “new cation” for the new ionic compound.

  46. Step 3: Take the anion out of the ionic compound, and place the anion to the right of the new cation. • Step 4: Make the new ionic compound neutral (….Do a swap/drop/and simplify). Practice – Predict the products and balance, CaSO4 + Na 

  47. You will be asked to predict the products for (and then balance) a given single-replacement equation which is given in the form of: ionic compound + halogen  ? (Remember: change places XY + A2Y2+ XA • ionic compound + diatomic halogen  new diatomic halogen + new ionic compound

  48. Step 1: Take the anion out of the ionic compound, and place the anion (as a diatomic atom) to the right of the arrow. Position the “+” sign after it. • Step 2: Take the cation from the ionic compound, and position it after the “+” sign.

  49. Step 3: The original diatomic halogen will now become the “new anion” for the new ionic compound. The halogen will not necessarily keep it’s original “2” subscript. • Step 4: Make the new ionic compound neutral (….Do a swap/drop/and simplify). • Practice – Predict the products and balance, NaI + Cl2

  50. You will be asked to predict the products for (and then balance) two given double-replacement equations. Ionic compound + ionic compound ? • Step 1: Exchange the two cations. • Step 2: Make each of the new ionic compounds neutral (….Do a swap/drop/and simplify). • Practice – Predict the products and balance, Ba(C2H3O2)2 + NaCl ?

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