Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Chemical Reactions - General Chemistry - Lecture Slides, Slides of Chemistry

Fundamental concepts of chemistry including atomic structure, history of the atom, development of the periodic table, nuclear chemistry, chemical nomenclature and formula, types of reactions, stoichiometry, gas laws, liquids and solids, thermodynamics, chemical equilibrium, acids and bases. This lecture includes: Chemical Reactions, Chemical Equations, Conservation of Mass, Combustion of Methane, Combustion of Methane, Balanced, Writing Equations, Symbols Used in Equations, Balancing by Inspecti

Typology: Slides

2012/2013

Uploaded on 12/31/2013

kamin
kamin 🇮🇳

4.2

(30)

195 documents

1 / 31

Toggle sidebar

Related documents


Partial preview of the text

Download Chemical Reactions - General Chemistry - Lecture Slides and more Slides Chemistry in PDF only on Docsity! Chemical Reactions: An Introduction docsity.com Chemical Reactions • Reactions involve chemical changes in matter resulting in new substances • Reactions involve rearrangement and exchange of atoms to produce new molecules – Elements are not transmuted during a reaction Reactants  Products docsity.com Figure 6.2: Hot and cold pack reactions docsity.com Figure 6.3 (a): Chemical reactions docsity.com Figure 6.3 (b): Chemical reactions docsity.com Chemical Equations • Shorthand way of describing a reaction • Provides information about the reaction – Formulas of reactants and products – States of reactants and products – Relative numbers of reactant and product molecules that are required – Can be used to determine weights of reactants used and of products that can be made docsity.com Figure 6.4: The reaction between methane and oxygen to give water and carbon dioxide docsity.com Conservation of Mass • Matter cannot be created or destroyed • In a chemical reaction, all the atoms present at the beginning are still present at the end • Therefore the total mass cannot change • Therefore the total mass of the reactants will be the same as the total mass of the products docsity.com Writing Equations • Use proper formulas for each reactant and product • proper equation should be balanced – obey Law of Conservation of Mass – all elements on reactants side also on product side – equal numbers of atoms of each element on reactant side as on product side • balanced equation shows the relationship between the relative numbers of molecules of reactants and products – can be used to determine mass relationships docsity.com Symbols Used in Equations • symbols used after chemical formula to indicate state – (g) = gas; (l) = liquid; (s) = solid – (aq) = aqueous, dissolved in water docsity.com Balancing by Inspection 1. Count atoms of each element • polyatomic ions may be counted as one “element” if it does not change in the reaction Al + FeSO4 Al2(SO4)3 + Fe 1 SO4 3 • if an element appears in more than one compound on the same side, count each separately and add CO + O2  CO2 1 + 2 O 2 docsity.com Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 3. count the number of atoms on each side – count polyatomic groups as one “element” if on both sides – split count of element if in more than one compound on one side Mg(s) + O2(g)  MgO(s) 1  Mg 1 2  O  1 docsity.com Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 4. pick an element to balance – avoid element in multiple compounds 5. find least common multiple of both sides & multiply each side by factor so it equals LCM Mg(s) + O2(g)  MgO(s) 1  Mg 1 1 x 2  O  1 x 2 docsity.com Examples • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 6. use factors as coefficients in front of compound containing the element  if coefficient already there, multiply them together Mg(s) + O2(g)  2 MgO(s) 1  Mg 1 1 x 2  O  1 x 2 docsity.com Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water 3. count the number of atoms of on each side – count polyatomic groups as one “element” if on both sides – split count of element if in more than one compound on one side NH3(g) + O2(g)  NO(g) + H2O(g) 1  N 1 3  H  2 2  O  1 + 1 docsity.com Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water 4. pick an element to balance – avoid element in multiple compounds 5. find least common multiple of both sides & multiply each side by factor so it equals LCM NH3(g) + O2(g)  NO(g) + H2O(g) 1  N 1 2 x 3  H  2 x 3 2  O  1 + 1 docsity.com Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water 6. use factors as coefficients in front of compound containing the element 2 NH3(g) + O2(g)  NO(g) + 3 H2O(g) 1  N 1 2 x 3  H  2 x 3 2  O  1 + 1 docsity.com Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water 10. Repeat – A trick of the trade, when you are forced to attack an element that is in 3 or more compounds – find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction! – We want to make the O on the left equal 5, therefore we will multiply it by 2.5 2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g) 2  N 2 6  H  6 2.5 x 2  O  2 + 3 docsity.com Examples • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water 11. Multiply all the coefficients by a number to eliminate fractions – 0.5 x 2, 0.33 x 3, 0.25 x 4, 0.67 x 3 2 x [2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)] 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) 4  N 4 12  H  12 10  O  10 docsity.com
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved