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Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 1 of 24

PHILIP DUTTONUNIVERSITY OF WINDSOR

DEPARTMENT OF CHEMISTRY AND BIOCHEMISTRY

TENTH EDITION

GENERAL CHEMISTRYPrinciples and Modern Applications

PETRUCCI HERRING MADURA BISSONNETTE

Chemical Reactions 4

Chemical Compounds

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 2 of 24

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 3 of 24

4-1 Chemical Reactions and Chemical Equations

As reactants are converted to products we observe:Color change

Precipitate formation

Gas evolution

Heat absorption or evolution

Chemical evidence may be necessary.

Precipitation of silver chromate

Figure 4-1

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 4 of 24

Evidence of a chemical reaction

Figure 4-2

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 5 of 24

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 6 of 24

Chemical Reactions

Nitrogen monoxide + oxygen → nitrogen dioxide

Step 1: Write the reaction using chemical symbols.

NO + O2 → NO2

Step 2: Balance the chemical equation.

2 1 2

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 7 of 24

Molecular Representation

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 8 of 24

Balancing Equations

Never introduce extraneous atoms to balance.

NO + O2 → NO2 + O

Never change a formula for the purpose of balancing an equation.

NO + O2 → NO3

An equation can be balanced only by adjusting the coefficients of formulas.

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 9 of 24

Balancing Equation Strategy

Balance elements that occur in only one compound on each side first.

Balance free elements last.

Balance unchanged polyatomics (or other groups of atoms) as groups.

Fractional coefficients are acceptable and can be cleared at the end by multiplication.

States of Matter

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 10 of 24

(g) gas (l) liquid (s) solid

Thus, the equation for combustion of triethylene glycol can be written as

2 C6H14O4(l) + 15 O2(g) 12 CO2(g) + 14 H2O(l)

Another commonly used symbol for reactants or products dissolved in water is

(aq) aqueous solution

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 11 of 24

4-2 Chemical Equations and Stoichiometry

Stoichiometry includes all the quantitative relationships involving:

atomic and formula masses

chemical formulas.

chemical equations

• Mole ratio or stoichiometric factor is a central conversion factor.

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 12 of 24

KEEP IN MIND

that it is important to include units and to work from a balanced chemical equation when solving stoichiometry problems.

A generalized stoichiometry diagram

Figure 4-3

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 13 of 24

The reaction of 2 Al(s) 6 HCl(aq) 2 AlCl3(aq) 3 H2(g)Figure 4-4

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 3Slide 14 of 24

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 15 of 24

4-3 Chemical Reactions in Solution

Close contact between atoms, ions and molecules necessary for a reaction to occur.

Solvent

We will usually use aqueous (aq) solution.

Solute

A material dissolved by the solvent.

Molarity

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 16 of 24

Molarity (M) = Volume of solution (L) Amount of solute (mol solute)

If 0.440 mol of urea is dissolved in enough water to make 1.000 L of solution the concentration is:

curea = 1.000 L

0.440 mol urea = 0.440 M CO(NH2)2

C = V n

Solution Dilution

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 17 of 24

Figure 4-6

Visualizing the dilution of a solution

Mi × Vi = ni

Mi Vi Mf Vf

= nf = Mf × Vf

M = n

V

Mi × ViMf = Vf

= Mi

Vi

Vf

Preparation of 0.250 M K2Cr2O7 —Example 4-9 illustratedFigure 4-5

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 18 of 24

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 19 of 24

4-4 Determining Limiting Reagent

The reactant that is completely consumed determines the quantities of the products formed.

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 20 of 24

Theoretical yield is the expected yield from a reactant.Actual yield is the amount of product actually produced.

Percent yield = 100%Actual yield

Theoretical Yield

4-5 Other Practical Matters in Reaction Stoichiometry

Theoretical, Actual and Percent Yield

Theoretical, Actual and Percent Yield

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 21 of 24

actual yield = theoretical yield

quantitative

Side reactions reduce the percent yield.

By-products are formed by side reactions.

Consecutive Reactions, Simultaneous Reactions and Overall Reactions

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 22 of 24

Multistep synthesis is often unavoidable.

Reactions carried out in sequence are called consecutive reactions.

When substances react independently and at the same time the reaction is a simultaneous reaction.

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 23 of 24

Overall Reactions and Intermediates

The Overall Reaction is a chemical equation that expresses all the reactions occurring in a single overall equation.

An intermediate is a substance produced in one step and consumed in another during a multistep synthesis.

Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 4Slide 24 of 24

End of Chapter Questions

Initial problem solving is linear and often based on memorizing solutions for particular situations.

Answer

a

bc

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