Chem Chapter 10, 11 Williams

Enthalpy:  amount of internal energy contained in a compound

Entropy:  amount of intrinsic disorder within the compound

• Enthalpy is zero for elemental compounds such as hydrogen gas and oxygen gas; therefore, enthalpy is nonzero for water

• IF precipitates form

• IF gasses are formed

• IF liquids are produced

substance is dissolved in water

Reactants: left side of reaction

Products: right side of reaction

it is a reversible reaction

• In theory all reactions are reversible - in practice, not really

  • Ex. burning paper produces CO2  and H2O but it is highly energetically unfavorable to form paper back out it

A reaction is BALANCED if it complies with the law of conservation of matter

• The amount and type of atoms of each element of the left side must be equivalent to the amount and type of atoms on the right side

Must have charge conservation: if the charge on the reactant side is neutral, the product side must be neutral too

Limiting reagent: reactant that runs out first

Excess reagent: if you have more reactant than can be consumed in the reaction (left over after reaction)

an oxygen combining with an organic molecule (ie. hydrocarbon), the result is always going to contain CO2 and H2O

show states of matter

Include stable atoms or molecules

• Ex. skeleton equation: C8H18 (g)  + O2 (g) → CO2 (g) + H2O(g)

  • Ex. balanced equation:  2 C8H18 (g)  + 25 O2 (g) → 16 CO2 (g) + 18 H2O(g)

  • CANNOT change subscript after figuring out molecule - can ONLY change coefficients

pH in bloodstream

• If H+ concentration goes up (more acidic) combines with bicarbonate to produce a weaker acid, carbonic acid

• If it is too basic, H2O will combine with CO2 and form carbonic acid

a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.

• Ex. catalytic converters in automobiles

1. Determine the correct formulas for all the reactants and products

2. Write the skeleton equation by placing the formulas for the reactants on the left and the formulas for the products on the right with a yield sign in between. If two or more reactants or or products are involved, separate their formulas with plus signs.

3. Determine the number of atoms of each element in the reactants and products. Count a polyatomic ion as a single unit if it appears unchanged on both sides of the equation.

4. Balance the elements one at a time by using coefficients. When no coefficient is written it is assumed to be 1. Begin by balancing elements that appear only once on each side of the equation. Never balance an equation by changing the subscripts in a chemical formula. Each substance has only one correct formula.

5. Check each atom or polyatomic ion to be sure that the number is equal on both sides of the equation.

6. Make sure all the coefficients are in the lowest possible ratio.

Two or more elements combine to form fewer products - “Many form few”

• Ex. Mg + O2 → MgO2

• Styrenes → polystyrene

A single compound breaks down into many new compounds or elements - “few create many”

• Ex. 2HgO → heat → 2Hg + O2

• Leaves decomposing in the forest

• H2CO3 → CO2 + H2O

• In a decomposition reaction one must be given products to balance or else there are too many options

A metal reacts with a “salt” - more “ionic” bonds, causing the cation in the salt to switch places with the metal

• 2K(s) + 2HOH (l) → 2KOH (g) + H2(g)

The cation of one “salt” switches places with the cation of the second “salt”

• ONLY possible if one of the product form a precipitate

  • Note: both products can form a precipitate

• 2KI(aq) + Pb(NO3)(aq) → PbI2(s) + 2KNO3(aq)

• Determine if product is precipitate using Table F

Assumed to be complete

Always produces CO2 and H2O

Always produces an “oxide” of the metal (ex. Magnesium → magnesium oxide)

Occurs with some reactions such as combustion, decomposition, some synthesis, single replacement

• Double replacement is NEVER a redox reaction

Individual atoms in the reaction change their oxidation state. One atom increases their oxidation value, one atom decreases their oxidation value

• Ex. Zn (s) + CuSO4 (aq) ↔  Cu (s) + ZnSO4 (aq)

• If the element is above the ion in the table, the reaction is favored. If it is below the ion, it is unfavorable

loss of electrons

gain of electrons

acid + base

• always double replacement and forms in water

No reaction takes place (all dissolved in water)

• any group 1 ion

• ammonium, nitrate, acetate, hydrogen carbonate, chlorate

• halides (Cl - , Br - , I -) EXCEPT when combined with Ag+, Pb 2+, or Hg2 2+

• sulfates (SO4 2-) EXCEPT when combined with Ag+, Ca 2+, Sr 2+, Ba 2+, or Pb 2+

Show all the reactants and products as molecules

• ex. Pb(NO3)2 (aq) + Na2SO4 (aq) ↔ PbSO4 (s)  + 2NaNO3 (aq)

Show all the ions, solids, liquids, and gasses in the equation

• ex. Pb2+ (aq) + 2NO3- (aq) + 2Na+ (aq)  + SO4 (aq) ↔ PbSO4 (s)  + 2Na+ (aq) + 2NO3 (aq)

Shows only the ions, solids, liquids, and gasses involved in the actual reaction. What’s left after the repeating compounds on both sides of the equation cancel out

• Ex. Pb2+ (aq) + SO42- (aq) ↔ PbSO4 (s)

ions that didn’t participate in the reaction, that cancel each other out in the net ionic equation

standard temperature pressure

1 atm (101.325 kPa, 14.696 psi) and 0 degrees Celsius (273.15 K), 10^5 Pa

Measurement unit designed to “count” very small particles

• 1 mole = 6.023 x 10^23 molecules

• provides a conversion factor from “amu/ atom” to “grams/ mole of atoms”

1000 kg

mol

1.66 x 10-24g

• Remember the “amu” is based upon the carbon 12 isotope

The molar mass of a substance makes it possible to know the number of atoms or molecules in a substance based upon its mass

• If you are not given any info on significant figures, take the molar mass out to 1 decimal place

1. Determine the molar mass of each element in the molecule

2. Multiply the molar mass of each element by the number of that element in the molecule

3. Add the products of each element

• Ex. 1 molecule SO3 = 80.1 amu → 1 mole SO3 = 80.1 g (80.1 amu x 6.023x1023)

*MOLE MAP

Determine the percent, by mass, for each element within a molecule

• Find the molar mass of the molecule

• Divide the mass of each element by the molar mass of the molecule

  • Convert to percent

*PRACTICE