Second+Semester+Review

[|Reaction Prediction (activity/solubility)] (GEN) [|Types or reaction and reaction prediction](GEN)
 * RXNS**

[|Stoichiometry] [|Percent Yield (stoich)] [|% yield and limiting reactant] [|% yield] %[|yield] [|Limiting Reactant] [|Limiting Reactant]
 * STOICH**

[|Types of heat transfer] [|Specific Heat] [|q=mCdeltaT] [|Q=mCAT] [|q=-q] [|All things deltaH (enthalpy) (thermo stoich, Hess, product-reactants)]
 * THERMO**

[|Unit Conversions] [|Unit Conversions] [|PV/T = PV/T] [|PV/T=PV/T] [|PV/T=PV/T] [|PV=nRT] [|PV=nRT] [|PV=nRT] [|Gas Stoich] [|Rates of a gas] [|Daltons law of partial pressures] [|Pt=Pa + Pb]
 * GAS**

Honors Only (leftover reactant, advanced thermo, solutions)

[|Boiling Point/Freezing point deltaT=ikm]

[|Gibbs Free Energy (HON)] [|Gibbs Free Energy = 0]

[|Vapor Pressure Lowering]

[|Osmotic Pressure]

[|henry's law (solubility of a gas based on pressure)]

[|Gas molar mass and density]

[|concentration units]

[|Saturated/Unsaturated/SuperSaturated]

[|PH Scale]


 * TOPICAL SUMMARY **

Balancing – same number of atoms on both sides Reaction prediction using solubility rules and activity series (rules are written on sheet)
 * __ Reactions (General Only) __**

1 mole = 6.02*1023 (Avogadro's number)
 * __ Stoichiometry __**

The //molar mass// is the number of grams per mole in a substance. ex) What is the molar mass of H2O? A hydrogen atom weighs 1 amu, so the molar mass of hydrogen is 1 gram. An oxygen atom weighs 16 amu, so the molar mass of oxygen is 16 grams. 1 gram + 1 gram + 16 grams = 18 grams.

The //mole to mole ratio// is the ratio of moles of one substance to moles of another. ex) 2H2 + O2 → 2H2O What is the mole to mole ratio of water and hydrogen?  There are two moles of hydrogen for every two moles of water, so the ratio is written:  2 mol H2 / 2 mol H2O

Stoichiometry with moles: ex) 2A + 3B2 → 2AB3 How many moles of AB3 can be made from 8 moles of A?  8 mol A * (2 mol AB3 / 2 mol A) = 8 mol AB3

ex) C3H8 + 5O2 → 4H2O + 3CO2 How many grams of H2O can be formed from 73.4 grams of O2? (assume excess C3H8)  73.4 g O2 * (1 mol O2 / 32 g O2) * (4 mol H2O / 5 mol O2) * (18 g H2O / 1 mol H2O) = 33.03 g H2O  molar mass of oxygen^ mole to mole ratio ^molar mass of water

How many grams of C3H8 are required to react with 73.4 grams of O2? 73.4 g O2 * (1 mol O2 / 32 g O2) * (1 mol C3H8 / 5 mol O2) * (44 g C3H8 / 1 mol C3H8) = 20.185 g C3H8

Remember dimensional analysis! Stoichiometry can be used to convert from:
 * grams to moles (periodic table, molar mass)
 * moles to moles (mole to mole ratio)
 * moles to atoms/molecules/formula units (Avogadro's number)
 * moles to Liters of gas (22.4 L = 1 mol of gas) ← ONLY WORKS AT STP (0oC)

ex) 2HF + Sn → SnF2 + H2 How many grams of SnF2 can be formed by reacting 7.42*1042 molecules of HF with tin?  7.42*1042 molc HF * (1 mol HF / 6.02*1023 molc HF) * (1 mol SnF2 / 2 mol HF) * (156.7 g SnF2 / 1 mol SnF2) = 965.7 g SnF2

2H2 + O2 → 2H2O If 20.0 grams of H2 and 150.0 grams of O2 react, how many grams of H2O are produced?
 * __ Limiting Reactants __**

In order to answer the question we must use stoichiometry twice: once going from grams of hydrogen to grams of water, and once going from grams of oxygen to grams of water. Whichever answer is //smaller// is the amount of water produced.

20 g H2 * (1 mol H2 / 2 g H2) * (2 mol H2O / 2 mol H2) * (18 g H2O / 1 mol H2O) = 180 g H2O 150 g O2 * (1 mol O2 / 32 g O2) * (2 mol H2O / 1 mol O2) * (18 g H2O / 1 mol H2O) = 168.75 g H2O

168.75 grams of water are produced.

Since the amount of oxygen limited how much water was produced, it is the //limiting reactant//. Not all of the hydrogen reacted with the oxygen when the reaction was finished. It is the //excess reactant.//

How many grams of the excess reactant remain (HONORS)?

To answer this question we use stoichiometry to go from grams of water produced to grams of hydrogen used, like so:

168.75 g H2O * (1 mol H2O / 18 g H2O) * (2 mol H2 / 2 mol H2O) * (2 g H2 / 1 mol H2) = 18.75 g H2

18.75 grams of hydrogen reacted.

20 g – 18.75 g = 1.25 grams of hydrogen remaining.

The //empirical formula// is the lowest ratio of atoms for any given formula. ex) What is the empirical formula of hydrogen peroxide? Hydrogen peroxide is H2O2. The lowest ratio of 2:2 is 1:1; therefore the empirical formula is HO.
 * __ Empirical Formula (HONORS): __**

ex) What is the empirical formula of octane? Octane is C8H18. The lowest ratio of 8:18 is 4:9; the empirical formula is C4H9.

Ionic compounds are written in empirical form. Empirical formulae are almost never used; they are used when deducing a formula from a percent mass.

ex) A common oxide of nitrogen contains 25.93% nitrogen. Deduce the empirical formula of the oxide. Before we begin solving, we must assume we have a certain amount of the compound. It is easiest to assume 100 grams. Therefore, the compound is comprised of 25.93 grams of nitrogen and 74.07 grams of oxygen.

25.93 g N * (1 mol N / 14 g N) = 1.852 mol N 74.07 g O * (1 mol O / 16 g O) = 4.629 mol O

N: 1.852/1.852 = 1 O: 4.629/1.852 = 2.5

NxOy → NO2.5 → N2O5

The empirical formula is N2O5.

If the molar mass of the molecular formula is 432 g/mol, what is the molecular formula? The molar mass of the empirical formula is 108 g/mol. 432/108 = 4.

2*4 = 8 5*4 = 20

The molecular formula is N8O20.

What is heat? Heat is an energy change indicated by a change in temperature.
 * __ Thermochemistry: __**

When a reaction loses heat (gives off heat), the reaction is known as an //exothermic// reaction. When a reaction gains heat (absorbs heat), the reaction is known as an //endothermic// reaction.

Always think of heat flow as warm to cold.

As things heat up, they tend to expand; as they cool down, they tend to contract. This is not always true.

__ Specific heat capacity: __ The //specific heat capacity// of a substance is the energy required to lower or raise 1 gram of a substance 1oC. It is measured in joules over grams of Celcius (j/g oC).

Water is equally hard to heat and cool because it requires a lot of energy.

Formula for calculating heat: q = mc∆T

q is the heat (in joules) m is the mass of the substance (in grams) c is the specific heat (in j/g oC) ∆T is the change in temperature (Tf – Ti, measured in Celsius or Kelvin)

(K = C + 273, C = K – 273)

q1 = mc∆T, q2 = mc∆T -q1 = q2 (q1 is exothermic, q2 is endothermic) -m1c1∆T1 = m2c2∆T2 (Tf is the same on both sides)

ex) How much heat is required to raise the temp of 654 grams of water from 34.5oC to 89.7oC? The specific heat of water is 4.184 j/g oC. The mass of water is 654 grams, and ∆T = 89.7 – 34.5 = 55.2oC.  q = 654 g * 4.184 j/g oC * 55.2oC = 151000 J

__ Enthalpy (∆H) __ Enthalpy is the thermal energy change associated with a chemical reaction or process expressed in kilojoules (kJ). If ∆H is negative, the reaction produces thermal energy (it is exothermic). If ∆H is positive, the reaction absorbs thermal energy (it is endothermic). The ∆H value represents the amount of heat transferred when the reactants react.

ex) 2S + 3O2 → 2SO3, ∆H = -792 kJ How much heat is produced when 85 grams of sulfur react?  85 g S * (1 mol S / 32 g S) * (-792 kJ / 2 mol S) = -1048.6 kJ

ex) What mass of propane, C3H8, must be burned in order to produce 76,000 kJ of energy? C3H8 + 5O2 → 3CO2 + 4H2O, ∆H = -2200 kJ

76000 kJ = (-2200 kJ / 1 mol C3H8) * (1 mol C3H8 / 44 g C3H8) * X g C3H8 76000 kJ = -50X X = 1520 g C3H8

__ Heat of Formation (∆Hf) __ // Heat of Formation // is the thermal energy associated with creating compounds or elements in their nonnatural sates. *Elements (including diatomic compounds) do not have heats of formation. ∆Hrxn = n∑∆Hf (products) - n∑∆Hf (reactants) n = number of moles

ex) N2 + 3H2 → 2NH3 N2: ∆Hf = 0  H2: ∆Hf = 0  NH3: ∆Hf = -46 kJ/mol

∆Hrxn = (2 mol * -46 kJ/mol) – (1 mol * 0 + 3 mol * 0) = -92 kJ

ex) H2SO4 + Ca → H2 + CaSO4 H2SO4: ∆Hf = -909 kJ/mol  Ca: ∆Hf = 0  H2: ∆Hf = 0  CaSO4: ∆Hf = -1433 kJ/mol

∆Hrxn = (1 mol * -1433 kJ/mol + 0) – (1 mol * -909 kJ/mol + 0) = -1433 + 909 = -524 kJ

__ Hess' Law: __ Hess' Law deals with combined heats of reaction; if given three or more equations and their heats of reaction, the heat of reaction for the combined equation can be found.

ex) B2H6 + 6Cl2 → 2BCl3 + 6HCl, ∆Hrxn = ?

BCl3 + 3H2O → H3BO3 + 3HCl, ∆Hrxn = -112.5 kJ B2H6 + 6H2O → 2H3BO3 + 6H2, ∆Hrxn = -493.4 kJ  ½ H2 + ½ Cl2 → HCl, ∆Hrxn = -92.3 kJ

In order to solve the problem, we must manipulate the three reactions given to us so that compounds common to both sides cancel each other out. This can be done in one of two ways; either by multiplying the reaction (and the heat of reaction) by a common coefficient, or by reversing the reaction (and the heat of reaction; for example, if the original heat of reaction is positive, it becomes negative and vice versa). Once the equations have been sufficiently manipulated, the compounds common to both sides will cancel out and the remaining reaction will be identical to the given one. The heat of reaction for that equation is then the sum of the heats of reaction for the manipulated equations.

To solve this equation, we must change: BCl3 + 3H2O → H3BO3 + 3HCl, ∆Hrxn = -112.5 kJ to: 3HCl + H3BO3 → BCl3 + 3H2O, ∆Hrxn = 112.5 kJ and then to: 6HCl + 2H3BO3 → 2BCl3 + 6H2O, ∆Hrxn = 225 kJ

We must also change: ½ H2 + ½ Cl2 → HCl, ∆Hrxn = -92.3 kJ to: 6H2 + 6Cl2 → 12HCl, ∆Hrxn = -1107.6 kJ

The equations are now: 6HCl + 2H3BO3 → 2BCl3 + 6H2O, ∆Hrxn = 225 kJ B2H6 + 6H2O → 2H3BO3 + 6H2, ∆Hrxn = -493.4 kJ  6H2 + 6Cl2 → 12HCl, ∆Hrxn = -1107.6 kJ

When the products common to both sides cancel out, we are left with: 6HCl + 2H3BO3 → 2BCl3 + 6H2O, ∆Hrxn = 225 kJ B2H6 + 6H2O → 2H3BO3 + 6H2 , ∆Hrxn = -493.4 kJ  6H2 + 6Cl2 → 6HCl + 6HCl , ∆Hrxn = -1107.6 kJ

B2H6 + 6Cl2 → 6HCl + 2BCl3 ∆Hrxn = 225 + -493.4 kJ + -1107.6 = -1376 kJ

__ Heating Curve of Water(HONORS) __ Specific heat: c cice = 2.050 J/g oC  cwater = 4.184 J/g oC  csteam = 2.080 J/g oC  ∆Hvap = 40.7 kJ/mol (heat of vaporization; boiling point) ∆Hfus = 6.01 kJ/mol (heat of fusion; freezing point)

ex) How much heat is gained when 45 grams of water at 20oC changes to steam at 115oC? To solve this, we must use stoichiometry and q=mc∆T.  q1 will represent the heat gained while the water boils. cwater = 4.184 J/g oC; ∆T = 100 – 20 = 80oC.  q2 will represent the heat gained while the water is at 100oC. To solve for it we will use stoichiometry.  q3 will represent the heat gained while the water vapor heats to 115oC. csteam = 2.080 J/g oC; ∆T = 115 – 100 = 15oC.  The total heat gained is the sum of q1, q2, and q3.
 * q1 = 45 g H2O * 4.184 J/g oC * 80oC = 15100 J
 * q2 = 45 g H2O * (1 mol H2O / 18 g H2O) * (40.7 kJ / 1 mol H2O) * (1000 J / 1 kJ) = 101750 J
 * q3 = 45 g H2O * 2.080 J/g oC * 15oC = 1404 J
 * q = 15100 + 101750 + 1404 = 118500 J

__ Entropy (∆S)(HONORS) __ // Entropy // is a measure of disorder. It cannot be negative, but ∆S **can** be negative. The lower the number, the less entropy there is and the more order there is. The higher the number, the more entropy there is and the less order there is. Entropy is expressed in J/mol K (joules over moles Kelvin).

An exothermic reaction has higher entropy. As temperature rises, so does entropy. Well mixed solutions also have higher entropy. Solids are more ordered than liquids, which are more ordered than gasses. Therefore, gasses have the highest entropy, followed by liquids, followed by solids.

∆S = n∑Sproducts - n∑Sreactants n is the number of moles; the value of S will be given.

__ Gibbs Free Energy (∆G) (HONORS) __ // Gibbs Free Energy // is the capacity to do work. It is measured in kJ. If ∆G is positive, a reaction will not be spontaneous (however, the reverse reaction will be spontaneous). If ∆G is zero, the reaction has reached equilibrium; it will continuously work forwards and backwards. If ∆G is negative, the reaction will be spontaneous.

∆G = ∆H – T∆G T = temperature (K)


 * __ Gases: __**

__ Kinetic Molecular Theory: __ Postulate 1 – Gases are tiny particles in constant random motion. Postulate 2 – Gases (i.e. gas particles) have practically no volume; the volume of a gas is the volume of the container. Postulate 3 – Gases have no attraction or repulsion (this is __false__ – they do have London forces; however, they are so weak it is insignificant). Postulate 4 – Gas collisions are perfectly elastic (that is, they bounce off each other with no energy lost). Postulate 5 – The kinetic energy of a gas is directly proportional to its temperature.

__ Pressure: __ Pressure is created from collisions (see Postulate 4). There is no such thing as negative pressure (i.e. suction). Pressure Units: 1 atmosphere (atm) = 760 mm Hg (torr) = 101.3 kilopascals (kPa) Standard Temperature and Pressure (STP) is 273 K and 1 atm.

__ Boyle's Law __ Pressure and Volume are inversely proportional (assuming temperature stays constant). P1V1 = P2V2

ex) A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg? 300 mL * 300 mm Hg = X mL * 500 mm Hg  X = 90000/500 = 180 mL

__ Charles' Law __ Volume and Temperature are directly proportional (assuming pressure stays constant). V1/T1 = V2/T2

ex) A gas has a volume of 4 liters at 323 K. What will its volume be (in liters) at 373 K? 4 L / 323 K = X L / 373 K  X = 1492/323 = 4.62 L

__ Gay-Lussac's Law __ Pressure and Temperature are directly proportional (assuming volume stays constant). P1/T1 = P2/T2

ex) The pressure of a gas at 300 K is 2.1 atm. What is the pressure of the gas when the temperature is lowered to 270 K? 2.1 atm / 300 K = X atm / 270 K  X = 567/300 = 1.89 atm

__ Combined Gas Law __ (P1V1)/T1 = (P2V2)/T2

ex) A gas has a volume of 39 liters at STP. What will its volume be at 4 atm and 298 K? (1 atm * 39 L) / 273 K = (4 atm * X L) / 298 K  X = 10.6 L

__ Ideal Gas Law __ PV = nRT P is pressure (atm) V is volume (L) n is number of moles R is a constant (.0821 atm L/mol K) ← R is __always__ this value T is temperature (K)

ex) 3.5 moles of oxygen are held at 3 atm of pressure and 300 K. What is the volume of the oxygen? 3 atm * V = 3.5 mol * .0821 atm L/mol K * 300 K  V = 28.7 L

__ Dalton's Law of Partial Pressures: __ Dalton's Law is used for finding the pressure of individual gases in a container where they are all together, or for finding the total pressure where the individual pressures are known.

PT = PA + PB + …. PA = PTX

PA and PB are the pressures of the individual gases A and B. PT is the total pressure. X is the ratio of moles of an individual gas to total moles (mol A/total mol).

ex) In a vessel there are 3 moles of Argon, 2 moles of Neon, and 6 moles of Xenon. If the total pressure is 28.3 atm, what is the pressure of the Neon? PNe = 28.3 atm * (2 mol/11 mol) = 5.15 atm

__ Graham's Law __ Graham's Law compares rates of diffusion of two gases.

r1/r2 = √(MM2/MM1)

MM2 is the molar mass of the heaver gas (MM1 is therefore the molar mass of the lighter gas).

ex) How many times faster does Helium diffuse than Xenon? Xenon's molar mass (131 g/mol) is greater than Helium's (4 g/mol)  √(131/4) = 5.7 times faster

There are two parts to a solution: a solute and a solvent. The solute is dissolved in the solvent. ex) salt water. Salt is the solute, water is the solvent, and salt water is the solution.
 * __ Concentration (HONORS) __**

Molarity (M) = moles of solute/liters of solution (mol/L) Molality (m) = moles of solute/kilograms of solvent (mol/kg) Percent mass = mass of solute/mass of solution Percent volume = volume of solute/total volume

__ Dilution Equation: __ M1V1 = M2V2 M = molarity (mol/L) V = volume (L)

ex) If you add 163 mL to a 1.10 L 6 M solution of HCl, what is the new molarity? 6 M * 1.10 L = M * (1.10 + .163) L  6.6 = 1.263 * M  M = 5.23 mol/L

ex) How would you dilute 233 mL of 5.3 M HNO3 solution to .34 M? 233 mL * 5.3 M = (x + 233) mL * .34 M  x = 3400 mL  To dilute 233 mL of 5.3 M solution to .34 M, you would need to add 3.4 L of water.

__ Colligative Properties __ // Colligative Properties // are specific properties that are true for all solutions, regardless of concentration.

Vapor Pressure Lowering (for a non-volatile solute): vapor = a gas that shouldn't be a gas but is at certain conditions volatile = liquid that readily evaporates into a vapor ◦ x = moles of solvent/total moles
 * solutions have a lower vapor pressure
 * that vapor pressure can be lowered by adding a substance with even lower vapor pressure
 * more solute = lower vapor pressure
 * Pnew vp = xPnormal vp

The //van t' Hoff Factor// is a constant that describes the number of particles a substance breaks into as it dissolves. Ionic compounds typically have a van 't Hoff factor greater than 2. Covalent and organic compounds have a van 't Hoff factor of 1. ex) NaCl; i = ? NaCl splits into two ions (Na+ and Cl-); therefore the van 't Hoff factor is 2.  ex) MgCl2; i = ? MgCl2 splits into three ions (Mg+, Cl- and Cl-); therefore the van 't Hoff factor is 3. ex) CO2; i = ? CO2 is a covalent compound; therefore the van 't Hoff factor is 1.  van 't Hoff factors are used to calculate Boiling Point Elevation, Freezing Point Depression, and Osmotic Pressure.

Boiling Point Elevation: ◦ i is the van 't Hoff factor; Kb is the boiling point elevation constant; m is the molality
 * If the solute boils at a higher temperature than the solvent, the solution's boiling point is raised by adding more solute.
 * ∆T = iKbm

Freezing Point Depression: ◦ i is the van t' Hoff factor; Kf is the freezing point depression constant; m is the molality
 * If the solute freezes at a higher temperature than the solvent, the solution's freezing point is lowered by adding more solute.
 * ∆T = iKfm

Osmotic Pressure: ◦ π is pressure (atm); i is the van 't Hoff factor; M is molarity; R = .0821; T is temperature (K)
 * pressure applied to cancel out osmosis
 * π = iMRT

A non-saturated solution can dissolve more solute. A saturated solution can dissolve no more solution. A super-saturated solution has dissolved more solute past its saturation point; this is achieved by raising the temperature of the saturated solution, dissolving more solute, and letting the solution cool.
 * __ Solutions (HONORS) __**

The higher the heat of a solution, the higher its saturation point.

__ Henry's Law: __ The concentration of a gas-liquid mixture can be determined by the pressure of a gas. P is pressure (atm) K is a constant (mol/L atm) ← will be given in the problem C is concentration (molarity, mol/L)
 * P = KC

ex) Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4 atm over the liquid at 25oC. The Henry's Law constant for CO2 in water at this temperature is .033 mol/L atm. 4 atm = .033C  C = 121.2 mol/L

__ pH __ pH is a scale from 1 to 14 that measures the concentration of H+ ions in a solution.

1--7--14

Low numbers are acids; high numbers are bases (or alkalis). Acids have a higher concentration of H+ ions, while bases have a higher concentration of OH- ions. pH's nearer to 7 are neutral; pure H2O (H + OH) is a 7.