Midterm 2 Flashcards
What are the two types of processes that are identified in heating curves
Two types of processes are identified in the heating curve:
- Phase change → added heat does not change temperatureq = n∆H → Number of moles * enthalpy of the phase change
- Heating of substance → added heat changes temperature
q = mc∆T → mass * specific hat capacity * change in temperature
what are the Enthalpies of reaction
The enthalpy change for a change is ∆H = Hf - Hi
so in chemical reactions ∆H rxn = ∑H products - ∑H reactants
∆H rxn is called the enthalpy of reaction
Like U absolute values of H are difficult to determine so we can’t calculate ∆H reaction from this equation.
what is the enthalpy change for endothermic vs exothermic processes
In endothermic processes ∆H rxn > 0
In exothermic processes ∆H rxn < 0
what are thermochemical equations
A thermochemical equation is the balanced equation that includes the value of ∆H
- Magnitude of ∆H depends on amounts of reactants and products
- ∆H forward = -∆H reverse
- ∆H depends on the physical state of reactants and products
what is the standard enthalpy change for a chemical reaction
standard enthalpy change for a chemical reaction. ∆H rxn
the enthalpy change for a reaction with reactants and products in their standard states
what is the thermodynamic standard state
Thermodynamic standard state
A particular conditions of T and P that is used as a reference for the calculations of thermodynamic properties: 1 atm and 25ºC/298 K
what are the standard enthalpies of formation
the enthalpy change for the hypothetical formation of 1 mol of substance in its thermodynamic standard state from its constituent elements in their standard states
for pure elements in their most stable forms at standard conditions ∆Hºf = 0 by definition
How do you calculate the enthalpy change for a reaction
Using enthalpies of formation to calculate enthalpies of reaction steps:
- Decompose reactants into elements
- recombine elements into productsNote: it doesn’t actually matter that this isn’t the pathway of formation since H is a state function
∆Hºrxn = ∑V(product)∆Hºf(products) - ∑V(reactants)∆Hºf(reactants)
where V(product) and V(reactant) are the unitless stoichiometric coefficients in the reaction equation
what is Hess’ law
Tabulated ∆H values can be used to calculated the enthalpy change of reactions. ∆H depends on amounts and initial and final states of reactants and products
An unknown ∆Hºrxn can be calculated by manipulating known equations to construct a pathway with the same initial and final states as the unknown equation.
how does chemistry protect us against solar rays
Reactions are an outer defence against radiation and high energy particles.
Chemical processes absorb solar radiation → photodissociation (break bonds) and photoionization (ionize)
what is photodissociation
Molecule absorbs photons and bonds are broken.
incoming energy must be sufficient to break the bond.
If you know the dissociation enthalpy for a chemical bond you can calculate what energy or wavelength is needed to break one bond with these formulas:
E = ∆H * (1 mol / 6.02 * 10^23)
wavelength = hc/E = (6.63 * 10^-34)(2.98 * 10^8) / E
what is photoionization
Molecules absorbs a photon and loses an electron → ions are formed (ionization)
occurs at high elevations (~90km) lower thermosphere
The amount of energy required to ionize one mol of a gas → enthalpy of ionization. (aka ionization energy)
Photon must have enough energy to remove an electron
are completely filtered out
Higher energy process than photodissociation, shorter wavelength photons filtered.
what is chemical equilibrium
Chemical reactions have forwards and reverse directions.
At equilibrium, concentrations of reactants and products become constants (but not equal). Reactions continue in both directions, it is a dynamic process.
what is the state of a system at equilibrium (5 points)
At equilibrium:
- Concentrations of reactants and Products do not change
- Reactions continue in both directions
- The rate of the reaction in the forward direction equals the rate of the reaction in the reverse reaction. Kf = Kr
- Can originate from a mixture of reactants and products, or pure reactants or pure products.
- The equilibrium constant K is a ration between the concentrations (or partial pressures) of reactants and the concentrations (or partial pressures) of products at equilibrium.
How do you calculate Kc and Kp
For a reaction: aA+bB<–> cC+dD
Kc=[C]^c[D]^d/[A]^a[B]^b
Kp=[PC]^c[PD]^d/[PA]^a[PB]^b
KP=KC(RT)^∆V
∆V=(c+d)-(a+b)
R = 0.082 Latm / molK
what is the state of a system away from equilibrium
When a chemical system is away from equilibrium:
- Concentrations of reactants and products change towards those at equilibrium
- The rate of the reaction in the forward direction is different from the rate of the reaction in the reverse direction. Kf doesn’t = Kr
- The reaction quotient Q is a ratio of the concentrations (or partial pressures) of reactants and products at each specific non-equilibrium condition. Q is not a constant.
- Before equilibrium is reached, the value of Q changes with time.
- Once equilibria is reached the value of K does not change with time.
what is Q
Q is a measure of the progression of a reversible reaction that is not at equilibrium. It is a ration of concentrations or a ration of partial pressures of reactants and products to the power of their stoichiometry.
how do you calculate Qc and Qp
For a reaction: aA+bB<–> cC+dD
Qc=[C]^c[D]^d/[A]^a[B]^b
Qp=[PC]^c[PD]^d/[PA]^a[PB]^b
QP=QC(RT)^∆V
∆V=(c+d)-(a+b)
R = 0.082 Latm / molK
what is K
K is the unique ration of concentrations or a ration of partial pressures of reactants and products at equilibrium to the power of their stoichiometry (under specified conditions).
what is happening when Q is >, <, and = to K
When Q > K, reaction proceeds towards reactants to establish equilibrium.
When Q < K, reaction proceeds towards products to establish equilibrium.
When K = Q the reaction is at equilibrium.
what is the Equilibrium constant
K
Equilibrium constants have no units and are dependent on temperature.
The concentrations or partial pressures are understood to be a ration with respect to a standard reference state known as activity.
what does it mean when K >, <, and = to 1
K < 1 at equilibrium = more reactants than products
K ~ 1 at equilibrium = similar amounts of products and reactants
K > 1 at equilibrium = more products than reactants.
what phases are involved in K
For heterogeneous mixtures (reactants and products are not in the same phase)
Pure solids, pure liquids and solvents are not included in the equilibrium constant expression. Their concentrations/activities are considered constant and equal to 1.
what is Le Châtelier’s principle
When a reaction mixture at equilibrium is disturbed to non-equilibrium concentrations, the concentrations will change to counteract the disturbance and re-establish equilibrium.
what happens when reactants or products are added to an equilibrium
When reactants are added, Q < K, reaction proceeds right, consumes reactants and produces products.
When products are added, Q > K, reaction proceeds left, consumes products and produces reactants.
what happens when reactant or products are removed from equilibrium
When reactants are removed, Q > K, reaction proceeds left, consumes products and produces reactants.
When products are remove, Q < K, reaction proceeds right, consumes reactants and produces products.
what happens when you change in the pressure in an equilibrium
Equilibrium can be disturbed by changing the pressure or volume for reactions that involve a change in the total number of molecules in the gas phase (the stoichiometry).
An increase in pressure causes the reaction to shift towards whichever side has less molecules.
A decrease in pressure causes the reaction to shift towards whichever side has more molecules.
why does changing pressure affect equilibrium
Ex: Doubling the pressure also doubles the concentrations
Let A be the of products at equilibrium and B be the of reactants at equilibrium.
$K=[A]/[B]^2$
Doubling the pressure:
[A] = 2[A]eq
[B] = 2[B]eq
$Q=2[A]eq/(2[B]eq)^2=2[A]eq/4[B]eq=1/2 *k$
Q < K, reaction proceeds towards products to re-establish equilibrium.
When the humber of moles of gas in reactants and products are the same, changing pressure has no effect.
how does added heat affect Le chateliers
Equilibrium can be disturbed by changing the temperature for reactions resulting in a new equilibrium position.
Temperature dependence of K is related to the sign of ∆Hº for the reaction
increases in T, adds heat, which is consumed to establish equilibrium.
what happens when heat is added to an endothermic reaction
for endothermic reactions, ∆Hº is positive. When a reaction at equilibrium is heated:
^heat + reactants → Products , ∆Hº > 0 , shifts towards products
rate to the right is faster (proceeds right) → K increases
what happens when heat is added to an exothermic reaction
for exothermic reactions, ∆Hº is negative, when a reaction at equilibrium is heated
Reactants ← Products + Heat^, ∆Hº < 0
rate to the left is faster (proceeds left) ← K decreases
what happens when a catalyst is added to an equilibrium
A catalyst increases the rate of the reaction by making a lower-energy pathway available for both forward and reverse reaction, equilibrium is achieved quickly but there is no change in equilibrium position.
how do you calculate K for the reverse reaction
The equilibrium constant in one direction is the inverse of that in the opposite direction
If $K_r=[C]^c[D]^d/[A]^a[B]^b$
then $K_p= 1/K_r=[A]^a[B]^b/[C]^c[D]^d$
how do you calculate K from multiple steps of reactions
For reactions added together to obtain an overall chemical equation the overall equilibrium constants is the product of the K values for the individual equilibrium steps
$K_C=K_1K_2$
how does K change when you multiply or divide the reaction
Multiply a reaction by 2 causes all the coefficients to double, so the new K values is the square of the old one.
$K_2=(K_c)^2$
When dividing a reaction by 2, all the coefficients half, so the new K value is the square root of the old one.
$K_2=√K_C$
how do you calculate equilibrium concentrations from initial amounts and K
Steps:
- Write the balanced chemical equation
- Make an ICE table of concentrations
- Insert Initial, Change, and Equilibrium concentrations or the mathematical expressions for them
- derive the equilibrium expression
- solve for x using the initial concentrations and k
what are Bronsted-Lowry acids and bases
Acids are proton (H+) donors (they donate protons to a base)
Bases are proton (H+) acceptors (they accept protons from an acid)
what is a standard B-L acid run
Standard Acid reaction: HA + H2O ←→ H3O+ + A-
where
HA = acid
H2O = Base
H3O+ = conjugate acid
A- = conjugate base
what is a standard base B-L run
Standard base reaction: B + H2O ←→ BH+ + OH-
B = base
H2O = acid
BH+ = conjugate acid
OH- = conjugate base
what does amphiprotic and amphoteric mean
water can donate and accept protons and is referred to as amphiprotic
Water and compounds that react with both acid and base are referred to as amphoteric
what is the Autoionization of water
Autoionization of water is a dynamic equilibrium process
2 H2O ←→ H3O+ + OH-
H3O+ = hydronium (conjugate acid)
OH- = hydroxide (conjugate base)
what are the stats for pure water at 25º C
Pure Water at 25ºC:
[H3O+] = 1.0 * 10^-7 M
[OH-] = 1.0 * 10^-7 M
[H2O] = 55 M
Kw = [H3O+][OH-] = (1 * 10^-7)^2 = 1 * 10^-14
[H2O] is not included in the equilibrium constant for water Kw
how do [H3O+] and [OH-] relate in acid, basic, and neutral solutions
For a neutral solution, [H3O+] = [OH-] = 1 * 10^-7
For acidic solutions, [H3O+] > [OH-], More conjugate acid (H3O+) in the solution
For basic solutions, [H3O+] < [OH-], More conjugate base (OH-) in the solution
what is pH
The universal measure of acidity is called pH, a unit less quantity that is expressed as
pH = -log[H3O+] or 10^-pH = [H3O+]
what is pOH
The universal measure of basicity is pOH, a unit less value that is expressed as
pOH = -log[OH-] or 10^-pOH = [OH-]
what is Pkw
For any aqueous solution (at 25ºC): Kw = (1 * 10^-14) = [H3O+][OH-]
-logKw = -log(1 * 10^-14) = -log([H3O+]) - log([OH-])
pKw = 14 = pOH + pH
how does the pH scale work
pH < 7 = Acidic
pH = 7 = neutral
pH > 7 = Basic
Going down 1 pH increases the [H3O+] by a factor of 10
what are strong acids
Strong Acid: Ionization reaction is complete in water
HCl + H2O → H3O+ + Cl-
Others include: HClO4, H2SO4, HNO3, HBr, HI
what are strong bases
Strong bases: Dissociation reaction is complete in water
NaOH → Na+ + OH-
Others include: LiOH, KOH, Ca(OH)2, Sr(OH)2,, Ba(OH)2
what are weak acids
Partially ionize in aqueous solutions and establish an equilibrium
HA + H2O ← → H3O+ + A-
Ka = [H3O+][A-] / [HA]
pKa = logKa
Equilibrium constant Ka is called the acid ionization constant. it is unique to an individual acid and is a measure of the strength of the acid
what are weak bases
Partially ionize in aqueous solutions and establish an equilibrium
B + H2O ← → BH+ + OH-
Kb = [BH+][OH-] / [B]
pKb = -logKb
Equilibrium constant Kb is called the base ionization constant. It is unique to an individual base and is a measure of the strength of the base
what is percent ionization
Is another measure of acid or base strength
Percent ionization of a weak acid or base varies with the initial concentration
how do you calculate percent ionization
For acids % ionization = [H3O+]equilibrium / [HA]initial * 100
For bases % ionization = [OH-]equilibrium / [BOH]initial * 100
[H3O+]equilibrium = √(Ka * [HA])
[OH-]equilibrium = √(Kb * [BOH])
for conjugate acid-base pairs how do Ka, Kb and pKa, and pKb relate
HA + H2O ←→ H3O+ + A-
A- + H2O ←→ HA + OH-
Adding the two we get the autoionization equilibrium for water:
2 H2O ←→ H3O+ + OH-
Ka * Kb = Kw
Kb = Kw / Ka
pKa + pKb = pKw = 14
pKb = 14 - pKa
The same relationship is obtained for a weak base and its conjugated acid.
Stronger acids have weaker conjugate bases and weaker acids have stronger conjugate bases
what happens when a salt with conjugates of strong acids and bases dissolves in water
Salts dissolve in water and dissociate into their ions. The acidity or basicity of the solution depends on the relative acidity and basicity of the ions.
NaCl + H2O: Na+ = conjugate acid of a strong base (NaOH), Cl- = conjugate base of a strong acid (HCl)
Since both the cation (Na+) and the anion (Cl-) are very weak, they will not react with water. And an NaCl solution has a pH of 7
what happens when a salt with the conjugate of one of either a strong acid or strong base dissolves
For something like NaNO2 +H2O: Na+ = conjugate acid of a strong base (NaOH), NO2- = conjugate base of a weak acid (HNO2)
The cation (Na+) is a very weak conjugate acid and will not react with water.
The anion (NO2-) is a weak conjugate base and will react with water.
NO2- + H2O ←→ HNO2 + OH-
Kb = 2.2 * 10^-11
The ionization of the nitrite ion will generate OH-. so the solution will have pH > 7
what happens when a salt of the conjugate of both a weak acid and base dissolves
For MeNH3CN + H2O: MeNH3 = conjugate acid of a weak base. CN- = conjugate base of a weak acid (HCN)
Both the cation and anion establish an equilibrium with water to from hydronium ions and hydroxide ions, respectively.
MeNH3+ + H2O ←→ MeNH2 + H3O+ Ka = 2.8 * 10^-11
CN- + H2O ←→ HCN + OH- Kb = 2.5 * 10^-5
since Kb > Ka solutions of MeNH3CN are basic, pH > 7
how do polyprotic acids and bases work
Three equilibria are established by the triprotic phosphoric acid
H3PO4 + H2O ←→ H3O+ + H2PO4- Ka1 = 7.5 * 10^-3, pKa1 = 2.2
H2PO4- + H2O ←→ H3O+ + HPO4 2- Ka2 = 6.2 * 10^-8, pKa1 = 7.2
HPO4 2- + H2O ←→ H3O+ + PO43 - Ka3 = 2.1 * 10^-13, pKa1 = 12.7
Generally Ka1»_space;> Ka2, Therefore, the acidity can be determined using only Ka1
Same for a base
why does CO2 affect pH in water
CO2 reacts with water to form carbonic acid, a weak acid KH
CO2 (g) ←→ CO2 (aq) , dependent on pressure and temperature
CO2 + H2O ←→ H2CO3 K = 1.7 * 10^-3
H2CO3 + H2O ←→ H3O+ + HCO3- , ka1 = 4.3 * 10^-7, pKa1 = 6.4
HCO3- + H2O ←→ H3O+ + CO3 2- , Ka2 = 5.6 * 10^-11, pKa2 = 10.3
Ka1»_space; Ka2 So the acidity can be determined using only Ka1. Also [Bicarbonate]»_space;> [Carbonate]
Overall, the process leads to an increase in [H3O+] → Lowers pH
Sumarize CO2 acidification
As amount of CO2 increase, the concentration of aqueous CO2 in natural waters increases
The concentration of carbonic acid increase and the concentration of hydronium ions increase (pH decreases)
This acidification of natural water has negative impacts on biosystems and minerals
what are buffers
pH < 1
A buffer solution presents small pH changes when concentrated strong acids or bases are added to it.
Buffer solutions contain similar concentrations of a weak acid and a salt of its conjugate base or similar concentrations of a weak base and a salt of its conjugate acid
[HA] ~ [A-]
[CH3COOH] ~ [CH3COO- Na+]
how do buffers work
Henderson-Hasselbalch Equation:
pH = pKa + log [A-]/[HA]
When [HA] ~ [A-] ⇒ pH = pKa
the pH will not significantly change as long as the ration of [HA] / [A-] does not change too much.
Buffer Capacity: The Amount of strong acid or base added to a buffer solution that will lead to pH change of 1 unit.
what buffers are in our body
Buffers keep you alive. Biological buffer systems can be very complex involving several acid/base pair…
A [H2CO3] / [HCO3-] buffer system keeps a constant pH in the blood. 7.45-7.35
A [H2PO4-] / [HPO4 2-] buffer system keeps a constant pH for intracellular fluid.
A [NH3] / [NH4 +] buffer system keeps a constant pH in urine.
