Molarity = M (Concentration of Solutions)

```Molarity (Concentration of Solutions) = M
M=
Moles of Solute =
Liters of Solution
mol
L
solute = material dissolved into the solvent
In sea water, water is the solvent, and NaCl, MgCl2, etc are the solutes.
In brass, copper is the solvent (90%), and zinc is the solute (10%).
Preparing a Solution - I
Example problem: A solution of sodium phosphate
is prepared by dissolving 3.95 g of sodium
phosphate in water and diluting it to 300.0 mL. What
is the molarity, M, of the salt and each of the ions?
Strategy
(1)Write chemical equation showing process.
(2)Calculate moles of each species.
(3)Divide # moles by # L water to obtain molarity.
(1)Na3PO4 (s)
H2O(solvent)
→
3 Na+ (aq) + PO4-3 (aq)
Dilution of Solutions
Dilute 25.00 mL of 0.0400 M KMnO4 to a final volume
of 500. mL. What is the resulting molarity (M) of the
diluted solution?
Another Strategy for calculating final concentration:
The number of moles of solute is the same before and
after dilution
V1 x M1 = moles solute = V2 x M2
V1M1 = V2 M 2
V1M1
M2 =
V2
25.00mL × 0.0400M
= 0.00200 M
M2 =
500.mL
TYPES of CHEMICAL REACTIONS
in LIQUID SOLUTIONS:
• Precipitation reactions
• Acid – Base reactions
• Reduction – Oxidation (REDOX) reactions
Table 4.1 Simple Rules for Solubility
of Salts in Water
1. Most nitrate (NO3-) salts are soluble.
2. Most salts of Na+, K+, and NH4+ are soluble.
3. Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl2, and Hg2Cl2.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4,
PbSO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important
soluble hydroxides are NaOH, KOH, and
Ca(OH)2 (marginally soluble).
6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-)
salts are only slightly soluble.
Predicting if a precipitate forms, and which?
Pb(NO3)2(aq) + NaCl(aq) →
Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + Cl-(aq)
RULE: If any of the possible new species formed by
combining anions with cations is insoluble,
then that precipitate will form.
USE TABLE 4.1
(I’ll give you this Table for the tests, but not on quiz.)
In this case, PbCl2 is insoluble, and a precipitate forms.
Note: Like the demo with PbI2(s) above.
Writing Equations:
a) Calcium Nitrate and Sodium Sulfate solutions are added together.
Molecular Equation
Ca(NO3)2 (aq) + Na2SO4 (aq)
CaSO4 (s) +2 NaNO3 (aq)
Complete Ionic Equation
Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq)
Net Ionic Equation
Ca2+(aq) + SO4-2(aq)
Spectator Ions are Na+ and NO3-
CaSO4 (s)
Quantitative Precipitation Problems:
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 0.125 M lead nitrate solution to precipitate all of the
The chemical equation for the reaction is:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
The moles of sodium iodide needed to precipitate PbI2 is twice the
lead ions. The number of moles of sodium iodide needed is:
2+
0.125
Mol
Pb
2
mol
I
2.50 L x
x
1.0 L soln.
1 mol Pb2+
The mass of sodium iodide is:
0.625 mol I- x 1 mol NaI x
1 mol I-
= 0.625 mol I-
149.9 g NaI
= 93.68 g NaI
1 mol NaI
Precipitation problem: When aqueous silver nitrate and sodium
chromate solutions are mixed, solid silver chromate forms in a
solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of
silver nitrate is added to 156.00 mL of a 0.0950 M solution of
sodium chromate, what mass of silver chromate (M = 331.8 g/mol)
will be formed?
This is a limiting-reactant problem because the amounts
of two reactants are given.
Strategy:
(1) Write the balanced equation.
(2) Calculate the number of moles of each reactant.
(3) Determine the limiting reactant.
(4) Calculate the moles of product.
(5) Convert moles of product to mass of the product using molar mass.
Selected Acids and Bases
Acids
Bases
Strong: H+(aq) + A-(aq)
Hydrochloric, HCl
Hydrobromic, HBr
Hydroiodoic, HI
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
Strong: M+(aq) + OH-(aq)
Sodium hydroxide, NaOH
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Strontium hydroxide, Sr(OH)2
Barium hydroxide, Ba(OH)2
Weak
Hydrofluoric, HF
Phosphoric acid, H3PO4
Acetic acid, CH3COOH
(or HC2H3O2)
Weak
Ammonia, NH3
accepts proton from water to make
NH4+(aq) and OH-(aq)
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose two protons to yield two Hydronium ions, and one sulfate ion.
What is the molarity of the sulfate and Hydronium ions in a solution
prepared by dissolving 155g of concentrate sulfuric acid into sufficient
water to produce 2.30 Liters of acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles
by the volume to get the molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the acid molarity.
Solution: Two moles of H+ are released for every mole of acid:
H2SO4 (l) + 2 H2O(l)
2 H3O+(aq) + SO4- 2(aq)
Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4
98.09 g H2SO4
-2
1.58
mol
SO
4
Molarity of SO4- 2 =
= 0.687 Molar in SO4- 2
2.30 L solution
Molarity of H+ = 2 x 0.687 M = 1.37 Molar in H+ (or H3O+)
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced chemical reactions (molecular, total ionic, and
net ionic) for the following Chemical reactions:
a) Calcium Hydroxide(aq) and Hydroiodic acid(aq)
b) Lithium Hydroxide(aq) and Nitric acid(aq)
c) Barium Hydroxide(aq) and Sulfuric acid(aq)
Plan: These are all strong acids and bases, therefore they will make
water and the corresponding salts.
Solution:
a)
Ca(OH)2 (aq) + 2HI(aq)
CaI2 (aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq)
2 H2O(l)
Like Example 4.10 (P 113)
What volume of 0.468 M H2SO4 is needed to neutralize 215.00 ml
of a 0.125 M LiOH solution?
Calculate the number of moles of base:
Vbase x Mbase = 0.21500 L x 0.125 M = 0.0268 mol LiOH
From the balance equation find the moles of acid needed:
2 LiOH(aq) + H2SO4 (aq)
2 H2O(l) + Li2SO4 (aq)
Since there are two protons per molecule, we will need half as much
sulfuric acid as we have lithium hydroxide: or 0.0134 mol H2SO4
Volume of acid:
Moles acid
0.0134 moles
= 0.468 Mol
= 0.0286 L H2SO4
Vacid =
Macid
L
EP 91: Aluminum hydroxide reacts with hydrochloric acid
According to the balanced equation
Al(OH)3 (s) + 3 HCl (aq)
→ 3 H2O(l) + AlCl3 (aq)
What volume of 1.50 M HCl(aq) is required to neutralize
10.0 g Al(OH)3(s)?
Strategy:
(1) Calculate moles of Al(OH)3(s).
(2) Calculate moles of HCl needed using balanced equation
(3) Calculate volume HCl from (2) and known molarity
Recognizing Oxidizing and Reducing Agents - I
Problem: Identify the oxidizing and reducing agent in each of the Rx:
a) Zn(s) + 2 HCl(aq)
ZnCl2 (aq) + H2 (g)
b) S8 (s) + 12 O2 (g)
8 SO3 (g)
c) NiO(s) + CO(g)
Ni(s) + CO2 (g)
Plan: First we assign an oxidation number (O.N.) to each atom (or ion)
based on the rules in Table 4.3. The reactant is the reducing agent if it
contains an atom that is oxidized (O.N. increased in the reaction). The
reactant is the oxidizing agent if it contains an atom that is reduced
( O.N. decreased).
Solution:
a) Assigning oxidation numbers:
-1
+1
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2 (aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
Problem: Calculate the mass of metallic Iron that must be
added to 500.0 liters of a solution containing
0.00040M of Pt2+(aq) ions in solution to reclaim all Pt
via: 2 Fe(s) + 3 Pt2+(aq) → 2 Fe3+(aq) + 3 Pt(s)
Solution:
V x M = # moles
500.0L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+
Fe(s) → Fe3+ + 3 ePt2+ + 2 e- → Pt(s)
Need 2 moles of Iron for every 3 moles of Platinum
0.20 mol
Pt2+
2 mol Fe
x
= 0.133 mol Fe
2+
3 mol Pt
0.133 mol Fe x
55.85 g Fe
= 7.4 g Fe
mol Fe
Balancing REDOX Equations:
The oxidation number method
Step 1) Assign oxidation numbers to all elements in the equation.
Step 2) From the changes in oxidation numbers, identify the oxidized
and reduced species.
Step 3) Compute the number of electrons lost in the oxidation and
gained in the reduction from the oxidation number changes.
Draw tie-lines between these atoms to show electron changes.
Step 4) Choose coefficients for these species to make the electrons lost
equal the electrons gained (or total increase in ON = total
decrease in ON)
Step 5) Complete the balancing by inspection.
REDOX Balancing Using Ox. No. Method - II
+2
-1e-
+3
Fe+2(aq) + MnO4-(aq) + H+(aq)
Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e+2
+7
Balance charge: Multiply Fe+2 & Fe+3 by five to correct for
the electrons gained by the Manganese:
5 Fe+2(aq) + MnO4-(aq) + H+(aq)
5 Fe+3(aq) + Mn+2(aq) + H2O(aq)
Balance O: Need 4 H2O on right to cancel O from the MnO4-.
Balance H: Need 8 H+ on the left to balance these water Hs.
5 Fe+2(aq) + MnO4-(aq) +8 H+(aq)
5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq)
BALANCED!
Note: Add 8 water molecules to both sides if you prefer to express the 8 H+ ions instead as H3O+ ions.
Balancing redox eqn using half-cell method in acidic solutions
Cu(s) +HNO3(aq) → Cu2+(aq) + NO(g)
Identify half-reactions, one is ox, other is red
Cu(s) → Cu2+(aq)
0 → +2
HNO3(aq) → NO(g)
+5 → +2
Balance all atoms that are neither H nor O
OK as is
Balance O by adding H2O to side deficient in O
Cu(s) → Cu2+(aq)
HNO3(aq) → NO(g) + 2H2O
Balance H by adding H+ to side deficient in H
Cu(s) → Cu2+(aq)
3H+ +HNO3(aq) → NO(g) + 2H2O
Balance charge by adding e- to side that has + charge
Cu(s) → Cu2+(aq) + 2e3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O
Multiply each eq by factors so electrons cancel out
3x(Cu(s) → Cu2+(aq) + 2e-)
2x(3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O)
3Cu(s)+ 6H+ + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
Balancing redox equations in basic solutions:
First balance in acidic solution, then add OH- to cancel H+
The following redox equation is balanced in acidic solution. Balance it
for a basic solution.
Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3- (aq)
+3H2O(l)
Add OH- equal to number of H+ on both sides
6OH-(aq) +Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3-+3H2O(l)
+6OH-(aq)
Combine OH- and H+ to form water to max extent possible
6H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq)
Cancel H2O on both sides to max extent possible
3H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3- +6OH- (aq)
Redox Titration- Calculation outline - I
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
b)
Molar ratio
Moles of CaC2O4
c)
Chemical Formula
Moles of Ca+2
Problem: Calcium Oxalate was
precipitated from blood by the
addition of Sodium Oxalate so that
calcium ion could be determined. In
the blood sample. The sulfuric acid
solution that the precipitate was
dissolved in required 2.05 ml of
4.88 x 10-4 M KMnO4 to reach the
endpoint.
a) calculate the amount (mol) of
Ca+2.
b) calculate the Ca+2 ion conc.
Plan: a) Calculate the molarity of
Ca+2 in the H2SO4 solution.
b) Convert the Ca+2 concentration
into units of mg Ca+2/ 100 ml blood.
Ch. 5: Pressure =
force per unit area
Density of Mercury = 13.6 g/cm3
760 mm column of 1 cm2 area
weighs 76 cm x 1 cm2 x 13.6 g/cm3
= 1030 g = 1.03 kg = 2.28 lbs
P = force / area = 2.28 pounds / cm2
= 14.7 pounds / in2
= 1.00 atm.
Force = weight = volume x density
= area x height x density
P = Force / area = height x density
IDEAL GAS LAW
• The ideal gas equation combines both Boyles Law and
Charles Law into one easy-to-remember law:
PV=nRT
•
•
•
•
n = number of moles of gas in volume V
R = Ideal gas constant
R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1
Later R = 8.314 J / (mol K) = 8.314 J mol-1 K-1
An ideal gas is one for which both the volume of
molecules and forces between the molecules are so small
that they have insignificant effect on its P-V-T behavior.
Independent of substance, in the limit that n/V →0,
all gases behave ideally. Usually true below 2 atm.
Variations on Ideal Gas Equation
•
•
During chemical and physical processes, any of the four variables in the
ideal gas equation may be fixed.
Thus, PV=nRT can be rearranged for the fixed variables:
– for a fixed amount at constant temperature
• P V = nRT = constant
Boyle’s Law
– for a fixed amount at constant pressure
• V / T = nR / P = constant
Charles’ Law
– for a fixed pressure and temperature
• V = n (RT/P) or V/n = constant Avogadro’s Law
– for a fixed amount at constant volume
• P / T = nR / V = constant
Amonton’s Law
JUST REMEMBER:
PV=nRT and rearrange as needed
Standard Temperature and Pressure (STP)
A set of Standard conditions have been chosen to make it easier to
quantify gas amounts (i,.e., “liters at STP”).
Standard Temperature = 00 C = 273.15 K
Standard Pressure = 1 atmosphere = 760 mm Mercury
At these “standard” conditions, if you have 1.0 mole of any ideal gas,
it will occupy a “standard molar volume”:
Standard Molar Volume = 22.414 Liters = 22.4 L
Gas Law: Solving for Pressure
Problem: Calculate the pressure in a container whose Volume is 87.5 L
and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC.
Plan: Convert all information into the units required, and substitute into
the Ideal Gas equation ( PV=nRT ).
Solution:
5038 g Xe
= 38.37014471 mol Xe
nXe =
131.3 g Xe / mol
T = 18.8 oC + 273.15 K = 291.95 K
PV = nRT
so P = nRT
V
P = (38.37 mol )(0.0821 L atm)(291.95 K) = 10.5108 atm = 10.5 atm
87.5 L (mol K)
Applying the Gas law to T changes
Problem: A copper tank is filled with compressed gas to a pressure of
4.28 atm at a temperature of 0.185 oF. What will be the pressure if the
temperature is raised to 95.6 oC?
Plan: The volume of the tank is not changed, and we only have to deal
with the temperature change, so convert to SI units,
and calculate the pressure ratio from the T ratio.
Solution:
P1 = P2 = nR/V
T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC
T1
T2
T = -17.68 oC + 273.15 K = 255.47 K
1
T2 = 95.6 oC + 273.15 K = 368.8 K
P2 = 4.28 atm x 368.8 K = 6.18 atm
255.47 K
P2 = P1 x T2 = ?
T1
Change of Conditions :Problem -I
• A gas sample in the laboratory has a volume of 45.9 L at
25 oC and a pressure of 743 mm Hg. If the temperature is
increased to 155 oC by pumping (compressing) the gas to a
new volume of 3.10 ml what is the pressure?
•
•
•
•
•
P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm
P2 = ?
V1 = 45.9 L
V2 = 3.10 ml = 0.00310 L
T1 = 25 oC + 273 = 298 K
T2 = 155 oC + 273 = 428 K
Example Problem: Molar Mass of a Gas
from its weight and P,V,T
Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml
flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr,
what is the molecular weight of the gas?
Plan: Use the Ideal gas law to calculate n, then calculate the molar mass.
Solution:
P = 550.0 Torr x 1mm Hg x 1.00 atm
= 0.724 atm
1 Torr
760 mm Hg
V = 250.0 ml x 1.00 L = 0.250 L
1000 ml
n =P V
RT
T = 25.0 oC + 273.15 K = 298.2 K
n = (0.724 atm)(0.250 L)
= 0.007393 mol
(0.0821 L atm/mol K) (298.2 K)
M = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol
Ideal Gas Mixtures
• Ideal gas equation applies to the mixture as a
whole and to each gas species (i) individually:
PtotalV = ntotalRT Ptotal is what is measured by
pressure guages. Σ ni = n1+n2+n3…= ntotal
i
PiV = niRT for all species i. Pi much harder to
measure.
Note: iΣ Pi = P1+P2+P3…= Ptotal
Chemical Equation Calc - III
Mass
Atoms (Molecules)
Number
6.02 x 1023
Reactants
Molarity
moles / liter
Solutions
Molecules
Moles
Molecular
g/mol
Weight
Products
PiV = niRT
Gases
Mole fraction of species i:
xi = ni/ntotal
• PiV = niRT for all i, and
Σ Pi = P1+P2+P3…= Ptotal
• Σ ni = n1+n2+n3…= ntotal
• Σ xi = x1+x2+x3…= 1.00000
• Pi = niRT/V = (ni/ntotal)ntotalRT/V = xi . Ptotal
Gas Law Stoichiometry - I - NH3 + HCl
Problem: A slide separating two containers is removed, and the gases
are allowed to mix and react. The first container with a volume of 2.79 L
contains Ammonia gas at a pressure of 0.776 atm and a temperature of
18.7 oC. The second with a volume of 1.16 L contains HCl gas at a
pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid
ammonium chloride will be formed, and what will be remaining in the
container, and what is the pressure?
Plan: This is a limiting reactant problem, so we must calculate the moles
of each reactant using the gas law to determine the limiting reagent. Then
we can calculate the mass of product, and determine what is left in the
combined volume of the container, and the conditions.
NH3
HCl
Solution:
Equation:
NH3 (g) + HCl(g)
TNH3 = 18.7 oC + 273.15 = 291.9 K
NH4Cl(s)
Velocity and Energy
• Kinetic Energy = KE = ½ mu2 for one molecule
• Average Kinetic Energy of a mole of gas (KEavg)
= 3/2 RT independent of gas identity
• Average Kinetic Energy of one molecule (KEavg)
= 3/2 RT/NA independent of mass, identity
= ½ mu2 (mean value of u2)
u2 = 3RT/(mNA) = 3RT/M
√(u2) = √(3RT/M)
“Root mean square velocity”
Must use R in J/mol K, and remember that 1 J = 1 kg m2/s2
Molecular picture of HEAT!
Diffusion Rates: proportional to
average velocity or to 1/√M
• Rate1/Rate2 = u1/u2 = (M2/M1)1/2
• HCl = 36.46 g/mol
NH3 = 17.03 g/mol
• RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2
• RateNH3 = RateHCl x 1.463
The Equilibrium Constant - Definition
Consider the generalized chemical reaction:
aA+bB = cC+dD
A, B, C and D represent chemical species and a, b, c, and d are
their stoichiometric coefficients in the balanced chemical
equation. At equilibrium,
C ] [ D]
[
K=
a
b
[ A] [ B ]
c
d
Note: The “units” for K are
concentration units raised to
some power = c+d–(a+b)
The square brackets indicate the concentrations of the species in
equilibrium and K is a constant called the equilibrium
constant. K depends only on T , and not on concentrations.
2
[
][
]
CO
SO
2
2
CO2(g) + 2 SO2(g)K =
1
3
[CS2 ][O2 ]
CS2(g) + 3 O2(g)R
• The equilibrium expression for a reaction written in
reverse is the reciprocal of that for the original
expression.
3
[
CS2 ][O 2 ]
CS2(g) + 3 O2(g)K 2 =
[CO 2 ][SO 2 ]2
CO2(g) + 2 SO2(g)R
• If the original reaction is multiplied by a factor n, the new
equilibrium constant is the original raised to the power n.
2 CS2(g) + 6 O2(g) = 2 CO2(g) + 4 SO2(g)
CO 2 ] [SO2 ]
[
K3 =
2
6
[CS2 ] [O2 ]
2
4
= ( K1 )
2
1
=
K1
Example: Calculation of the Equilibrium
Constant from equilibrium amounts
At 454 K, the following reaction takes place:
3 Al2Cl6(g)
R 2 Al3Cl9(g)
At this temperature, the equilibrium concentration of Al2Cl6(g) is
1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x
10-2 M. Compute the equilibrium constant at 454 K.
Strategy: Substitute values into K
[
Al 3Cl9 ] (1.02 × 10 M )
K=
=
3
3
(1.00 M )
[Al2Cl6 ]
2
−2
2
−4
= 1.04 × 10 M
−1
Determining Equilibrium Concentrations from K
Example: Methane can be made by reacting carbon disulfide with
hydrogen gas. K for this reaction is 27.8 (L/mol)2 at 900°C.
CS2 (g) + 4 H2 (g)
CH4 (g) + 2 H2 S(g)
At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol
CS2, 1.10 mol of H2, and 0.45 mol of H2S. How much methane was
formed?
Strategy:
(1)Calculate the equilibrium concentrations from the moles given
and the volume of the container.
(2)Use the value of K to solve for the concentration of methane.
(3) Calculate the number of moles of methane from M and V.
Equilibrium Expressions Involving
Pressures
For a reaction of the type
aA+bB = cC+dD
It is sometimes convenient to write the equilibrium
expression in terms of partial pressures, e.g.
PC ) ( PD )
(
KP =
a
b
( PA ) ( PB )
c
d
= K ( RT )
P indicates the partial pressures of the species in equilibrium
and KP is a constant called the equilibrium constant in
terms of partial pressures. KP depends only on T , and
not on pressure.
Δn
Equilibrium involving pure solids or liquids:
set their activity = 1
Example: NH4NO2(s) = N2(g) + 2 H2O(g)
The equilibrium constant for this reaction would normally
be expressed as:
2
N 2 ][ H 2 O ]
[
K'=
[ NH 4 NO2 ]
However, a pure solid or liquid retains the same activity
during the reaction. Thus we set the activity of NH 4 NO 2 (s)
to one.
K = [ N 2 ][ H 2 O ]
2
1.0000
and
( )( P )
K p = PN2
H2O
1.0000
2
The Reaction Quotient, Q
Consider the reaction: a A(g ) + b B(g ) = c C( g ) + dm D( g )
The reaction quotient, Q is defined as
C ]t [ D ]t
[
Q=
a
b
[ A]t [ B ]t
c
d
where the subscripts t indicate momentary concentrations
at some time t before (or after) equilibrium has established.
Q has same form as K, but the concentrations are
the actual rather than the equilibrium
concentrations
Example: Calculating Equilibrium
Pressures and Concentrations from K
and initial conditions
Consider the equilibrium: CO(g) + H2O(g) = CO2(g) + H2(g)
0.250 mol CO and 0.250 mol H2O are placed in a 125 mL flask
at 900 K. What is the composition of the equilibrium mixture if
K = 1.56?
The original reactant concentrations are:
[CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M
Q = 0. Therefore, Q < K, so reactants are consumed and
CO(g) + H2O(g) = CO2(g) + H2(g)
Conc.
(M)
CO(g)
H2O(g)
CO2(g)
H2(g)
Init.
2.00
2.00
0
0
Change -x
-x
+x
+x
Equil.
2.00 - x
x
x
2.00 - x
```