AP/ECE Chemistry
Chapter 12 Gaseous Equilibrium Notes
Section 12.1 & 12.2- Gaseous Equilibrium Systems & The Equilibrium Constant Expression
Chemical Equilibrium is defined as a state in which the concentrations of reactants and products in a balanced chemical equation undergo NO NET CHANGE over time
Trial 1: Consider the equation that describes the reactions that occur when 0.200 atm hydrogen iodide is sealed in a reaction vessel at 520 oC.
2HI (g) ↔ H2 (g) + I2 (g)
Initial (P atm) 0.200 0.00 0.00
Change (P atm) -0.040 +0.02 +0.02
Equilibrium (P atm) 0.160 0.02 0.02
The forward reaction proceeds at the same rate as the reverse reaction when equilibrium is reached and notice that ΔP hydrogen = ΔP iodine = - ½ ΔP hydrogen iodide. The changes in amounts follow the coefficient ratios.
Trial 2: To further study this reaction, 0.100 atm of each gas was seated in a reaction vessel at 520oC.
2HI (g) ↔ H2 (g) + I2 (g)
Initial (P atm) 0.100 0.100 0.100
Change (P atm) +0.140 -0.070 -0.070
Equilibrium (P atm) 0.240 0.030 0.030
Again, the changes in amounts follow the coefficient ratios (ΔP hydrogen = ΔP iodine = - ½ ΔP hydrogen iodide).
Many other experiments as such were conducted and for each reaction at equilibrium, it was found that the product (multiplied) of the partial pressures of the products raised to their coefficients, divided by the product of the partial pressures of the reactants raised to their coefficients gave a constant value called the equilibrium constant (K). In this specific case the K was Kp for the equilibrium constant associated with the partial pressures of gases.
KP = (Phydrogen)1 x(Piodine)1
(Phydrogen iodide)2
Take the equilibrium conditions established in trial 1 and 2 above. Use the Kp expression and the known equilibrium concentrations of the reactants and products to calculate the value of Kp.
Did you get a unitless value of 0.016?
Much in the same way that chemical equations could become thermochemical equations by including the change in enthalpy (ΔH), equilibrium equations can be improved by the inclusion of the equilibrium constant value:
2HI (g) ↔ H2 (g) + I2 (g) KP = 0.016
Large values for KP indicate product favored equilibrium reactions while small values of KP indicate reactant favored equilibrium reactions.
Equilibrium Law:
K = Product of the partial pressures of products/Product of the partial pressures of reactants
where K is the equilibrium constant at a particular temperature.
A more general written expression for the equilibrium constant for gases with a given partial pressure is:
a A (g) + b B (g) c C (g) + d D (g) KP = (PC)c x (PD)dProducts
(PA)a x (PB)bReactants
Writing and Interpreting Equilibrium Constant Expressions (Kc)
Consider some reaction with stoichiometric coefficients as follows:
A + 2B 2C + D
The reaction will proceed until there is no longer any net reactants or products being formed. This tells us nothing about the rate of the reaction (kinetics or how fast a reaction proceeds).
Back and forth reactions continue to occur at the microscopic level but macroscopically the system is in equilibrium.
Equilibrium Law:
K = Product of the concentration of products/Product of the concentrations of reactants
where K is the equilibrium constant at a particular temperature.
For concentrations expressed in molarity (M) units, the symbol is Kc
Given aA + bB cC + dD then (small case letters being the coefficients of the balanced equation):
Kc = [C]c [D]d
[A]a [B]b
Values in brackets represent the concentration of that particular substance expressed in molarity units.
Note: Substances with constant concentrations (i.e. solids, pure liquids and solvents when solute < 1M) are not included in the formula since their concentrations change very little.
Example with aqueous species (dissolved ions or molecules in the solution): the solutes appear as molar (M) concentrations. This type of equilibrium expression is denoted as KC (equilibrium constant for concentration). Imagine that Lead (II) Nitrate and Sodium Iodide solutions were mixed. A net ionic equation to describe this reaction is written below. The sodium and nitrate ions act as spectators.
Pb2+ (aq) + 2 I- (aq) PbI2 (s)
This is written as an equilibrium expression because at some point an equilibrium will be achieved where for every formula unit of PbI2 (s) that is dissociated, a different pairing of Pb2+ (aq) and 2 I-(aq) will create a new formula unit of PbI2 (s). Thus, a dynamic equilibrium is established. The brackets represent molar concentration.
KC = 1 _
[Pb2+] [I-]2
Heterogeneous Equilibria: when more than one state of matter is present
Example with solids or liquids: you can ignore the solids or liquids as adding or removing these solids or liquids* does not affect the equilibrium position because it does not impact the partial pressures of the gases (* as long as some amount of solid or liquid is present)
Pure liquids or solids do not appear in equilibrium constant expressions for KP
CaCO3 (s) CaO (s) + CO2 (g) KP = (PCO2)1 x CaO The CaO and CaCO3 do not appear because CaCO3 they are solid.
Example:
Write the equilibrium expressions for the following reversible reactions.
a)4NH3(g) +5O2(g) 4NO(g) + 6H2O(g)
KP = (P NO)4 x (P H2O)6
(P NH3)4 x (P O2)5
b)BaO(s) + CO2(g) BaCO3(s)
KP = ___1__
(P CO2)
Modifying the equilibrium constant- Math Relationships between equilibrium constants:
1) Reversing the reaction- take the reciprocalK’ = 1/K = K-1
Reactions that are spontaneous in one direction (K>1) are nonspontaneous in the reverse direction.
2HI (g) ↔ H2 (g) + I2 (g) KP = 0.016 therefore,
H2 (g) + I2 (g) ↔ 2HI (g) KP’ = 1/ 0.016 = 62
2) Multiplying coefficients in a reaction equation by some number, z- raise K to that power.
K’ = Kz
This is useful for finding K values for unknown reactions using known reaction values, as we shall see.
2HI (g) ↔ H2 (g) + I2 (g) KP = 0.016 therefore,
HI (g) ↔ ½ H2 (g) + ½ I2 (g) (where z= ½) KP’ = (0.016)1/2 = 0.13
3) Adding chemical reactions together
Koverall = K1 x K2 x …
Ex.aA + bB cC K1
cC + dD bB + fFK2
aA + dD fFKoverall = K1 x K2
Doesn’t this remind you of Hess’s Law?
Example:
Given the two reactions below, what is the equation for the addition of these two reactions, and what is the final Kc value?
SO2(g) + ½ O2(g) SO3(g)K1 = 20.
NO2(g) NO(g) + ½ O2(g)K2 = 0.012
NO2(g) + SO2(g) NO(g) + SO3(g)K3= K 1 x K2 = 20 x 0.012 = 0.24
Example 2:
Determine KP of a combination reaction when given KP values for related reactions:
Given reaction 1: C (s) + CO2 (g) 2 CO (g) K1 = 2.4 x 10-9
reaction 2: COCl2 (g) CO (g) + Cl2K2 = 8.8 x 10-13
Find K3= ? for C(s) + CO2 (g) + 2 Cl2 (g) 2 COCl2 (g)
Reverse and double reaction 2, which means take the reciprocal of the square of K2
K3 = K1 x 1/ K22 = 3.1 x 1015
Section 12.1 & 12.3 Determination of K
Establishing Equilibrium Over Time
If the partial pressures of the hypothetical equation are monitored over time, we can use logic to determine the concentration of a given component at a given time.
3 A (g) + B (g) 2 C (g)
For every 3 units of A that are used, 1 unit of B will be used and 2 units of C will be produced.
Time (min) / 0 / 1 / 2 / 3 / 4 / 5 / 6PA / 1.00 / 0.778 / 0.580 / 0.415 / 0.355 / 0.325 / 0.325
PB / 0.400 / 0.326 / 0.260 / 0.205 / 0.185 / 0.175 / 0.175
PC / 0.000 / 0.148 / 0.280 / 0.390 / 0.430 / 0.450 / 0.450
To help organize yourself, use an ICF (or ICE) table. The ICF stands for initial, change and final (ICE-initial, change and equilibrium). Use an ICF table when you aren’t certain the conditions have reached equilibrium and an ICE table when you are certain they have.
For example from time 0 to time 1: t 0 min 1 min
3 A (g) + B (g) 2 C (g)
Initial (P atm) 1.00 0.400 0.000
Change (P atm) -3 x -x +2x
Final (P atm) 0.778 0.326 0.148
1.0– 3 x = 0.778 therefore x = 0.074 and the final pressure of B drops to 0.326, and the final pressure of C increases to 0.148.
Another example, from time 0 to time 2: t 0 min 2 min
3 A (g) + B (g) 2 C (g)
Initial (P atm) 1.00 0.400 0.000
Change (P atm) -3 x -x +2x
Final (P atm) 0.580 0.260 0.280
0.400– x = 0.260 therefore x = 0.140 and the final pressure of A drops to 0.580, and the final pressure of C increases to 0.280.
An additional example, from time 0 to time 3: t 0 min 3 min
3 A (g) + B (g) 2 C (g)
Initial (P atm) 1.00 0.400 0.000
Change (P atm) -3 x -x +2x
Final (P atm) 0.415 0.205 0.390
0.00 + 2 x = 0.390 therefore x = 0.195 and the final pressure of A drops to 0.415, while the final pressure of B drops to 0.205.
Given the information in the above table determine x for t 0 min 4 min, t 0 min 5 min and t 0 min 6 min. Then, use these x values to fill in the rest of the table above.
x
t 0 min 4 min 0.215
t 0 min 5 min 0.225
t 0 min 6 min. 0.225
How can you tell when the system is at equilibrium?
There should be no net change in concentration of reactants and products. Does this happen?
What is KP for this hypothetical system?
3 A (g) + B (g) 2 C (g)
When you are certain the system has achieved equilibrium, you can write the KP expression and plug in the values of the partial pressures of the gases.
KP = (P C)2 = (0.450)2 = 33.7
(P A)3 x (P B)1 (0.325)3(0.175)
If the pressures of these gases are graphed, it appears like below:
3 A (g) + B (g) 2 C (g)
Notice that C increases 2 units while A decreases3 units and
B decreases 1 unit. Over time, dynamicequilibrium is established
asthe concentrations remain constant.
Again if K is large the reaction is product favored and a general graph of concentration versus time would appear like that depicted below:
Again if K is small the reaction is reactant favored and a general graph of concentration versus time would appear like that depicted below:
Example:
Determine KP from initial pressures and one given value of equilibrium pressure at a given temperature of 1000 K.
2 SO2 (g) + O2 (g) 2 SO3 (g)K= ?
Initial P (atm):1.00 1.000.00
Change P (atm):-0.925 -0.463+0.925
Equilibrium P (atm):0.075 0.5370.925
Use an ICE box here because you are told the final value is the pressure at equilibrium.
1.00 – 2x = 0.075 therefore -0.925 = -2x and x = 0.463
KP = (P SO3)2 _ = (0.925)2 _ = 2.83 x 10 2
(P O2) (P SO2)2 (0.537)(0.075)2
Example 2:
2.0 moles of I2 and 4.0 moles of Br2 are placed in a 2.0L reactor at 150oC, and reaction occurs until equilibrium is reached:
I2 (g) + Br2 (g) 2IBr (g)
Analysis then shows that the reactor contains 3.2 moles of IBr. What is the value of the equilibrium constant Kc for the reaction?
I2 (g) + Br2 (g) 2IBr (g) Kc= ?
Initial [ ] (M):1.00 2.000.00Kc = [IBr]2_= [1.60]2 = 10.7
Change [ ] (M):-0.80 -0.80+1.60[I2] x [Br2] [0.20] X [1.20]
Equilibrium [ ] (M):0.20 1.201.60
In general:
-A very large equilibrium constant is one where the value of Kc is at least 100 times greater than the initial concentrations.
-The value of Kc is considered to be small when the equilibrium constant is less than one one-hundredth (<0.01) of the initial concentrations.
-If Kc is neither very large nor very small, assumptions cannot be used and analytical solutions are then sought.
Section 12.4- Application of Equilibrium Constant
Reaction progress: Comparing the Values of Q and K
The Reaction Quotient, Q
The reaction quotient identifies the current state of the concentrations of a reaction. Q can have any value, while K has a fixed and therefore, constant value at a given temperature. When the system is in equilibrium the Q = K. The equation for Q is set up in exactly the same manner as that for K.
If Q<K then reactant product If Q>K then product reactant If Q=K the equilibrium is established
In order to solve an equilibrium expression, you must determine:
1) The initial conditions of the system (Pressure in atm or concentration in molarity),
2) what pressure changes will take place and
3) the pressure of the system at equilibrium.
Often it is easiest to determine these values with an equilibrium table.
Solving equilibrium problems given initial concentration and the value of K.
If solving for unknown equilibrium concentrations
1. Set up K expression
2. write reaction table
3. substitute values in expression
4. solve for x
Sometimes the quadratic formula will need to be used. Simplify perfect squares if possible
If reaction direction is not obvious, first compare Q with K
If K is small and the reactant concentration is large, then x may be ignored (5% change in concentration as a rule of thumb.)
Ex. At 800 K, Kc = 62.5 for the equilibrium
H2(g) + I2(g) 2HI(g)
1.0mole of H2 and 2.00 moles of I2 are introduced into an empty 0.50 L reaction vessel. Find the equilibrium concentrations of all components at 800K
Kc = [HI]2 =62.5
[H2][I2]
Concentration / H2 / I2 / 2HIInitial / 2.0M / 4.0M / 0
Change / -x / -x / 2x
Equilibrium / 2.0-x / 4.0-x / 2x
Kc = (2x)2 /((2.0-x)(4.0-x)) = 62.5
Rearranging and solving the quadratic formula gives 1.9 and 4.2 mol/L of which only 1.9 mol/L is a plausible answer. This makes [H2]= 0.1 M, [I2]= 2.1 M, [HI]= 3.8 M.
Example:
At a certain temperature, the reaction
CO(g) + Cl2(g) COCl2(g) has an equilibrium Kc = 13.8. Is the following mixture in equilibrium and if not, in which direction will the reaction proceed? [CO]o = 2.5M, [Cl2]o = 1.2M, [COCl2]o = 5.0M
CO(g) + Cl2(g) COCl2(g) Kc= [COCl2] / [CO] [Cl2] = 13.8
Q= [COCl2] / [CO] [Cl2] = [5.0] / [2.5] [1.2] = 1.7
Since Q < K, the reaction should proceed forward.
Example Two:
For the system described by the following equilibrium equation PCl5 (g) PCl3 (g) + Cl2 (g) the KP = 26 at 300. oC. If the initial partial pressures are 0.01200 atm PCl5 , 0.9000 atm PCl3 and 0.4500 atm Cl2
a) is the reaction at equilibrium,
b) if not, which way will it shift,
c) if not what are the final concentrations of the reactant and products.
a) KP= (P PCl3) (P Cl2) = (0.9000 ) (0.4500) = 34 Since 34 26 this system is
(P PCl5) (0.01200) not at equilibrium.
b) Q > KP (34 > 26) therefore the reaction should proceed backwards.
c) PCl5 (g) PCl3 (g) + Cl2 (g) KP = 26
Initial P (atm): 0.01200 0.9000 0.4500
Change P (atm):+x -x -x
Equilibrium P (atm): ?? ?
Find x:
KP= (P PCl3) (P Cl2) = 26 = (0.9000- x )(0.4500 - x)
(P PCl5) (0.01200 + x)
0.312 + 26x = 0.405 – 1.35x + x2 therefore x2 – 27.35 x + 0.093
Use the quadratic formula x = -b (b2 – 4ac)
2a
a=1, b= -27.35 and c= 0.093 therefore x = 0.0034, 27.353 and only the x= 0.0034 is reasonable
PCl5 (g) PCl3 (g) + Cl2 (g) KP = 26
Initial P (atm): 0.01200 0.9000 0.4500
Change P (atm):+x -x -xx = 0.0034
Equilibrium P (atm):0.01540 0.89660.4466
Sometimes equilibrium expressions are solved by methods of successive approximations or basic assumptions about concentrations.
Equilibrium Constant for Gas-Phase Reactions Kp can be compared with Kc
From PV = nRTn/V = P/RT, at constant temperature where n/V = concentration
For the balanced chemical equation aA(g) + bB(g) cC(g)
Kp = PCc/(PAa PBb)
Relationship between Kp and Kc
Consider aA(g) bB(g) Kc = [B]b/[A]aKp = PBb/PAa
PA = nART/VPB = nBRT/V
Kp = (nBRT/V)b/(nART/V)a= ((nB/V)b/(nA/V)a) (RT)b-arecall n/V = molarity (M)
Kp = [B]b(RT)n(gas)/[A]a= Kc(RT)n(gas) where ng = Total moles of gas product – Total moles of gas reactant
Units for equilibrium constants
Normally considered to be dimensionless
Units for Kp = (atm)n Kc = (M)n = (mol/L)n
Other Equilibrium Constants:
KPequilibrium constant with Partial Pressures
KC equilibrium constant with molar concentrations
Kspsolubility product equilibrium constant (Chapter 16). Deals with how much of an insoluble compound or slightly soluble compound actually dissolves.
ABx (s) Ay+ (aq)+ x Bz- (aq) Ksp= [Ay+] [Bz-]x
With the common ion effect, the solubility of a compound can be decreased by the presence of an ion already in solution that is “common” to one of the ions in the compound under consideration. In this case, the initial concentrations of the dissolved ions must be added to the Ksp equation. Be mindful of simplifications where applicable, as these equations can quickly become complex as the order of the equation increases.
In truth, solutes are considered insoluble is a saturated solution has < 0.1 M concentration of the substance.
Ka acid dissociation constant for a weak acid (meaning it has not completely dissociated and therefore is not one of the strong acids) (Chapters 13 & 14)
HCH3O2 (aq) H + (aq) + CH3O2- (aq)
Kbbase dissociation constant for weak base (meaning it has not completely dissociated and therefore is not one of the strong bases) (Chapters 13 & 14)
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Ammoniaammonium
Kb = [NH4+] [OH-] Liquid water is omitted.
[NH3]
Kwionization constant for water (Chapters 13 & 14)
H2O (g) H+ (aq) + OH- (aq)
Kw= [H+ ] [OH-] = 1.0 x 10-14
Kf formation constant (for the formation of complex ions through complexation reactions;complex ions are metal atoms surrounded by other ions or compounds called ligands; complex ions are found in solution)
Cu2+ (aq) + 4 NH3 (aq) Cu(NH3)42+ (aq)Kf = [Cu(NH3)42+ ]
[Cu2+] [NH3]4
For the reverse reaction, the equilibrium expression is flipped and the expression is known as the dissociation constant, Kd Kf = 1/Kd
Keqany of the previous equilibrium constants (a general K)
Example Problem:
A Rocket blasts off using hydrogen and oxygen gas as fuel. If the reaction takes place at 3000 K and the equilibrium pressures are 1.0 atm H2O, 1.5 x 10-3 atm H2 , and 7.6 x 10-4 atm O2 , what is Kc ?
2 H2 (g) + O2 (g) 2 H2O (g)
KP = (P H2O)2 = (1.0 atm)2____ _ = 5.8 x 108
(P H2)2 (P O2) (1.5 x 10-3 atm)2 (7.6 x 10-4 atm)
The change in number of moles of gas is -1.
Kc = KP/ (RT)Δn = (5.8 x 108) / (0.0821 x 3000)-1 = 1.4 x 1011
Example Problem Two:
At 400oC, Kc = 64 for the equilibrium
H2(g) + I2(g) 2HI(g)
1.00 mol of H2 and 2.00 mol of I2 are introduced into an empty 0.50L reaction vessel. Find the equilibrium concentrations of all components at 400oC.
Kc = [HI]2=64
[H2][I2]
Concentration / H2 / I2 / 2HIInitial / 2.0M / 4.0M / 0
Change / -x / -x / 2x
Equilibrium / 2.0-x / 4.0-x / 2x
Kc = (2x)2 /((2.0-x)(4.0-x)) =64
Rearranging and solving the quadratic formula gives 1.9 and 4.5 mol/L of which only 1.9 mol/L is a plausible answer. Therefore, [H2] = 0.1 M, [ I2] = 2.1 M, [HI] = 3.8 M
Example Problem Three:
Nitrogen and oxygen can combine to form either NO(g) or N2O(g)
Equation 1: N2(g) + O2(g) 2NO(g)Kc1 = 4.1x10-31
Equation 2: N2(g) + ½ O2(g) N2O(g)Kc2 = 2.4x10-18
Also N2O(g) can react with oxygen to produce NO(g)
Equation 3: N2O(g) + ½ O2(g) 2NO(g)
What is Kc for this reaction? What is Kp (assume 500oC)?
Reverse Kc of equation 2, so take the reciprocal of that value and multiply it by Kc of equation 1.
KC3= Kc1 x 1/Kc2 = (4.1x10-31) x 1/(2.4x10-18)= 1.7 x 10-13
KP = Kc(RT)Δng = 1.7 x 10-13( 0.0821. 773.15)1/2 = 1.35 x 10-12
Example Problem Four:
At 850oC, the equilibrium constant Kp for the reaction
C(s) + CO2(g) 2CO(g)
has a value of 10.7. If the total pressure in the system at equilibrium is 1.000 atm, what is the partial pressure of carbon monoxide?
KP = (P CO)2= 10.7
(P CO2)
P CO + P CO2 = 1.000 atm therefore P CO2 = 1.000 atm – P CO
Substitute and solve for P COusing the quadratic equation
P CO 2 + 10.7 P CO – 10.7 gives P CO= 0.92.
Section 12.5- Effect of Changes on Conditions on an Equilibrium System
Stresses on a system: Le Châtelier’s Principle
Stresses on a system can come in the form of changes in temperature, concentration, and in the case of gases, pressure. Henri Louis Le Châtelier (1850-1936) in researching flames in hopes of preventing mine explosions, observed the following:
If a change in conditions (stress) is imposed on a system at equilibrium, the equilibrium position will shift in the direction that tends to reduce that change in conditions. (Systems can be homogeneous or heterogeneous)
As an example, if a particular reaction is exothermic and heat is added to the system, the system assumes that too much product has been formed and will force the reaction back towards reactants.
If reactants are added, this will ‘tip’ the see-saw up in the products direction, meaning more products will be produced after equilibrium is re-established. The same general logic applies for all the specific cases described below.
Effect of Changing Concentrations
Concentration Change / Observed EffectIncrease reactant / Favors products
Decrease reactant / Favors reactants
Increase product / Favors reactants
Decrease product / Favors products
Effect of Pressure on Gaseous Reactions
Value of ng / Increasing Pressure / Decreasing PressurePositive / Favors reactants / Favors products
Zero / No effect / No effect
Negative / Favors products / Favors reactants
ng = Total moles of product – Total moles of reactant
Effect of Temperature Changes
Temperature Change / Reaction Type / Effect On Reaction / Effect On KIncrease / Exothermic / Favors reactants / Decrease
Increase / Endothermic / Favors products / Increase
Decrease / Exothermic / Favors products / Increase
Decrease / Endothermic / Favors reactants / Decrease
* The only experimental variable that has any effect on the value of the equilibrium constant is temperature.
Examples:
- For the reaction at equilibrium
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)Ho = 128kJ
state the effects (increase, decrease, or no change) of the following stresses on the number of moles of sodium carbonate (Na2CO3) at equilibrium in a closed container. (Note that Na2CO3 is a solid; its concentration will remain constant, but the amount can change.)
- Removing CO2(g) INCREASE
- Adding H2O(g)DECREASE
- Raising the temperatureINCREASE
- Adding NaHCO3(s)INCREASE
- Some hydrogen and iodine are mixed at 526.85 oC in a 1.00 liter container, and when equilibrium is established the following concentrations are present: [HI] = 0.490M, [H2] = 0.250M, and [I2] = 0.015M. If an additional 0.300 moles of HI is added, what concentrations will be present when the new equilibrium is established?
Use the given equilibrium values to calculate KcKc = [HI]2 = [0.490]2=64 almost equals 62.5
[H2][I2] [0.250][0.015] (KC value from earlier)
Concentration / H2 / I2 / 2HIInitial / 0.250M / 0.015M / 0.790
Change / +x / +x / -2x
Equilibrium / 0.250+x / 0.015+x / 0.790- 2x
58.5 x2+19.72 x – 0.389725 and then x = -0.36 , 0.019 so [H2] = 0.269 M , [I2] = 0.034 M ,[HI]= 0.752 M
- At 25oC the equilibrium constant, Kc, for the following reaction is 4.66x10-3. If 0.800 moles of N2O4 is injected into a closed 1.00 L glass container at 25oC, what will the equilibrium concentrations of the two gases be? What will the concentrations be at equilibrium if the volume is suddenly halved at constant temperature? N2O4(g) 2NO2(g)
Kc = [NO2]2 = 4.66 x 10-3
[N2O4]
Concentration / N2O4 / NO2Initial / 0.800M / 0.00M
Change / -x / +2x
Equilibrium / 0.800-x / 2x
Kc = [2x]2 = 4.66 x 10-3 4x2+ 0.00466 x – 0.00368 x = -0.031, 0.030
[0.800 -x]
[NO2] = 0.060 M [N2O4] = 0.77 M
After the volume is halved:
Concentration / N2O4 / NO2Initial / 1.60M / 0.00M
Change / -x / +2x
Equilibrium / 1.60-x / 2x
Kc = [2x]2 = 4.66 x 10-3 4x2 + 0.00466 x – 0.00736 x = -0.04, 0.04
[1.60 -x]
[NO2] = 1.560 M [N2O4] = 0.08 M
1