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A. REVERSIBLE REACTIONS
REVERSIBLE REACTION = reactions that proceed in the forward (R  P) and
reverse (P  R) direction SIMULTANEOUSLY.
REACTION EQUILIBRIUM
1. In most spontaneous reactions the formation of products is greatly favoured over the
reactants and the reaction proceeds to completion (one direction). In some reactions
the product is only slightly favoured overthe reactants and can proceed in both
directions.
Reversible reactions involve two competing reactions that occur at the same time:
 FORWARD REACTION – the conversion of reactants to products
 REVERSE REACTION – the conversion of products to reactants
2. Consider the following reversible reaction:
2SO2(g) + O2(g) 
 SO3(g)
 The DOUBLE ARROW ( 
 ) indicates that the reaction is reversible.
 When a reversible reaction begins, only the forward reaction occurs (if starting
from the reactants).
 As the reaction proceeds the forward reaction slows and the reverse reaction
speeds up.
 Eventually the system reaches equilibrium.
ver. 10/15/2010

2 | Chemistry12
A reversible reaction is said to be at EQUILIBRIUM when the rate of the forward
reaction EQUALS the rate of the reverse reaction.
Rate(forward) = Rate(reverse)
3. A reversible reaction will achieve equilibrium regardless of the direction from whichit
is approached.
 These graphs show the variation of reactant and product concentration with time.
 At equilibrium, the concentrations of reactant and product donot change and the
relative amounts of reactants and products are the same regardless of whether the
reaction began from the reactants or products.

Unit 2 ReactionEquilibrium | 3
4. When a reaction reaches equilibrium all MACROSCOPIC properties (measurable) are
constant.
 These properties include colour intensity, concentration, and pressure.
 Note that when a system reaches equilibrium, the rates of the forward and reverse
reactions are equal but the concentrations of reactants and products are not.
5. Systems at equilibrium are characterized by the following:
a) the system is closed, no material enters or leaves;
CaCO3(s) 
 CaO(s) + CO2(g)
b) opposite reactions occur at the same rate;
c) equilibrium was reached by starting with either reactants or products;
d) the temperature is constant;
e) no change in macroscopic properties.
Chemical equilibrium are said to be:
DYNAMIC = forward and reverse reactions continue to occur although
there is no net change in MACROSCOPIC properties.

4 | Chemistry12
EXOTHERMIC reactions (forward or reverse) are favoured because of the
tendency to move towards MINIMUM ENTHALPY (ΔH).
Reactions that produce the greatest amount of randomness are favoured because of
the tendency to move towards MAXIMUM ENTROPY (∆S).
1. A SPONTANEOUS reaction is a reaction that will occur by itself without outside
assistance. Spontaneous reactions can occur when the activation energy barrierislow.
 Reactions tend to favour the side of the reaction having lower energy
(ENTHALPY, ∆H).
 EXOTHERMIC reactions result in a decrease in energy, products arefavoured.
 ENDOTHERMIC reactions result in an increase in energy, reactants arefavoured.
2. From energy stand point, exothermic reactions are favoured; however, some
endothermic reactions, such as chemical ice packs, will occur spontaneously.
 Tendency for reactions to increase disorder or randomness (ENTROPY, ΔS).
 When entropy increases in the forward direction, products are favoured.
 When entropy decreases in the forward direction, reactants are favoured.
 Entropy can be predicted by examining the phases of the reactants and products:
GASES (g) >> SOLUTIONS (aq) > LIQUIDS (l) >> SOLIDS (s)
B. PREDICTING SPONTANEOUS REACTIONS

There are two “drives” or “tendencies” for reactions:
a) tendency for a reaction to increase RANDOMNESS (MAXIMUM ENTROPY, ΔH).
b) tendency for a reaction to lower ENERGY (MINIMUM ENTHALPY, ΔS).
 In general, highly random states are more probable than highly ordered states.
Endothermic reactions can be spontaneous when the difference in randomness
between reactants and products is so great that it overcomes the tendency towards
minimum enthalpy — these reactions are said to be “DRIVEN” by the entropy of the
system.
EXAMPLE 2.1 – PREDICTING SPONTANEOUS REACTIONS
Based on changes in enthalpy and entropy, predict whether each of the following reactions will
be spontaneous, non–spontaneous, or reach equilibrium.
a
) Zn(s) + 2HCl(aq) 
 ZnCl2(aq) + H2(g) ∆H = –152 kJ
b) 3C(s) + 3H2(g) 
 C3H6(g) ∆H = +20.4 kJ
c) 2Pb(NO3)2(s) + 597 kJ 
 2PbO(s) + 4NO2(g) + O2(g)
a) Enthalpy decreases (exothermic)  favours products
Entropy increases (gas on product side)  favours products
SPONTANEOUS
b) Enthalpy increases (endothermic)  favours reactants
Entropy decreases (3 moles gas vs. 1 mole gas)  favours reactants
NON–SPONTANEOUS
c) Enthalpy increases (endothermic)  favours reactants
Entropy increases (5 moles gas on product side)  favoursproducts
EQUILIBRIUM

6 | Chemistry12
SAMPLE 2.1 – PREDICTING SPONTANEOUS REACTIONS
Based on changes in enthalpy and entropy, predict whether each of the following reactions will
be spontaneous, non–spontaneous, or reach equilibrium.
a) 6CO2(g) + 6H2O(l) + ENERGY  C6H10O6(s)+6O2(g)
b) CH4(g) + H2O(g) + 49.3 kJ  CO(g) +3H2(g)
c) N2(g) + 3H2(g)  2NH3(g) + 92.4kJ
a) Enthalpy increases (endothermic)  Reactants
Entropy decreases (6 moles liquid form 1 mole solid) Reactants
Non-spontaneous
b) Enthalpy increases (endothermic)  Reactants
Entropy increase (2 moles gas form 4 moles gas) Products
Equilibrium
c) Enthalpy decreases (exothermic)  Products
Entropy decreases (4 moles gas form 2 moles gas) Reactants
Equilibrium

Factors that upset an equilibrium system are referred to as STRESSES
LE CHATELIER'S PRINCIPLE
when a stress is applied to a system at equilibrium, the systemreadjusts to relieve or
offset the stress and the system reaches a new state of equilibrium.
1. Changes in factors such as temperature, pressure, concentration, and catalyst can upset
the balance in rates of the forward and reverse reactions of a systemin equilibrium.
 Stresses cause changes to the reactant and product concentrations.
 Net increase in [product] is called a “shift to theright”.
 Net increase in [reactant] is called a “shift to the left”.
 Henry Louis Le Châtelier (1850–1936), studied the effects of changing conditions on
equilibrium systems.
 In other words, whatever we do to an equilibrium, the equilibrium will try toundo.
2. Consider the following reaction:
2NO(g) + Cl2(g) 
 2NOCl(g) + 76 kJ
The effects of various stresses on a systemat equilibrium can be summarized as follows:
a) TEMPERATURE
 If the temperature of the equilibrium is decreased, Le Chatelier’s Principle
predicts that the equilibrium shifts to the right.
2NO(g) + Cl (g)  2NOCl(g) +
2 

 Exothermic reaction shifts” to the right to produce more heat.
 In terms of reaction rate, the reverse rate initially experiences a greater
decrease because heat is a “reactant” in the reverse reaction.
 As equilibrium is re-established, the reverse rate increases and the forward
rate decreases. The net change however, is that both forward and reverse
rates will decrease.
C. LE CHATELIER’S PRINCIPLE
76 kJ 

8 | Chemistry12
b) CONCENTRATION
 If the concentration of Cl2 increases, Le Chatelier’s Principle predicts that the
equilibrium will shift to the right to use up the added Cl2.
2NO(g) + 
 2NOCl(g)+ 76 kJ
 In terms of reaction rate, Cl2 is a reactant in the forward reaction. Increase
[Cl2] increases forward reaction rate initially.
 As equilibrium is re-established, forward rate decreases and reverse rate
increases. Overall, both rates will increase.
Cl2(g)

c) PRESSURE
 Increasing the partial pressure of gas has the same effect as increasing its
concentration.
 Pressure can also be increased by decreasing the volume of thecontainer.
 A decrease in volume simultaneously increases the partial pressure and
concentration of ALL gases present in the system.
 Shift to reduce the overall pressure — this results in a shift towards the side
of the reaction with the fewest moles of gas present.
2NO(g) + Cl (g) 2NOCl(g) + 76 kJ
2 

 The direction that involves the greater number of gas will experience a
greater increase in rate initially. For this reaction, the forward reaction rate
would increase initially. As equilibrium is re-established, forward rate
decreases and reverse rate increases.
 When the number of MOLES OF GAS are equal on both sides of the
equilibrium, no shift is observed.
d) CATALYST
 A catalyst lowers the activation energy for a reaction, however it decreases
the activation energy of both the forward and reverse reactions and speeds
up the forward and reverse rate by an equivalent amount.
 Adding a catalyst to a reaction already at equilibrium will increase the rates
of both forward and reverse reactions but will have no effect on reactant
and product concentrations.
 Adding a catalyst to a reaction that is not at equilibrium will allow it to be
reached faster.

10 | Chemistry12
SAMPLE 2.2 – USING LE CHATELIER’S PRINCIPLE TO PREDICT SHIFTS IN EQUILIBRIUM
Consider the following reaction:
N2(g) + 2H2(g) 
 2NH3(g) + 92 kJ
Predict the direction of shift and the effect on the amount of H2(g) resulting from the following
stresses:
a) increase [N2]
b) increase [NH3]
c) increase temperature
d) increase volume
a) right, H2 decreases
b) left, H2 increases
c) left, H2 increases
d) right, H2 increases

1. When a systemis at equilibrium, the [reactants] and [products] remain constant.
Experimentally, it is found that the ratio of [product] to [reactant] is constant at a
particular temperature.
 Even though the [reactants] and [products] may change as result of a shift in
equilibrium, the ratio remains constant.
 For an EQUILIBRIUM EQUATION with the general form:
aA + bB 
 eE + fF
experimentally, it is found that:
[E]e[F]f
Keq =
[A]a
[B]b = a constant
 This expression of concentrations is called the EQUILIBRIUM EXPRESSION and its
numerical value, Keq, is called the EQUILIBRIUM CONSTANT.
2. The equilibrium constant is the ratio of product concentration terms to reactant
concentration terms.
[PRODUCTS]
Keq = [REACTANTS]
 In the Keq expression, the exponent to which each of the concentrations is raised is
equal to its coefficient in the balanced equation.
 The units for Keq vary depending on the number of concentration terms in the
numerator as compared to the denominator.
 These units do not have any particular importance so units for Keq are generally not
shown.
D. EQUILIBRIUM EXPRESSIONS AND EQUILIBRIUM CONSTANT

12 | Chemistry12
EXAMPLE 2.2 – WRITING EQUILIBRIUM EXPRESSIONS
Write the Keq expression for the following equilibrium:
2HI(g) 
 H2(g) + I2(g)
The equilibrium expression is:
[H2][I2]
Keq =
[HI]2
EXAMPLE 2.3 – CALCULATING KEQ FROM EQUILIBRIUM CONCENTRATION
Consider the following equilibrium
2HI(g) 
 H2(g) + I2(g)
What is the value of Keq if the equilibrium concentrations of HI, H2, and I2 are 0.250 M, 0.120 M,
and 0.120 M respectively
[H2][I2]
Keq =
[HI]2
Substituting into Keq expression:
(0.120)(0.120)
Keq =
(0.250)2
Keq = 0.230

EXAMPLE 2.4 – CALCULATING EQUILIBRIUM CONCENTRATION GIVEN KEQ
Consider the following equilibrium
2HI(g) 
 H2(g) + I2(g)
If the value of Keq = 0.230 at a particular temperature, what is the equilibrium [HI] if the [H2] =
0.075 M and [I2] = 0.320 M?
[H2][I2]
Keq =
[HI]2
Substituting into Keq expression:
(0.075)(0.320)
0.230 =
[HI]2
(0.075)(0.320)
[HI] =
(0.230)
[HI] = 0.323 M
3. The following substances are not included in the equilibrium expressionbecause their
concentrations are essentially constant:
a) SOLIDS cannot be compressed appreciably and hence their concentrations cannot
be changed. The concentration of a solid is determined by the density of thesolid.
b) PURE LIQUIDS cannot be compressed appreciably and hence their concentrations
cannot be changed. However, if there are two or more liquids present they may mix
and are no longer pure and the concentrations may be change due to dilution.
Liquids are only pure if there is only one liquid present in the equilibrium equation.

14 | Chemistry12
Keq > 1 then products GREATER THAN reactants
Keq < 1 then products LESS THAN reactants
Keq = 1 products EQUAL reactants
SAMPLE EXERCISES 2.3 – WRITING EQUILIBRIUM CONSTANT EXPRESSIONS
Write the Keq expression for the following equilibrium:
a) N2(g) + 3H2(g) 
 2NH3(g)
b) NH3(g) + 5O2(g) 
 4NO(g) + 6H2O(g)
c
) Si(s) + O2(g) 
 SiO2(s)
d
) CH3COCH3(l) + Cl2(g) 
 CH3COCH2Cl(l)+ HCl(g)
4. The equilibrium constant, Keq, is a ratio of
[PRODUCTS]
[REACTANTS] , so numerically, if Keq is large
the products are favoured and if Keq is small the reactants are favoured.
Large Keq =
[PRODUCTS]
[REACTANTS]
[PRODUCTS]
Small Keq =
[REACTANTS]
SAMPLE EXERCISES 2.4 – RELATING EXTENT OF REACTION TO THE VALUE OF THE
EQUILIBRIUM CONSTANT, KEQ
Predict the relative amount of reactants and products at equilibrium.
a) H2(g) + I2(g) 
 2HI(g) Keq = 6.3 × 10–11
b) 4NH3(g) + 5O2(g) 
 4NO(g) + 6H2O(g) Keq = 5 × 103
c) Si(s)+ O2(g) 
 SiO2(s) Keq = 2 × 10–9

Varying temperature is the only factor that changes the value of Keq
Keq increases when the temperature of an endothermic reaction isincreased
Keq decreases when the temperature of an exothermic reaction is increased
1. The value of Keq remains constant even when the concentration, pressure, or surface
area is changed; however, changing temperature affects the value of Keq.
 Changes in concentration, pressure, or surface area result in change in the number
of reacting molecules per liter. The equilibrium counteracts these changes (stresses)
by shifting to re-distribute the change and re-establish a newequilibrium.
The ratio of [PRODUCT]/[REACTANT] and the value of Keq remains the same.
 Changing temperature causes a shift in the equilibrium; however, the number of
reacting molecules does not change. As a result, the ratio of
[PRODUCT]/[REACTANT] changes and the value of Keq changes.
2. Consider the reaction:
REACTANTS 



PRODUCTS +  76 kJ
 For an exothermic reaction, an increase in temperature will cause a shift to the left.
The [PRODUCTS] decreases while the [REACTANTS] increases, and since
[PRODUCTS]
Keq = [REACTANTS] , the value of Keq decreases.
 For an endothermic reaction, an increase in temperature will cause a shift to the
right. The [PRODUCTS] increases while the [REACTANTS] decreases, and the value
of Keq would increase.
 76 kJ + REACTANTS

 PRODUCTS
 If the value of Keq is provided for two different temperatures at which the reaction
occurs, it is possible to predict whether a reaction is endothermic or exothermic.
E. DEPENDENCE OF KEQ ON TEMPERATURE

16 | Chemistry12
EXAMPLE 2.3 – PREDICTING ENERGY CHANGES FROM KEQ
Consider the following reaction:
PCl5(g) 
 PCl3(g) + Cl2(g)
Is the above reaction endothermic or exothermic if Keq is 2.24 at 227°C and 33.3 at 487°C?
The increase in Keq indicates that the equilibrium is shifted to the right by an
increase in temperature and hence the reaction is endothermic.

1. Determining the value of Keq requires information about the concentrations of the
reactants and products when the system is at equilibrium. Such information can only be
obtained by actually performing the experiment.
The following examples illustrate the various type of calculations which can be
performed based on the equilibrium expression and some experimental data.
EXAMPLE 2.4 – CALCULATING KEQ FROM EQUILIBRIUM CONCENTRATIONS
Consider the following equilibrium system:
N2(g) + 3H2(g) 
 2NH3(g)
At 200°C, the concentration of nitrogen, hydrogen, and ammonia at equilibrium are measured
and found to be [N2] = 2.12, [H2] = 1.75, and [NH3] = 84.3. Determine the value of Keq.
Determine the equilibrium expression for the reaction is:
[NH3]2
Keq =
[N2][H2]3
Substitute the equilibrium concentrations into the equilibrium expression:
(84.3)2 (7.11 × 103)
Keq =
(2.12)(1.75)3 =
(2.12)(5.36)
Keq = 626
*Note: there are no units are no units associated with Keq because
they depend on the equilibrium system and are of little or no significance.
F. EQUILIBRIUM CONSTANT CALCULATIONS

18 | Chemistry12
EXAMPLE 2.5 - CALCULATING CONCENTRATIONS FROM KEQ
The equilibrium concentrations of SO2 and O2 are each 0.0500M and Keq = 85.0 at 25°C for the
reaction
2SO2(g) + O2(g) 
 2SO3(g)
Calculate the equilibrium concentration for SO3 at this temperature.
The equilibrium expression for the reaction is
[SO3]2
Keq =
[SO2]2[O2]
Let x equal the concentration of SO3 at equilibrium.
Substituting into the value of Keq given and the equilibrium concentrations you have
(x)2
85.0 =
(0.0500)2(0.0500)
x2 = 0.0106
x = 0.103
therefore [SO3]eq is 0.103 M.

EXAMPLE 2.6 – CALCULATING KEQ FROM INITIAL CONCENTRATIONS
Suppose that 4.00 moles of HI(g) is placed into a 2.00 L flask at 425°C react to produce H2 and I2.
according to the equation:
2HI(g) 
 H2(g) + I2(g)
At equilibrium the concentrations of H2 and I2 are found to each be 0.214 mol/L. Calculate the
value of Keq.
First, determine the equilibrium expression:
Keq =
[H2][I2]
[HI]2
Next, calculate the concentrations of the reactants and products:
[HI(g)]initial =
4.00 mol
2.00 L = 2.00 mol/L
[H2(g)]equil’m = [I2(g)]equil’m = 0.214 mol/L
Initial concentration is involved an “Initial – Change – Equilibrium” (ICE) table must be used.
Variable “x” to represent the change in concentration of H2 and setup an ICE table as follows:
2HI(g) 
 H2(g) + I2(g)
initial 2.00 M 0.00 M 0.00 M
change – 2x + x + x
equilibrium 2.00 – 2x x x
At equilibrium,
[H2] = [I2] = x = 0.214 mol/L
[HI] = 2.00 – 2x = 2.00 – 2(0.214) = 1.57 mol/L

20 | Chemistry12
Finally, use the equilibrium concentrations to calculate the value of Keq:
[H2] = 0.214 mol/L; [I2] = 0.214mol/L; [HI] = 1.57mol/L
Keq =
(0.214)(0.214)
(1.57)
2 = 1.86 × 10−2

EXAMPLE 2.7 – CALCULATING INITIAL CONCENTRATIONS FROM KEQ
A certain amount of NO2(g) was placed into a 5.00 L bulb and reacted according to the
equation: 2NO(g) + O2(g) 
 2NO2(g). When equilibrium was reached, the concentration of
NO(g) was 0.800 M. If Keq has a value of 24.0, how many moles of NO2 were originally placed
into the bulb.
Since an initial concentration is involved we need to setup an ICE table.
[NO]equil’m = 0.800 M and [NO2]initial = y
2NO(g) + O2(g) 
 2NO2(g)
initial 0.00 M 0.00 M y
change +2x +x –2x
equilibrium 2x x y – 2x
At equilibrium,
[NO] = 2x = 0.800 M [O2] = × =
0.800 M
2 = 0.400 M
[NO2] = y – 2x = y – 0.800 M
The equilibrium expression for the reaction is
[NO2]2
Keq =
[NO]2[O2]
Substituting into the equilibrium expression gives:
(y - 0.800)2
24.0 =
(0.800)2(0.400)
24.0 =
(y - 0.800)2
0.256
(y – 0.800)2 = 6.144
y – 0.800 = 2.479

22 | Chemistry12
y = 3.279 M = [NO2]initial
The problem requires number of moles of NO2 which were originally placed into the 5.00 L bulb,
so moles of NO2 = 3.279 mol/L × 5.00 L = 16.4 mol

EXAMPLE 2.8 – DETERMINING [EQUILIBRIUM] FROM [INITIAL] KEQ
Keq = 3.5 for the reaction: SO2(g) + NO2(g) 
 SO3(g) + NO(g). If 4.0 mol of SO2(g), 4.0 mol of
NO2(g) are put into a 5.0 L bulb and allowed to come to equilibrium, what concentration of all
species will exist at equilibrium?
First, write out the equilibrium expression:
[SO3][NO]
Keq =
[SO2][NO2]
Next, setup an ICE table,
SO2(g) + NO2(g) 
 SO3(g) + NO(g)
initial 0.80 M 0.80 M 0.00 M 0.00 M
change – x – x + x + x
equilibrium 0.80 – x 0.80 – x x x
Since we don’t know any equilibrium concentrations we need to
use the variables and substitute them into the equilibrium expression:
(x) (x)
Keq =
(0.80 - x) (0.80 -x)
x2
3.5 =
(0.80 - x)2 (this is a perfect square)
x
1.87 =
(0.80 - x)
x = 0.52 M
[SO3] = [NO] = × = 0.52 M
[SO2] = [NO2] = 0.80 – × = 0.80 – 0.52 = 0.28 M

24 | Chemistry12
EXAMPLE 2.9 – DETERMINING [EQUILIBRIUM] AFTER A SHIFT
A 1.0 L reaction vessel contained 1.0 mol of SO2, 4.0 mol of NO2, 4.0 mol of SO3, and 4.0 mol of
NO at equilibrium according to the reaction:
SO2(g) + NO2(g) 
 SO3(g) + NO(g)
If 3.0 mol of SO2 is added to the mixture what will the new concentration of NO be when
equilibrium is re–established?
The unknown is the final [NO] when equilibrium is re–established.
All other concentrations can be easily calculated and the value of Keq is needed.
Keq is not given but we know the concentrations of each gas previously at
equilibrium, so Keq can be calculated.
[SO3][NO] (4.0) (4.0)
Keq =
[SO2][NO2]
=
(1.0) (4.0)
= 4.0
Since the temperature does not change, the value of Keq will remain the same.
Next, setup an ICE table,
SO2(g) + NO2(g) 
 SO3(g) + NO(g)
initial 1.0 + 3.0 4.0 4.0 4.0
change – x – x + x + x
equilibrium 4.0 – x 4.0 – x 4.0 + x 4.0 + x
Substituting into the equilibrium expression:
(4.0 + x)2
Keq = (4.0 - x)2 = 4.0
(4.0 + x)
2.0 =
(4.0 -x)
x = 1.33
[NO] = 4.0 + x = 4.0 + 1.33 = 5.3 M

If Q = Keq, then the system is at EQUILIBRIUM and no shift will occur
[PRODUCTS]
If Q < Keq, then [REACTANTS] is TOO SMALL and shift right, more PRODUCTS
[PRODUCTS]
If Q > Keq, then [REACTANTS] is TOO BIG and shift left, more REACTANTS
2. For some problems, a numerical answer is NOT required but rather a DECISION must
be made.
 For example, which way will the reaction shift in order to reach equilibrium or how
will the concentration of reactants and product change in order to reach
equilibrium.
 For these decision type problems a REACTION QUOTIENT (Q) or Trial Keq, is used.
 Consider the following equation:
2NO(g) + O2(g) 
 2NO2(g)
the reaction quotient is:
[NO2]2
Q =
[NO]2[O2]
 Notice that the reaction quotient is the same as the equilibrium expression except
that we will use initial concentrations to solve for Q instead of equilibrium
concentrations which are used for Keq.
 Our decision can be based upon comparing the reaction quotient, Q, to the
equilibrium constant, Keq.

26 | Chemistry12
EXAMPLE 2.10 – DETERMINING THE DIRECTION OF SHIFT, TRIAL KEQ
Keq = 49 for the equilibrium:
2NO(g) + O2(g) 
 2NO2(g).
If 2.0 mol of NO(g), 0.20 mol of O2(g), and 0.40 mol of NO2(g) are put into a 2.0 L bulb, which
way will the reaction shift in order to reach equilibrium? Support your answer with the
appropriate calculations.
First of all, this question is NOT asking for a numerical value. Asking for a DECISION.
We will base our decision on comparing the Reaction Quotient, Q, to the Keq.
[NO2]2
Q =
[NO]2
[O2]
(where Q = the REACTION QUOTIENT)
Calculate the initial concentrations:
2.0 mol 0.20 mol
[NO]initial = 2.0 L = 1.0 M [O2] initial= 2.0 L = 0.10 M
0.40 mol
[NO2]initial = 2.0 L = 0.20 M
Now substitute the initial concentrations into reaction quotient:
(0.20)2
Q =
(1.0)2(0.10)
= 0.40
Since Q < Keq, the reaction must shift right to produce more PRODUCTS.

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