X Chemistry Model Test Questions with Answers and Important Numericals with Solutions
Chapter # 1
Chemical Equilibrium
Short Questions
Q1.Write
an equilibrium equation of monoatomic carbon and a molecule of oxygen as
reactant and carbon monoxide as product.
Q2. State
law of mass action. What is active mass? How is the active mass is represented?
Q3. Define chemical
equilibrium state with examples.Why chemical equilibrium is dynamic?
Q4. Outline the
characteristics of reversible reaction.
Q5. Distinguish
between reversible and irreversible reactions.
Q6. Why
equilibrium constant may or may not have unit? Justify with example.
Q7. How
direction of a reaction can be predicted if Kc is known to you.
Q8. Write
equilibrium constant expression for the following equations:
N2(g) +
2O2(g)⇌ 2NO2(g)
2NO2(g) ⇌ N2O4(g)
PCl3(g) + Cl2(g) ⇌ PCl5(g)
2SO2(g) + O2(g) ⇌ 2SO3(g)
N2(g) + 3H2(g) ⇌ 2NH3(g)
H2(g) + Br2(g) ⇌ 2HBr(g)
H2(g) + I2(g) ⇌ 2HI(g)
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
Q9. Point out the
coefficients of each in the following hypothetical reactions:
2A + 3B ⇌ 4C + 2D
2M + 4N ⇌ 5O
4X ⇌ 2Y + 3Z
Q10. What do
you mean by the equilibrium constant, reaction quotient, extent of a reaction,
forward reaction, backward reaction?
Long Questions
Q1. Describe
dynamic equilibrium with two examples
Q2. State
law of mass action. Derive an expression for equilibrium constant for a general
reaction.
Q3. How can you predict the following stages of a reaction by comparing the values of Kc and Qc
(i) Net reaction proceeds in forward direction
(ii)Net reaction proceeds in reverse direction
Q4. Describe
the characteristics of equilibrium constant
Q5.Predict which system at equilibrium will contain maximum amount of product and which system will contain maximum amount of reactant?
(a)2CO2(g)⇌2CO(g)+ O2(g) Kc (927oC) = 3.1 x 1018
(b) 2O3(g) ⇌ 2O2(g) Kc (298 K) = 5.9 x 1055
Q6. Write
down the macroscopic characteristics of dynamic equilibrium.
Numericals
1. For
the reaction, H2(g) + I2(g)
⇌ 2HI(g) (Kc = 57.0); the
concentrations of H2(g), I2(g) and HI(g) at
time t are:
[H2]t = 0.10 mol dm−3,
[I2]t = 0.20 mol dm−3,
[HI]t = 0.40 mol dm−3
(Answer;
Qc (8.0) is less than Kc (57.0), reaction will move in
the forward direction)
2. Equilibrium occurs when nitrogen monoxide gas
reacts with oxygen gas to form nitrogen dioxide gas
2NO(g) + O2(g) ⇌ 2NO2(g)
At equilibrium at 230oC, the concentrations are measured to be
[NO] = 0.0542 mol dm−3,
[O2] = 0.127 mol dm−3
and
[NO2] = 15.5 mol dm−3
Calculate the equilibrium
constant at this temperature. (Book problem 1; page 8)
(Answer; 6.44 x 105 mol-1 dm3)
3. A reaction takes place between iron
and chloride ion as
Fe3+ + 4Cl− ⇌ FeCl4−
At equilibrium the concentrations are measured to be
Fe3+ = 0.2 mol dm−3,
Cl− = 0.28 mol dm−3
and
FeCl4−
= 0.95 x 10−4 mol dm−3
Calculate equilibrium constant Kc for given
reaction. (Book
problem 1; page 9)
(Answer;
)
4. Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperature. At 2000oC, the value of the equilibrium constant for the given reaction is 4.1 x 10−4
N2(g) + O2(g) ⇌ 2NO(g)
Find the concentration of NO in an equilibrium mixture
at 1 atmosphere at 2000oC. In air, [N2] = 0.036 mol/L and
[O2] = 0.0089 mol/L. (Book problem 3; page 10)
5. The
value of Kc for the reaction is 1 x10−4
2HI ⇌ H2
+ I2
At a given temperature, the molar
concentration for action mixture is HI = 2 x 10−5 mol dm−3,
H2 = 1 x 10−5 mol dm−3 and I2 = 1 x
10−5 mol dm−3. Predict the direction of the reaction. (Book problem 3; page 18)
1. Which one of the
following statements is false about dynamic equilibrium?
(a) Concentration of reactant and products are not changed
(b) It takes place in a close container
(c) Rate of forward reaction is equal to rate of reverse reaction
(d) Equilibrium cannot be disturbed by any external stress.
2.When the magnitude of Kc is small, indicates
(a) Reaction mixture contains most of the reactant
(b) Reaction mixture contains most of the product
(c) Reaction mixture contains almost equal amount of reactant and product
(d) Reaction goes to completion
3. Qc can be defined as
(a) ratio of product and reactant
(b) ratio of molar concentration of product and reactant at specific time
(c) ratio of molar concentration of product and molar volume of reactant
(d) ratio of molar concentration of product and reactant raised to the power of coefficient
4. Consider
the following reaction and indicates which of the following best describe
equilibrium constant expression Kc
4NH3(g)
+ 5O2(g) ⇌ 4NO(g)
+ 6H2O(g)
5. For which system
does the equilibrium constant, Kc has units of concertation?
(a) N2(g) + 3H2(g) ⇌ 2NH3(g)
(b) N2(g) + O2(g) ⇌ 2NO(g)
(c) H2(g) + I2(g) ⇌ 2HI(g)
(d) CO2(g) + H2(g) ⇌ CO(g) + H2O(g)
6.A reaction which is never goes to completion is known as reversible reaction. Reversible reaction is represented by
(a) Double straight line
(b) Single arrow
(c) Double arrow
(d) Doted lines
7. The unit of Kc for reaction; N2(g) + O2(g) ⇌ 2NO(g)
(a) mol dm−3
(b) mol−2 dm6
(c) mol−1 dm3
(d) no unit
8. The system is stable in equilibrium when:
(a) Qc = Kc
(b) Qc > Kc
(c) Qc < Kc
(d) None of these
9. The value of Kc increases when
(a) [Product] is less
(b) [Product] is more
(c) [Reactant] is more
(d) [Reactant] = [product]
10.Which of the
following represent backward reaction?
(a) (i) and (ii)
(b) (ii) only
(c) (i) only
(d) (iii) only
Solved
Short Questions
Q1.Write an equilibrium equation of
monoatomic carbon and a molecule of oxygen as reactant and carbon monoxide as
product.
Answer
2C(g) + O2(g) ⇌ 2CO(g)
Q2. State
law of mass action. What is active mass? How is the active mass is represented?
Answer
Statement of LMA
The rate of chemical reaction is
directly proportional to the product of the active masses of its reacting
substances at constant temperature.
Rate of reaction a Active mass of reacting substance
Active Mass
The Molar Concentration in mole/dm3 of substances is
termed as active mass.
Representation of Active Mass
active mass is represented by square brackets; [ ]. e.g. Active mass of H2O = [H2O]
Q3.Define chemical
equilibrium state with examples. Why chemical equilibrium is dynamic?
Answer
Definition
of Chemical Equilibrium
Chemical Equilibrium is the state of a
reversible reaction (in a closed vessel) at which there is no observable change in the concentrations
of reactants and products with time and rate of forward reaction is exactly
equal to the rate of reverse reaction Thus at equilibrium state:
Rate of forward reaction = Rate
of backward reaction
Example
of attainment of equilibrium for
hydrogen iodide formation form hydrogen and iodine
An
example of reaction at equilibrium is a reaction of hydrogen and iodine vapours
in a closed container to produce hydrogen iodide. At the start of reaction,
there is a higher concentration of hydrogen and iodine and after that the
concentration of decreases as hydrogen iodide is formed. The concentration of
hydrogen iodide increases as the forward reaction proceeds. As hydrogen iodide
is formed, the reverse reaction is then able to occur. consequently the rate of the forward reaction will go on
decreasing and the reverse reaction will go on increasing and ultimately the
two rates will become equal to each other. Thus, the equilibrium will set up
and concentration of various species (H2, I2, HI) becomes
constant. It is represented as
Q4. Outline the
characteristics of reversible reaction.
Answer
A few
important characteristic features of dynamic equilibrium are given below:
1. A dynamic equilibrium can only exist
in a closed system (neither reactants nor products can leave or enter the system).
2. At
equilibrium, the concentrations (amount) of reactants and products remains
constant.
3. At equilibrium, forward and reverse
reactions are taking place at equal and opposite rate.
4. Equilibrium can be approached from
either side of the reaction equation.
5. An equilibrium state can be disturbed and
again achieved under the given conditions of concentration, pressure and temperature.
Q5.Distinguish
between reversible and irreversible reactions.
Answer
Difference between reversible and irreversible reactions
Q6. Why
equilibrium constant may or may not have unit? Justify with example.
Answer
1. Kc has no units in reactions with equal number of moles on
both sides of the equation. This is because concentration
units cancel out in the expression for Kc.
e.g.
For the reaction in which the number of moles
of reactants and products are not equal in the balanced chemical
equation, Kc has some unit.
For the reaction; CO2(g) + H2(g)
⇌ CO2(g) + H2O(l),
the unit of Kc is derived as
2. For reactions when the number of
moles of reactants and product are not equal, Kc has a unit.
e.g.
For the reaction; N2(g) + 3H2(g)
⇌ 2NH3(g); the unit of Kc
is derived as
Q7. How
direction of a reaction can be predicted if Kc is known to you.
Answer
The
reaction quotient, Qc help us to predict the direction of reaction.
It has the same mathematical structure as Kc but Qc is a
ratio of initial concentration. Comparing Kc and Qc values predicts
the direction of reaction. Direction of a reaction at a particular moment can
be predicted by inserting the concentration of the reactants and products at
that particular moment in the equilibrium expression.
Thus, we
can make the following generalization about the direction of the reaction.
1. if Qc = Kc, the
actual product and reactant concentrations are equal to the equilibrium concentration,
and the system is stable.
2. If
Qc < Kc, there is increase in product concentration
for equilibrium. So the forward reaction occurs, forming additional products.
3. If
Qc < Kc, there is decrease in product concentration
for equilibrium. So the reverse reaction occurs, forming more reactants.
Q8.Write
equilibrium constant expression for the following equations:
N2(g) +
2O2(g)⇌ 2NO2(g)
2NO2(g) ⇌
N2O4(g)
PCl3(g) + Cl2(g) ⇌ PCl5(g)
2SO2(g)
+ O2(g) ⇌ 2SO3(g)
N2(g) + 3H2(g) ⇌ 2NH3(g)
H2(g) + Br2(g) ⇌ 2HBr(g)
H2(g) + I2(g) ⇌ 2HI(g)
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
Answer
Q9. Point out the coefficients of each in the following hypothetical reactions:
(i) 2A + 3B ⇌ 4C + 2D
(ii) 2M + 4N ⇌ 5O
(iii) 4X ⇌ 2Y + 3Z
Answer
(i) 2A(g)
+ 3B(g) ⇌ 4C(g)
+ 2D(g)
Coefficient of A
= 2
Coefficient of B
= 3
Coefficient of C
= 4
Coefficient of D
= 2
(ii) 2M(g) + 4N(g) ⇌ 5O(g)
Coefficient of M
= 2
Coefficient of N = 4
Coefficient of O
= 5
(iii) 4X(g) ⇌ 2Y(g)
+ 3Z(g)
Coefficient of X
= 4
Coefficient of Y
= 2
Coefficient of Z
= 3
Q10. What do you mean by the equilibrium constant, reaction quotient,
extent of a reaction, forward reaction, backward reaction?
Answer
forward Reactions
It is a reaction in which reactants react to form products. It
takes place from left to right.
reverse Reactions
It is a reaction in which products react to produce reactants. It
takes place from right to left.
Equilibrium constant
Equilibrium constant is a ratio of the product of the molar
concentrations (active masses) of products to the product of molar concentrations
(active masses) of reactants with each concentration term raised to the power
of coefficient as expressed in the balanced chemical equation at constant
temperature.
reaction quotient
reaction quotient (Qc) has the same mathematical structure
as Kc but Qc is a ratio of initial concentration of
products and reactants. Comparing Kc and Qc values
predicts the direction of reaction.
Extent of a Reaction
The limit up to which reactants are converted into products is
called extent of reaction. It indicates to which extent reactants are converted
to products. In fact, it measures how far a reaction proceeds before
establishing equilibrium state
Long
Questions
Q1. Describe
dynamic equilibrium with two examples
Answer
Example of attainment of equilibrium
for ammonia formation form nitrogen and hydrogen gas
An example of reaction at equilibrium is a reaction of nitrogen and
hydrogen gas in a closed container to produce ammonia gas. At the start of
reaction, there is a higher concentration of nitrogen and hydrogen and after
that the concentration of decreases as ammonia is formed. The concentration of ammonia
increases as the forward reaction proceeds. As ammonia is formed, the reverse
reaction is then able to occur. consequently
the rate of the forward reaction will go on decreasing and the reverse reaction
will go on increasing and ultimately the two rates will become equal to each
other. Thus, the equilibrium will set up and concentration of various species (N2,
H2, NH3) becomes constant. It is represented as
N2(g) + 3H2(g)
→ 2NH3(g)
N2(g) + 3H2(g)
← 2NH3(g)
N2(g) + 3H2(g)
⇌ 2NH3(g)
Example of attainment of equilibrium
for hydrogen iodide formation form hydrogen and iodine
An example of reaction at equilibrium is a reaction of hydrogen and
iodine vapours in a closed container to produce hydrogen iodide. At the start
of reaction, there is a higher concentration of hydrogen and iodine and after
that the concentration of decreases as hydrogen iodide is formed. The
concentration of hydrogen iodide increases as the forward reaction proceeds. As
hydrogen iodide is formed, the reverse reaction is then able to occur. consequently the rate of the forward
reaction will go on decreasing and the reverse reaction will go on increasing
and ultimately the two rates will become equal to each other. Thus, the
equilibrium will set up and concentration of various species (H2, I2,
HI) becomes constant. It is represented as
Q2. State
law of mass action. Derive an expression for equilibrium constant for a general
reaction.
Answer
Statement of LMA
The rate at which
a substance reacts is directly proportional to its active mass and the rate
of a chemical reaction is directly proportional to the product of the active
masses of its reacting substances at constant temperature.
Rate of reaction a Active mass of reacting substance
The law of mass action also suggest
that the ratio of the reactant concentration and the product concentration at
equilibrium state is constant.
Derivation of Equilibrium Constant (Kc) Expression for General Reversible Reaction
Let us apply the law of Mass Action
to derive equilibrium constant (Kc), for a general hypothetical reversible
reaction in which reactants A and B combine to form products C and D where a,
b, c and d are numbers of moles needed to balance a chemical equation. At
equilibrium state, the concentrations of A, B, C and D become constant. Let
[A], [B], [C] and [D] are the active masses or molar concentrations in mole/dm3
of A, B, C and D at equilibrium state respectively.
According to Law of Mass Action,
rate of forward reaction and backward reactions are given as:
Rate of forward reaction a [A]a [B]b
OR
Rf = Kf [A]a[B]b
(Kf=specific
rate constants for forward reaction)
Rate of reverse reaction a [C]c[D]d
OR
Rr=Kr [C]c[D]d
(Kr=specific
rate constants for reverse reaction)
Since, at the equilibrium state Rate of Forward and reverse
reaction becomes equal:
Rf = Rr
Kf [A]a
[B]b = Kr [C]c
[D]d
(on re-arranging i.e. By taking the
constants on L.H.S side and the variables on R.H.S of the equation, the above
equation becomes)
The ratio Kf/Kr
is a constant quantity at any specific temperature, which is known as
Equilibrium Constant for the Reaction denoted by Kc (or simply K)
where subscript ‘c’ indicates concentrations in mole/dm3.
Q3. How can you predict the following stages of a reaction by comparing the values of Kc and Qc
(i)Net reaction proceeds in forward direction
(ii)net reaction proceeds in reverse direction
Answer
It is important to determine the reaction’s direction at any given
time in a reversible reactions. The reaction quotient, Qc help us to
predict the direction of reaction. It has the same mathematical structure as Kc
but Qc is a ratio of initial concentration. Comparing Kc and Qc
values predicts the direction of reaction. Direction of a reaction at a
particular moment can be predicted by inserting the concentration of the
reactants and products at that particular moment in the equilibrium expression.
(i)Net
reaction proceeds in forward direction
If Qc < Kc, there is increase in product
concentration for equilibrium. So the forward reaction occurs, forming
additional products.
(ii)Net
reaction proceeds in reverse direction
If Qc < Kc, there is decrease in product
concentration for equilibrium. So the reverse reaction occurs, forming more
reactants.
Q4. Describe
the characteristics of equilibrium constant
Answer
The equilibrium constant (Kc)
of a reversible reaction is a constant ratio of Kf/Kr
(specific rate constant for forward reaction/specific rate constant for reverse
reaction) at constant temperature.
Equilibrium constant is a ratio of the product of the molar
concentrations (active masses) of products to the product of molar concentrations
(active masses) of reactants with each concentration term raised to the power
of coefficient as expressed in the balanced chemical equation at constant
temperature.
Thus Kc is directly
proportional to molar equilibrium concentrations (active masses) of products
and inversely proportional to molar equilibrium concentrations (active
masses) of reactants.
Unit of Kc
The value of equilibrium constant
does not depend on the initial concentrations of the reactants and the products.
The value of Kc depends only on temperature i.e.
1. Kc has no units in reactions with
equal number of moles on both sides of the equation. This is because
concentration units cancel out in the expression for Kc.
e.g.
For the reaction in which the number
of moles of reactants and products are not equal in the balanced
chemical equation, Kc has some unit.
For the reaction; CO2(g)
+ H2(g) ⇌ CO2(g) + H2O(l),
the unit of Kc is derived as
2. For
reactions when the number of moles of reactants and product are not equal, Kc
has a unit.
e.g.
For the reaction; N2(g) +
3H2(g) ⇌ 2NH3(g); the unit of Kc
is derived as
Importance of Kc
Kc determines which in greater concentration at
equilibrium – the products or the reactants. In general:
Kc > 1); equilibrium lies to the right and
favours the product.
Kc < 1); equilibrium lies to the left and
favours the reactants.
Q5. Predict which system at equilibrium will contain maximum amount of product and which system will contain maximum amount of reactant?
(a)2CO2(g)⇌2CO(g)+ O2(g) Kc (927oC) = 3.1 x 1018
(b)2O3(g)⇌ 2O2(g) Kc
(298 K) = 5.9 x 1055
Answer
Numerical value of the equilibrium constant predicts the extent or
scope of a chemical reaction. It indicates to which extent reactants are
converted to products. In general, there are three possibilities of predicting
extent of reactions as magnitude of Kc may be very high, very low or
moderate, so can be extent of reaction
(a)2CO2(g)
⇌ 2CO(g) +
O2(g) Kc
(927oC) = 3.1 x 1018
(b)2O3(g)
⇌ 2O2(g) Kc
(298 K) = 5.9 x 1055
Both reactions have high value of Kc. Reactions with high Kc values are virtually complete. High Kc indicates maximum product concentration and minimum reactant t concentration. This type of reaction is known as forward reaction. The reaction has almost gone to completion.
2H2(g) + O2(g) ⇌ 2H2O(g), Kc = 2.4 x 1047 At 227oC
Reactions with low Kc value never finish. Low Kc indicates maximum reactant concentration and minimum product concentration. These are called reverse or backward responses. Such type of reactions never go to completion.
F2(g) ⇌ 2F(g), Kc = 7.4 x 10−13 At 227oC
2HF(g) ⇌ H2(g) + F2(g) Kc = 10−13
Reactions which have moderate value of Kc are considered to be at equilibrium. The concentration of reactants and products is almost same
N2O4(g) ⇌ 2NO2(g) , Kc = 0.3 At 25oC
Q6.Write
down the macroscopic characteristics of dynamic equilibrium.
Answer
Macroscopic characteristics
of forward and reverse reactions in dynamic
equilibrium
A few important characteristic features of dynamic equilibrium are
given below:
1. A dynamic
equilibrium can only exist in a closed system (neither reactants nor products
can leave or enter the
system).
2. At
equilibrium, the concentrations (amount) of reactants and products remains
constant.
3. At equilibrium, forward and reverse
reactions are taking place at equal and opposite rate.
4. Equilibrium can
be approached from either side of the reaction equation.
5. An equilibrium state can be disturbed and
again achieved under the given conditions of concentration, pressure and temperature.
Writing Equilibrium Constant
Expression for Reaction along with Rate of forward and reverse Reactions
Q1.Write
an equilibrium equation of monoatomic carbon and a molecule of oxygen as
reactant and carbon monoxide as product.
Answer
2C(s) + O2(g)
⇌ 2CO(g)
The rate of forward reaction = Rf = Kf [O2]
The rate of reverse reaction = Rr = Kr [CO]2
Q2.Write
down the Equilibrium Constant expression for the reversible reaction of sulphur
dioxide with oxygen to
form sulphur trioxide.
Answer
For the reversible reaction of sulphur dioxide with
oxygen to form sulphur trioxide, the Equilibrium Constant expression is derived
as follows:
2SO2(g) + O2(g) ⇌ 2SO3(g)
The rate of forward reaction = Rf
= Kf [SO2]2[O2]
The rate of reverse reaction= Rr = Kr [SO3]2
Q3. Write down the
Equilibrium Constant expression for reversible reaction of nitrogen with oxygen
to form nitrogen monoxide.
Answer
For the reversible reaction of nitrogen with oxygen
to form nitrogen monoxide, the Equilibrium Constant expression is derived as
follows:
N2(g) + O2(g)
⇌ 2NO(g)
The rate of forward reaction = Rf
= Kf [N2][O2]
The rate of reverse reaction = Rr = Kr [NO]2
Q4.Write down the
Equilibrium Constant expression for the reversible reaction of nitrogen with
hydrogen to form ammonia,
Answer
For the reversible reaction of nitrogen with
hydrogen to form ammonia, the Equilibrium Constant expression is derived as
follows:
N2(g) +3H2(g)
⇌ 2NH3(g)
The rate of forward reaction = Rf
= Kf [N2][H2]3
The rate of reverse reaction = Rr = Kr [NH3]2
Q5. Write down the Equilibrium Constant
expression for the reversible reaction of combination of nitrogen dioxide into its
dimer dinitrogen tetraoxide
Answer
2NO2(g) ⇌ N2O4(g)
The rate of forward reaction = Rf
= Kf [NO2]2
The rate of reverse reaction = Rr = Kr [N2O4]
Finding Out Coefficients for given
Hypothetical and Real Reactions
Q. Figure out coefficients of each
in the following hypothetical reactions
(i) 9X(g) + Y3(g) ⇌ 3X3Y(g)
(ii) 4X(g) ⇌ 2Y(g) + 3Z(g)
(iii) 2A(g) + 3B(g) ⇌ 4C(g) + 2D(g)
(iv) 2M(g) + 4N(g) ⇌ 5O(g)
General Consideration
In Chemistry, the coefficient is
the number in front of the formulae or symbols to balance the chemical
equations. The coefficient tells us how many molecules or atoms of a given
formula are present. coefficient
is a number used in chemical equation, just as a prefix of chemical formula or
symbol to define the number of molecules or atoms reacting and producing in a
reaction.
they multiply all the atoms in a
formula.
e.g.
(i) Thus
the symbol “2 NaHCO3” indicates two units of sodium bicarbonate
which contain 2 Na atoms 2 H atoms
2 C atoms and 6 O atoms (2 x 3 = 6 the coefficient times the subscript for O).
(ii) 2H2O means
we have 2 molecules of water
(iii) In the
balanced chemical equation for Ammonia
N2
+ 3H2 → 2NH3
We know we have coefficients of
1 for N2
3 for H2
and
2 for NH3
It takes 1 molecule of N2 with 3 molecules
of H2 to produce 2 molecules of NH3
Solution
(i) 9X(g)
+ Y3(g) ⇌ 3X3Y(g)
Coefficient of X = 9
Coefficient of Y3 = 1
Coefficient of X3Y = 3
(ii) 4X(g) ⇌ 2Y(g) + 3Z(g)
Coefficient of X = 4
Coefficient of Y = 2
Coefficient of Z = 3
(iii) 2A(g) + 3B(g) ⇌ 4C(g) + 2D(g)
Coefficient of A = 2
Coefficient of B = 3
Coefficient of C = 4
Coefficient of D = 2
(iv) 2M(g) + 4N(g) ⇌ 5O(g)
Coefficient of M = 2
Coefficient of N = 4
Coefficient of O = 5
Writing Forward and Reverse
Reactions for reversible Reactions
Q1.Write down forward and reverse reactions for the following:
(i) N2(g) + O2(g) ⇌ 2NO(g)
(ii) 2SO2(g) + O2(g) ⇌ 2SO3(g)
(iii) H2(g) + I2(g) ⇌ 2HI(g)
(iv) 2NO(g) + O2(g) ⇌ 2NO2(g)
(v)2NH3(g) ⇌ N2(g) + 3H2(g)
(vi) PCl5(g) ⇌ PCl3(g) + Cl2(g)
(vii) 2N2O(g) ⇌ 2N2(g) + O2(g)
(viii)COCl2(g) ⇌ CO(g) + Cl2(g)
Solution
(i) N2(g) + O2(g)
⇌ 2NO(g)
Forward Reaction; N2(g) + O2(g)
→ 2NO(g)
Reverse Reaction; N2(g) + O2(g)
← 2NO(g)
(ii) 2SO2(g)
+ O2(g) ⇌ 2SO3(g)
Forward Reaction; 2SO2(g) + O2(g)
→ 2SO3(g)
Reverse Reaction; 2SO2(g)
+ O2(g) ← 2SO3(g)
(iii) H2(g) + I2(g) ⇌ 2HI(g)
Forward Reaction; H2(g) + I2(g) → 2HI(g)
Reverse Reaction; H2(g) + I2(g) ← 2HI(g)
(iv) 2NO(g) + O2(g) ⇌ 2NO2(g)
Forward Reaction; 2NO(g) + O2(g)→ 2NO2(g)
Reverse Reaction; 2NO(g) + O2(g) ← 2NO2(g)
(v) 2NH3(g) ⇌ N2(g) +
3H2(g)
Forward Reaction; 2NH3(g) → N2(g) +
3H2(g)
Reverse Reaction; 2NH3(g) ← N2(g) +
3H2(g)
(vi) PCl5(g) ⇌ PCl3(g)
+
Cl2(g)
Forward Reaction; PCl5 → PCl3 + Cl2
Reverse Reaction; PCl5 ← PCl3 + Cl2g)
(vii) 2N2O(g) ⇌ 2N2(g) + O2(g)
Forward Reaction; 2N2O(g) → 2N2(g) + O2(g)
Reverse Reaction; 2N2O(g) ← 2N2(g) + O2(g)
(viii) COCl2(g)
⇌ CO(g) + Cl2(g)
Forward Reaction; COCl2(g) → CO(g) + Cl2(g)
Reverse Reaction; COCl2(g) ← CO(g) + Cl2(g)
Calculating Equilibrium
Constant
Q1.Equilibrium occurs when nitrogen monoxide gas
reacts with oxygen gas to form nitrogen dioxide gas
2NO(g) + O2(g) ⇌ 2NO2(g)
At equilibrium at 230oC, the concentrations are measured to be
[NO] = 0.0542 mol dm−3,
[O2] = 0.127 mol dm−3
and
[NO2] = 15.5 mol dm−3
Calculate the equilibrium constant at this temperature.
(Book problem 1; page 8)
Solution
Given
Given equilibrium concentrations
are;
[NO] = 0.0542 mol dm−3
[O2] = 0.127 mol dm−3
[NO2] = 15.5 mol dm−3
Required
equilibrium
constant (Kc) = ?
equilibrium constant (Kc) Expression
Calculation
Q2. A reaction takes place between iron and chloride ion as
Fe3+ + 4Cl− ⇌ FeCl4−
At equilibrium the concentrations are measured to be
Fe3+ = 0.2 mol dm−3,
Cl− = 0.28 mol dm−3
and
FeCl4− = 0.95 x 10−4 mol dm−3
Calculate equilibrium constant Kc for given reaction.
(Book problem 1; page 9)
Solution
Given
Given equilibrium concentrations
are;
[Fe3+] = 0.2 mol dm−3
[Cl−] = 0.28 mol dm−3
[FeCl4−] = 0.95
x 10−4 mol dm−3
Required
equilibrium
constant (Kc) = ?
equilibrium constant
(Kc) Expression
Q3. When hydrogen reacts with iodine at 25°C to form hydrogen iodide by a reversible reaction as follows:
The equilibrium concentrations are:
[H2] = 0.05 mol dm−3
[I2] = 0.06 mol dm−3
and
[HI] = 0.49 mol dm−3.
Calculate the equilibrium constant for this reaction.
Solution
Given
Given equilibrium concentrations
are;
[H2] = 0.05
mol dm−3
[I2] = 0.06
mol dm−3
[HI] = 0.49 mol
dm−3
Required
equilibrium
constant (Kc) = ?
Q4. For the formation of ammonia by Haber’s process, hydrogen and nitrogen react reversibly at 500°C as follows
2NH3(g) ⇌ N2(g) + 3H2(g), Kc = 3.0 x 10−9
The equilibrium concentrations of these gases are:
nitrogen 0.602 mol dm−3;
hydrogen 0.420 mol dm−3
and
ammonia 0.113 mol dm−3.
What is value of Kc?
Solution
Given
Given equilibrium concentrations
are;
[NH3] = 0.113 mol dm−3
[N2] = 0.602
mol dm−3
[H2] = 0.420
mol dm−3
Required
equilibrium
constant (Kc) = ?
No comments:
Post a Comment