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questions
List of practice Questions
The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10
–19
M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO
4
, MnCl
2
, ZnCl2 and CdCl
2
. in which of these solutions precipitation will take place?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
At 700 K, equilibrium constant for the reaction:
\(H_2 (g) + I_2 (g) ⇋ 2HI (g)\)
is 54.8. If 0.5 mol L
–1
of HI(g) is present at equilibrium at 700 K, what are the concentration of H
2
(g) and I
2
(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
What is K
c
for the following equilibrium when the equilibrium concentration of each substance is: [SO
2
]= 0.60 M, [O
2
] = 0.82 M and [SO
3
] = 1.90 M ?
\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Write the expression for the equilibrium constant, K
c
for each of the following reactions:
\(2NOCl (g) ⇋ 2NO (g) + Cl_2 (g)\)
\(2Cu(NO_3)_2 (s) ⇋ 2CuO (s) + 4NO_2 (g) + O_2 (g)\)
\(CH_3COOC_2H_5(aq) + H_2O(l) ⇋ CH_3COOH (aq) + C_2H_5OH (aq)\)
\(Fe^{3+} (aq) + 3OH^– (aq) ⇋ Fe(OH)_3 (s)\)
\(I_2 (s) + 5F_2 ⇋ 2IF_5\)
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Find out the value of
\(K_c\)
for each of the following equilibria from the value of
\(K_p\)
:
\(2NOCl (g) ⇋ 2NO (g) + Cl_2 (g); \ K_p= 1.8 × 10^{–2}\ at \ 500\ K\)
\(CaCO_3 (s) ⇋ CaO(s) + CO_2(g); \ K_p= 167 \ at \ 1073\ K\)
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
For the following equilibrium,
\(K_c = 6.3 × 10^{14}\)
at
\(1000\ K\)
\(NO (g) + O_3 (g) ⇋ NO_2 (g) + O_2 (g)\)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is
\(K'_c\)
for the reverse reaction?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Reaction between
\(N_2\)
and
\(O_2^–\)
takes place as follows:
\(2N_2 (g) + O_2 (g) ⇋ 2N_2O (g)\)
If a mixture of
\(0.482 \ mol\)
\(N_2\)
and
\(0.933\ mol\)
of
\(O_2\)
is placed in a
\(10\ L\)
reaction vessel and allowed to form
\(N_2O\)
at a temperature for which
\(K_c = 2.0 × 10^{–37}\)
, determine the composition of equilibrium mixture.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Nitric oxide reacts with Br
2
and gives nitrosyl bromide as per reaction given below:
\(2NO (g) + Br_2 (g) ⇋ 2NOBr (g)\)
When 0.087 mol of NO and 0.0437 mol of Br
2
are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br
2
.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
At
\(450\ K\)
,
\(K_p= 2.0 × 10^{10}/bar\)
for the given reaction at equilibrium.
\(2SO_2(g) + O_2(g) ⇋ 2SO_3 (g)\)
What is
\(K_c\)
at this temperature ?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
What is the minimum volume of water required to dissolve 1g of calcium sulfate at 298 K? (For calcium sulfate, K
sp
is 9.1 × 10
–6
).
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10
–18
).
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed. Will it lead to the precipitation of copper iodate? (For cupric iodate K
sp
= 7.4 × 10
–8
).
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
The equilibrium constant expression for a gas reaction is,
\(K_c = \frac {[NH_3]^4 [O_2]^5 }{[NO_4] [H_2O_6]}\)
Write the balanced chemical equation corresponding to this expression.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
One mole of H
2
O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
\(H_2O (g) + CO (g) ⇋ H_2 (g) + CO_2 (g)\)
Calculate the equilibrium constant for the reaction.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of
\(ICl\)
was
\(0.78 \ M\)
?
\(2ICl (g) ⇋ I_2 (g) + Cl_2 (g); \ K_c = 0.14\)
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\(CH_3COOH (l) + C_2H_5OH (l) ⇋ CH_3COOC_2H_5 (l) + H_2O (l) \)
Write the concentration ratio (reaction quotient), Q
c
, for this reaction (note: water is not in excess and is not a solvent in this reaction)
At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
A sample of pure PCl
5
was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl
5
was found to be
\(0.5×10^{–1}\ mol L^{–1}\)
. If value of K
c
is
\(8.3×10^{–3}\)
, what are the concentrations of PCl
3
and Cl
2
at equilibrium?
\(PCl_5 (g) ⇋ PCl_3 (g) + Cl_2(g)\)
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Equilibrium constant, K
c
for the reaction
\(N_2 (g) + 3H_2 (g) ⇋ 2NH_3 (g)\)
At 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L
–1
N
2
, 2.0 mol L
–1
H
2
and 0.5 mol L
–1
NH
3
. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
\(2BrCl (g) ⇋ Br_2 (g) + Cl_2 (g)\)
for which K
c
= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10
–3
mol L
–1
, what is its molar concentration in the mixture at equilibrium?
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO
2
in equilibrium with soild carbon has 90.55% CO by mass
\(C(s) + CO_2(g) ⇋ 2CO(g)\)
Calculate K
c
for this reaction at the above temperature.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Calculate:
∆G
Θ
and
the equilibrium constant for the formation of NO
2
from NO and O
2
at 298 K
\(NO (g) + \frac 12O_2 (g) ⇋ NO_2 (g)\)
where,
∆
f
G
Θ
(NO
2
) = 52.0 kJ/mol
∆
f
G
Θ
(NO) = 87.0 kJ/mol
∆
f
G
Θ
(O
2
) = 0 kJ/mol.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
\(PCl_5 (g) ⇋ PCl_3 (g) + Cl_2 (g)\)
\(CaO (s) + CO_2 (g) ⇋ CaCO_3 (s)\)
\(3Fe (s) + 4H_2O (g) ⇋ Fe_3O_4 (s) + 4H_2 (g)\)
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO
2
, CN
–
, HClO
4
,F
–
,OH
–
,CO and S
2
–
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
Which of the following are Lewis acids? H
2
O, BF
3
, H
+
, and NH
4
+
.
CBSE Class XI
Chemistry
Law Of Chemical Equilibrium And Equilibrium Constant
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