Test: 35 Year JEE Previous Year Questions: Equilibrium
52 Questions MCQ Test Chemistry for JEE | Test: 35 Year JEE Previous Year Questions: Equilibrium
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PARAGRAPH 1
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Q. Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt.2 is
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Let the heat capacity of insulated beaker be C.
Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g
⇒ ± Total heat capacity = (C + 200 × 4.2) J/K Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1
⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ = 5.7 × 1000 J
⇒ 5.7 × 1000 = (C + 200 × 4.2) × Δ´T.
5.7 × 1000 = (C + 200 + 4.2) × 5.7
⇒ (C + 200 × 4.2) = 1000
In second experiment,
Total mass of aqueous content = 200 g
⇒ Total heat capacity = (C + 200 × 4.2) = 1000
⇒ Heat released = 1000 × 5.6 = 5600 J.
Overall, only 0.1 mol of CH3COOH undergo neutralization.
⇒ Δ Hneutralization of CH3COOH = -5600/0.1
= – 56000 J/mol
= – 56 KJ/mol.
⇒ Δ Hneutralization of CH3COOH = 57 – 56 = 1 KJ/mol
PARAGRAPH 1
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt.1).
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Q. The pH of the solution after Expt. 2 is
Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured. (Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)
Final solution contain 0.1 mole of CH3COOH2 and CH3COONa each.
Hence it is a buffer solution.
PARAGRAPH 2
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:
The standard reaction Gibbs energy, ΔrG°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
(Given R = 0.083 L bar K–1 mol–1)
Q. The equilibrium constant Kp for this reaction at 298 K, in terms of βequilibrium, is
PARAGRAPH 2
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:
The standard reaction Gibbs energy, ΔrG°, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
(Given R = 0.083 L bar K–1 mol–1)
Q. The INCORRECT statement among the following, for this reaction, is
(A) Correct statement.
As on decrease in pressure reaction move indirection where no. of gaseous molecules increase.
(B) Correct statement At the start of reaction QP < KP so dissociation of X2 take place spontaneousely.
(C) Incorrect statement as
Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 5 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
Q.
Statement -1 The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature.
Statement -2 When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change.
The statement-1 is clearly wrong in context to LeChateliers principle, which states that “increase in temperature shifts the equilibrium in the forward direction of those reactions which proceed with absorption of heat (endothermic reactions), and in the backward direction of those reactions which proceed with the evolution of heat (exothermic reactions).” Statement -2 is clearly true again according to Lechatelier principle.
Read the following statement and explanation and answer as per the options given below :
Statement -1 HNO3 is a stronger acid than HNO2
Statement -2 In HNO3 there are two nitrogen-to-oxygen bonds whereas in HNO2 there is only one.
TIPS/Formulae :
Among oxyacids, the acidic character increases with increase in oxidation state of the central atom.
O.S. of N in HNO3 = + 5
O.S.of N in HNO2 = + 3
thus HNO3 stronger than HNO2. Hence assertion is correct.
The assertion is true but the reason is wrong as can be clearly seen from the above structures.
Read the following statement and explanation and answer as per the options given below :
Statement -1 For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero.
Statement -2 At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.
We know that for every chemical reaction at equilibrium, Gibb's free energy (ΔG = 0) is zero.
However standard Gibb's free energy (ΔG°) may or may not be zero. Thus statement 1 is False.
For a spontaneous reaction, at constant temperature and pressure, the reaction proceeds in the direction in which ΔG is < 0 i.e. in the direction of decreasing Gibb's energy (G) so the statement 2 is True.
Thus the only such option is (d) which is correct answer.
1 M NaCl and 1 M HCl are present in an aqueous solution.
The solution is
NOTE : A buffer is a solution of weak acid and its salt with strong base and vice versa.
HCl is strong acid and NaCl is its salt with strong base. pH is less than 7 due to HCl.
Species acting as both Bronsted acid and base is
(HSO4)– can accept and donate a proton
(HSO4)– + H+ → H2SO4 (acting as base)
(HSO4)– – H+ → SO42–. (acting as acid)
Let the solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is
Change in volume of the system does not alter which of the following equilibria?
In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles.
For the reaction CO (g) + (1/2) O2 (g) = CO2 (g), Kp / Kc is
Kp = Kc (RT)Δn;
Which one of the following statements is not true?
pH of an acidic solution should be less than 7. The reason is that from H2O. [H+] = 10–7M which cannot be neglected in comparison to 10–8M. The pH can be calculated as.
from acid, [H+] = 10–8M.
from H2O, [H+] = 10–7M
∴ Total [H+] = 10–8 + 10–7 = 10–8 (1 + 10) = 11× 10–8
∴ pH = – log [H+] = –log 11×10–8 = –[log11 + 8 log 10]
= –[1.0414 – 8] = 6.9586
The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10–5 mol L–1. Its solubility product number will be
For the reaction equilibrium
the concentrations of N2O4 and NO2 at equilibrium are 4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of Kc for the reaction is
Consider the reaction equilibrium
On the basis of Le Chatelier’s principle, the condition favourable for the forward reaction is
Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles.
∴ High pressure will be required.
When rain is accompanied by a thunderstorm, the collected rain water will have a pH value
The rain water after thunderstorm contains dissolved acid and therefore the pH is less than rain water without thunderstorm.
NOTE : Conjugate acid-base differ by H+
What is the equilibrium expression for the reaction 
The solids have con centration unity
For the reaction, 
is equal to
The equilibrium constant for the reaction 
at temperature T is 4×10–4.
The value of Kc for the reaction
at the same temperature is
The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is ‘s’. The corresponding solubility product is Ksp. ‘s’ is given in term of Ksp by the relation :
If α is the degree of dissociation of Na2SO4 , the Vant Hoff’s factor (i) used for calculating the molecular mass is
The solubility product of a salt having general formula MX2 , in water is : 4 × 10 -12. The concentration of M2+ ions in the aqueous solution of the salt is
Where s is the solubility of MX2 then Ksp = 4s3; s × (2s)2 = 4×10–12 = 4s3; s = 1 × 10–4
∴ [M++] = s = 1[M++] = 10 × 10– 4
The exothermic formation of CIF3 is represented by the equation :
Which of the following will increase the quantity of CIF3 in an equilibrium mixture of Cl2 , F2 and ClF3 ?
The reaction given is an exothermic reaction thus accordingly to Lechatalier’s principle lowering of temperature, addition of F2 and or Cl2 favour the for ward direction and hence the production of ClF3.
For the reaction 
(Kc = 1.8 x 10-6 at 184°C) (R = 0.0831 kJ/ (mol. K))
When Kp and Kc are compared at 184°C, it is found that
For the reaction:- 
Given Kc = 1.8 × 10–6 at 184ºC
R = 0.0831 kj/mol. k
Kp = 1.8 × 10–6 × 0.0831 × 457 = 6.836 × 10–6
[∵ 184°C = (273 + 184) = 457 k, Δn = (2 + 1, –1) = 1]
Hence it is clear that Kp > Kc
Hydrogenion concentration in mol/L in a solution of pH = 5.4 will be :
On solving, [H+] = 3.98 × 10–6
What is the conjugate base of OH - ?
Conjugate acid-base pair differ by only one proton.
Conjugate base of OH– is O2–
An amount of solid NH4 HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm? The equilibrium constant for NH4HS decomposition at this temperature is
Then 0.5 + x + x = 2x + 0.5 = 0.84 (given) ⇒ x = 0.17 atm.
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be
Total moles after dissociation 1 – x + x + x = 1 + x
= mole fraction of PCl3 × Total pressure
The equilibrium constant for the reaction
is Kc = 4.9 × 10–2. The value of Kc for the reaction
will be
On taking the square of the above reaction
Given the data at 25ºC
What is the value of log Ksp for AgI? (2.303 RT/F = 0.059 V)
From (i) and (ii) we have,
The first and second dissociation constants of an acid H,A are 1.0 × 10–5 and 5.0 × 10–10 respectively. The overall dissociation constant of the acid will be
(Given)
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffer solution of HA in which 50% of the acid is ionized is
Given pKa = 4.5 and acid is 50% ionised.
[HA] = [A–] (when acid is 50% ionised)
∴ pH = pKa + log 1
∴ pH = pKa = 4.5
pOH = 14 – pH = 14 – 4.5 = 9.5
In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (molecular mass = 283) the equilibrium which sets in is 
If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 × 10–8, what is the mass of AgIO3 contained in 100 ml of its saturated saolution?
Let s = solubility
= 1.0 × 10–4 mol/lit = 1.0 × 10–4 × 283 g/lit
(∵ Molecular mass of AgIO3 = 283)
The equilibrium constants Kp1 and Kp2 for the reactions 
respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressures at these equilibria is
Let the initial moles of X be ‘a’ and that of Z be ‘b’ then for the given reactions, we have
Total no. of moles = a (1 – α) + 2aα
= a – aα + 2aα = a (1 + α)
Total no . of moles = b(1 – α) + bα + bα
= b – bα + bα + bα = b(1 + α)
For the following three reactions a, b and c, equilibrium constants are given:
(i) CO(g) + H2O(g) 
CO2 (g)+ H2 (g); K1
(ii) CH4 (g) + H2O(g) CO(g)+ 3H2 (g); K2
(iii) CH4 (g) + 2H2O(g) CO2 (g)+ 4H2 (g); K3
Reaction (c) can be obtained by adding reactions (a) and (b) therefore K3 = K1. K2
Hence (c) is the correct answer.
Four species are listed below:
i. 
ii. H3O+
iii. 
iv. HSO3F
Which one of the following is the correct sequence of their acid strength?
The correct order of acidic strength of the given species in
or (i) < (iii) < (ii) < (iv)
It corresponds to choice (c) which is correct answer.
The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the correspondng salt, BA, will be
In aqueous solution BA(salt) hydrolyses to give
Now pH is given by
Substituting given values, we get
Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10– 4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form? (KSP for for BaCO3 = 5.1 × 10–9)
Three reactions involving H2PO4– are given below :
(i) H3PO4 + H2O → H3O+ + H2PO4–
(ii) H2PO4– + H2O → HPO42– + H3O+
(iii) H2PO4– + OH– → H3PO4 + O2–
In which of the above does H2PO4– act as an acid ?
Hence only in (ii) reaction H2PO4– is acting as an acid.
In aqueous solution the ionization constants for carbonic acid are
K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11.
Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
As H2CO3 is a weak acid so the concentration of H2CO3 will remain 0.034 as 0.034 >> x.
As 
is again a weak acid (weaker than H2CO3) with x >> y.
Note : [H3O+] = H+ from first step (x) and from second step (y) = (x + y)
So the concentration of 
concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations.
Thus the final concentrations are
Solubility product of silver bromide is 5.0 × 10–13. The quantity of potassium bromide (molar mass taken as 120 g mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is
For precipitation to occur Ionic product > Solubility product
i.e., precipitation just starts when 10–11 moles of KBr is added to 1ℓ AgNO3 solution
∴ Number of moles of Br– needed from KBr = 10–11
∴ Mass of KBr = 10–11 × 120 = 1.2 × 10–9 g
At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm.
Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
Total P at equilibrium = 0.5 – x + 2x = 0.5 + x atm
0.8 = 0.5 + x
∴ x = 0.8 – 0.5 = 0.3 atm 
= 1.8 atm
The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 × 10–4. Thevalue of Kc for the reaction
at the same temperature is:
For the reaction
Hence for the reaction
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of the acid is :
Ka = Cα2 = 0.1x 10-2x10-2 = 10–5
How many litres of water must be added to 1 litre an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
Q pH = 1 ; H+ = 10–1 = 0.1 M
pH = 2 ; H+ = 10–2 = 0.01 M
∴ M1 = 0.1 V1 = 1; M2 = 0.01 V2 = ?
From
M1V1 = M2V2; 0.1 × 1 = 0.01 × V2
V2 = 10 litre
∴ volume of water added = 10 – 1 = 9 litre.
For the reaction 
if KP = KC (RT)x where the symbols have usual meaning then the value of x is (assuming ideality):
where x = Δng = number of gaseous moles in product – number of gaseous moles in reactant
The standard Gibbs energy change at 300 K for the reaction 
At a given time, the composition of the reaction mixture is 
The reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]
R = 8.314 J/K/mol. e = 2.718
ΔG° = – 2.303 RT log KC.
2494.2 J = – 2.303 × (8.314 J/K/mol) × (300K) logKC
The following reaction is performed at 298 K.
The standard free energy of formation of NO(g) is 86.6 kj/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (Kp = 1.6 × 1012)
ΔG°NO(g)= 86.6k J/mol = 86600 J/mol; 
T = 298, KP = 1.6 × 1012
ΔG° = – RT ln KP
Given equation,
The equilibrium constant at 298 K for a reaction A +B 
C+D is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L–1) will be :
Given,
On solving
a = 0.81
[Δ]At eq = 1 + a = 1 + 0.81 = 1.81
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