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All questions of Chemical Equilibrium for EmSAT Achieve Exam

Ca(HCO3)2 is strongly heated and after equilibrium is attained, temperature changed to 25° C.

Kp = 36 (pressure taken in atm)
Thus, pressure set up due to CO2 is
  • a)
    36 atm
  • b)
    18 atm
  • c)
    12 atm
  • d)
    6 atm
Correct answer is 'D'. Can you explain this answer?

Mira Joshi answered
The reaction is as follow:-
Ca(HCO3)2(s)⇌CaO(s) + 2CO2 (g) + H2O(g)
At 25° C H2O goes in liquid state
Kp = (PCaO)1×(PCO2)2
(PCa(HCO3)2)
Since, Ca(HCO3)2, CaO and H2O are not in gaseous state, so their partial pressure is taken 1.
Putting all values, we have
36 = (PCO2)2 
Or PCO2 = 6 atm

Following equilibrium is set up at 298 K in a 1 L flask.

If one starts with 2 moles of A and 1 mole of B, it is found that moles of B and D are equal.Thus Kc is 
  • a)
    9.0
  • b)
    15.0
  • c)
    3.0
  • d)
    0.0667
Correct answer is option 'B'. Can you explain this answer?

Sushil Kumar answered
For the equilibrium reaction:
A+2B ⇌ 2C+D
volume of flask = 1L
Initial moles of A = 2 mol
initial concentration of A=[A]i = 2 M
initial mole of B = 1 mol 
[B]i = 1 M
[A]eq = 2-x, [B]eq = 1-2x, [C]eq = x, [D]eq = 3x
Given [D]eq = 1 * 1L
= 1 M
Thus x = 1M
[A]eq = 1, [B]eq = -1, [C]eq = 1, [D] = 3
Kc = {([D]eq)3 * ([C]eq)}/{[A]eq * ([B]eq)2
= Kc = {(3)3*1}/{1*(-1)2}
= 27/1
= 27

 In which of the following reaction can equilibrium be attained
  • a)
    Reversible reaction
  • b)
    Cyclic reaction
  • c)
    Decomposition reaction
  • d)
    Irreversible reaction
Correct answer is option 'A'. Can you explain this answer?

Rajat Kapoor answered
Reversible Reaction
The common observation for any reactions when they are reacted in closed containers would not go to completion, for some given conditions like temperature and pressure.
For all those cases, only the reactants are found to be present in the intial stages, but with the progress of reaction, the reactants concentration decreases and to that of the products increases. A stage is finally reached where there is no more change of reactants and products concentration is observed. The state where the reactants and products concentrations do not show any visible change within a given period of time is better known as the state of chemical equilibrium. 
The reactant amount that remains unused depends upon the experimental conditions like concentration of components, temperature of the system, pressure of the system and the reaction nature.

Following equilibrium is set up at 1000 K and 1 bar in a 5 L flask,

At equilibrium, NO2 is 50% o f the total volume. Thus, equilibrium constant Kc is 
  • a)
    0.133
  • b)
    0.266
  • c)
    0.200
  • d)
    0.400
Correct answer is option 'A'. Can you explain this answer?

The correct answer is Option A.    
                N2O4  ⇌  2NO2
Initial            1                 0           
Equilibrium  1−x             2x
Total moles = 1 - x + 2x 
NO2 is 50% of the total volume when equilibrium is set up.
Thus, the volume fraction (at equilibrium) of NO2 = 50/100 = 0.5 = ½
So,    2x / (1+x) = ½
     => x = ⅓
For 1 litre;
Kc = [NO2] / [N2O4]
    = [4*(1/9)] / [⅔]
    = 0.66; 
For 5 litres; 
Kc = 0.66 / 5
= 0.133
Thus, option A is correct.
 

Assume following equilibria when total pressure set up in each are equal to 1 atm, and equilibrium constant (Kp) as K1; K2 and K3


Thus,
  • a)
     K1 = K2 = K3
  • b)
    K1 < K2 < K3
  • c)
    K3 < K2 < K1
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

The correct answer is option C
CaCO3 ​→ CaO + CO2​
Kp​ = k1 ​= Pco2​​
total pressure of container P
k1​ = p
NH4​HS → NH3 ​+ H2​S
PNH3​​ = PH2​S ​= P0​
P0​ + P0​ = p (total pressure)
P0 ​= p/2
k2​ = kp ​= [PNH3​​][PH2​s​] p24
NH2​CoNH2 ​→ 2NH3 ​+ CO2​
PNH3​​ = 2P0​        PCO2​ ​= P0​
2P0​ + P0 ​= P

For HF, pKa = 3.45. What is the pH of an aqueous buffer solution that is 0.1M HF (aq) and 0.300 M KF (aq)?
a)11.03
b)2.97
c)10.07
d)3.93 
Correct answer is option 'D'. Can you explain this answer?

Suresh Iyer answered
Given, pKa = 3.45
Concentration of HF = 0.1 M, concentration of KF = 0.300 M
For acidic buffer;
pH = pKa + log [salt of weak acid]/[weak acid]
= 3.45 + log0.3/0.1
= 3.45 + 0.48
= 3.93

H2O (l) H2O(s) ; ΔH = -q
Application of pressure on this equilibrium
  • a)
    cause formation of more ice
  • b)
    cause fusion of ice
  • c)
    has no effect
  • d)
    lower the melting point
Correct answer is option 'B,D'. Can you explain this answer?

Krishna Iyer answered
The correct answers are Options B and D. 
 
As we know that reaction is exothermic it means heat is released in the reaction so, if we apply pressure then reaction will proceed in backward direction but if there is gas phase equilibrium the reaction will shift in that direction in which less number of moles are present. If pressure increases then the ice will melt and ice gets more energy at low temp. To melt ,so it’s melting point decreases.
 

The [Ag+(aq)] = 10-5 in a solution .The [Cl(aq)] to precipitate AgCl having Ksp of 1.8×10-10 M2 is — M
  • a)
    10-7
  • b)
    10-8
  • c)
    10-9
  • d)
    10-5
Correct answer is option 'D'. Can you explain this answer?

Hansa Sharma answered
For precipitation, Qrkn > Ksp, then tto establish equilibrium, ions will combine to give molecule as ppt.
So applying the above concept,
 1.8×10-10 < [Ag+][Cl-]
 1.8×10-10 < 10-5 × [Cl-]
Or [Cl-] > 1.8×10-5
By seeing the option, only option d satisfies this condtion.

For the following electrochemical cell reaction at 298 K,
 
E°cell = 1.10 V
  • a)
    0.027
  • b)
    37.22
  • c)
    0.012
  • d)
    85.73
Correct answer is option 'B'. Can you explain this answer?

Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq),

E= +1.10V

∴ Eo = 0.0591/n log10Keq
because at equilibrium, 
Ecell = 0

(n = number of electrons exchanged = 2)

1.10 = 0.0591/2 log10Keq
2.20/0.0591 = log10Keq
Keq = antilog37.225

Solubility of BaCl2 if Ksp is 10-6 at 25°C is
  • a)
    Cannot be predicted
  • b)
    6.3 10-3 M
  • c)
    10-6 M
  • d)
    10-3 M
Correct answer is option 'B'. Can you explain this answer?

Ram Mohith answered
BaCl₂ ⇔ Ba²⁺ + 2Cl⁻ Ksp = (s)(2s)² = 4s³ Given, 4s³ = 10⁻⁶ ⇒s³ = 0.25 ⨯ 10⁻⁶ ⇒s = (0.25)³ ⨯ 10⁻² ⇒s = 0.629 ⨯ 10⁻² ⇒s = 6.3 ⨯ 10⁻³

We know that the relationship between Kc and Kp is Kp = Kc (RT)Δn
What would be the value of Δn for the reaction NH4Cl (s) ⇔ NH3 (g) + HCl (g)
a)1
b)0.5
c)1.5
d)2
Correct answer is option 'D'. Can you explain this answer?

Nandini Patel answered
The answer is d.
The relationship between Kp and Kc is
Kp = Kc (RT) ∆n
Where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants)
For the reaction,
NH4C1(s) ⇆ NH3(g) + HCl(g)
∆n = 2 – 0 = 2 

Kc forthe decomposition of NH4HS(s) is 1.8x 10-4 at 25°C.

If the system already contains [NH3] = 0.020 M, then when equilibrium is reached, molar concentration are
  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'B'. Can you explain this answer?

Sushil Kumar answered
 NH4HS (s)  ⇋ NH3 (g) + H2S (g)
Initial    1                   -               -
At eqm     1-x                 x+0.02     x
Kc = [NH3][H2S]   (Since NH4HS is solid, we ignore it.)
1.8×10-4    = (x+0.02)(x)
x2+0.02x-1.8×10-4 = 0
Applying quadratic formula; x = -0.02+√{(0.02)2-4×1.8×10-4}
= 0.033-0.020/2 = 0.0065
Therefore, concn of NH3 at equilibrium = x+0.020 = 0.0265
concn of H2S at equilibrium = x = 0.0065
So, option b is correct

For the following gaseous phase equilibrium,

Kp is found to be equal to Kx (Kx is equilibrium constant when concentration are taken in terms of mole fraction. This is attained when pressure is 
  • a)
    1 atm
  • b)
    0.5 atm
  • c)
    2 atm
  • d)
    4 atm
Correct answer is option 'A'. Can you explain this answer?

Preeti Iyer answered
The correct answer is Option A.
Kp = Equilibrium constant in terms of partial pressure
Kc = Equilibrium constant in terms of concentration
Kx = Equilibrium constant in terms of mole fraction
             Kp = KcRTΔn ---(1)
           Kp = K * (Pt)Δn ---(2)
a)   1 atm
Given PCl5 (g) ---> PCl3 (g) + Cl2 (g)
  Δn = 2 – 1
Given Kp = Kx  
From (2)
          Kp = Kx when PT = 1

Variation of log Kp with temperature. 1/T is given by for the equilibrium.
NH4HS (s)   NH3(g) + H2S
Q. The equilibrium is displaced in forward side on
  • a)
    increasing temperature and decreasing pressure
  • b)
    increasing temperature and pressure both
  • c)
    decreasing temperature and pressure both
  • d)
    decreasing temperature and increasing pressure
Correct answer is option 'A'. Can you explain this answer?

Sushil Kumar answered
The correct answer is Option A.
When ammonia is added after equilibrium is established, the partial pressure of ammonia will increase.When the temperature of an endothermic reaction is increased, the equilibrium will shift in the forward direction so that the heat is absorbed which will nullify the effect of increased temperature. Hence, the partial pressure of ammonia will increase.When the volume of the flask is increased, the pressure will decrease.

Passage I
Solid ammonium chloride is in equilibrium with ammonia and hydrogen chloride gases

0.980 g of solid NH4CI is taken in a closed vessel of 1 L capacity and heated to 275° C.
Q. Percentage decomposition of the original sample is
  • a)
    24.81%
  • b)
    6.24%
  • c)
    3.12%
  • d)
    12.13%
Correct answer is option 'D'. Can you explain this answer?

Knowledge Hub answered
The state of HCl is given wrong. It will be in gaseous state.
So, the reaction be like;-
NH4Cl(s)  ⇌  NH3(g) + HCl(g)        kp = 1.00×10-2 at 275° C
Kp = kc(RT)2
1.00×10-2 = kc(0.0821×548)2
Or kc = 4.94×10-6
                          NH4Cl(s)  ⇌  NH3(g) + HCl(g)
Initial  1                     -             -
At eqm 1-x                  x            x 
Kc = x2
x = √(4.94×10-6)
=  2.22×10-3
Therefore, NH4Cl dissociated at eqm = 2.22×10-3 × 53.5 = 0.118
%age decomposition = 0.118/0.980×100 = 12.13%

Which of the following on the addition will cause deep red colour to disappear?
  • a)
    AgNO3
  • b)
    HgCI2
  • c)
    H2C2O4
  • d)
    H2O
Correct answer is option 'A,B,C'. Can you explain this answer?

Neha Joshi answered
The correct answers are option A,B,C
Because of silver salt ie . silver nitrate with powerful germicidal activity
Ag+ + SCN → AgSCN↓
Hg2+ + SCN− → Hg(SCN)2
4Fe3+ + 3(COO)2−2 → 2Fe2(COO)3

Direction (Q. Nos. 21) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)
Q. For the equilibrium in gaseous phase in 2 L flask we start with 2 moles of SO2 and 1 mole of O2 at 3 atm, 
When equilibrium is attained, pressure changes to 2.5 atm. Hence, equilibrium constant Kc is
    Correct answer is '4'. Can you explain this answer?

    Om Desai answered
    The correct answer is 4
    2SO2(g) + O2(g) ⇋ 2SO3
    Initial moles      2            1
    At equilibrium 2 - 2x     1 - x    2x
    Net moles at equilibrium  =  2 - 2x + 1 - x + 2x
    =(3 - x)moles
    Initial:
             moles = 3, 
        Pressure = 3 atm,
          Volume = 2L,
                 PV = nRT
              3 x 2 = 3RT  -------- 1
    At equilibrium
         Moles = 3 - x,
    Pressure = 2.5 atm
      Volume = 2L
            P‘V = n’RT ---------- 2
    Divide eqn  2 by 1

    ⇒2.5 = 3 - x
    ⇒x = 0.5

    For the equilibrium,

    at 1000 K. If at equilibrium pCO = 10 then total pressure at equilibrium is 
    • a)
      6.30 atm
    • b)
      0.63 atm
    • c)
      6.93 atm
    • d)
      69.3 atm
    Correct answer is option 'C'. Can you explain this answer?

    Lavanya Menon answered
    C(s) + CO2(g) <=========> 2CO(g)
    Kp = pCO2/pCO2
    GIven Kp = 63 and pCO = 10pCO2
    Putting the value of pCO in above equation,
    63 = 100(pCO2)2/pCO2
    Or pCO2 = 0.63
    pCO = 6.3
    Therefore, total pressure = 6.3+0.63 = 6.93 atm

    Equilibrium can be attained i
    • a)
      all types of system
    • b)
      closed system
    • c)
      open system
    • d)
      isolated system
    Correct answer is option 'B'. Can you explain this answer?

    Preeti Iyer answered
    The equilibrium state can only be reached if the chemical reaction takes place in a closed system. Otherwise, some of the products may escape, leading to the absence of a reverse reaction. (Note that in the diagrams under "Characteristics of Chemical Equilibrium," all reactions are in closed systems.)

    Can you explain the answer of this question below:

    Equilibrium reactions are found in large scale in production of

    • A:

      ammonia

    • B:

      sulfuric acid

    • C:

      lactic acid

    • D:

      both A and B

    The answer is d.

    Shreya Gupta answered
    An understanding of equilibrium is important in the chemical industry. Equilibrium reactions are involved in some of the stages in the large-scale production of ammonia, sulfuric acid and many other chemicals. 

    Direction (Q. Nos. 21) Choice the correct combination of elements and column I and coloumn II  are given as option (a), (b), (c) and (d), out of which ONE option is correct.
    The progress of the reaction  with time t is shown below.


    Match the parameters in Column l with their respective values in Column II.


    Codes
          
    • a)
      a
    • b)
      b
    • c)
      c
    • d)
      d
    Correct answer is option 'A'. Can you explain this answer?

    Rajesh Gupta answered
    The correct answer is Option A.
    Loss in concentration of A in I hour = = 0.1
    Gain in concentration of B in I hour =0.2
    (i) ∵0.1 mole of A changes to 0.2 mole of B in a given time and thus, n=2
    (ii) Equilibrium constant,
    = 1.2mollitre−1
    (iii) Initial rate of conversion of A = changes in conc. of A during I hour = 
    = 0.1 mol litre−1hour−1
    (iv) ∵ Equilibrium is attained after 5 hr, where [B]=0.6 and [A]=0.3

    Direction (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four
    choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
    Q. For the following equilibrium,

    Vapour density is found to be 100 when 1 mole of PCI5 is taken in 10 dm3 flask at 300 K. Thus, equilibrium pressure is
    • a)
      1.00 atm
    • b)
      4.92 atm
    • c)
      2.46 atm
    • d)
      2.57 atm
    Correct answer is option 'D'. Can you explain this answer?

    Knowledge Hub answered
    Given, vapour pressure = 100
    So, molecular weight = 200
    Degree of dissociation;α = 1/(n-1)[Mtheo-Mobs/Mobs]

    α   =   1/(2-1)×[208.5-200/208.5]
    α  = 0.0425
    PCl5(s)    ⇌    PCl3(g) + Cl2(g)
    At start      1                 -           -     -
    At eqm.  (1-0.0428)   (0.0428)  (0.0428)
    Total moles at eqm. = 1.0428
    Total pressure at equilibrium; pV = nRT
    p = 1.0428×0.0821×300/10
    p  = 2.5676 atm

    At 700 K and 350 bar, a 1 : 3 mixture of N2(g) and H2(g) reacts to form an equilibrium mixture containing X (NH3)= 0.50. Assuming ideal behaviour Kp for the equilibrium reaction, 
    • a)
      2.03x 10-4
    • b)
      3.55x 10-3
    • c)
      1.02 x 10-4
    • d)
      3.1 x 10-4
    Correct answer is option 'D'. Can you explain this answer?

    Knowledge Hub answered
    The correct answer is option A
    2.03x 10-4
    The given equation is :-
     N2​(g)+3H2​(g) ⇌ 2NH3​(g)
    Initial moles : 1             3         0
    At eqm ;       (1−x)    (3−3x)   (2x)               
    (let)
    Total moles of equation
     =1 − x + 3 − 3x + 2x = (4−2x)
    Now, X(NH3​) = 
    ⇒ 2x = 2 − x
    ⇒ 3x = 2 ⇒ x = 0.66 = 
    32​
    Now, at equation, moles of N2​= 1/3, moles of NH3​ = 4/3
                 moles of H2 ​ =3 − 2 = 1

     

    The solubility product expression for silver(I) sulphide, using x to represent the molar concentration of silver(I) and y to represent the molar concentration of sulphide, is formulated as:
    • a)
      x2y2
    • b)
      xy3
    • c)
      x2y
    • d)
      xy2
    Correct answer is option 'C'. Can you explain this answer?

    Om Desai answered
    The correct answer is Option C.
    The ionization equilibrium of silver (I) sulfide is 
    Ag2S⇌2Ag+ +S2
    The solubility product expression for silver (I) sulfide is KSP[Ag+]2 [S2].
    But [Ag+]2 = x and [S2−] = y.
    Hence the expression for the solubility product becomes KSP[Ag+]2 [S2−] = x2y.

    Role of NH4Cl in qualitative analysis of third group cations
    • a)
      to provide basic medium
    • b)
      to provide acidic medium
    • c)
      to increase the degree of dissociation of NH4OH
    • d)
      to suppress the degree of dissociation of NH4OH
    Correct answer is option 'D'. Can you explain this answer?

    Raghav Bansal answered
    The correct answer is option D
    Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte, which provides an ion common to that provided by weak electrolyte.
    The NH4OH is a weak base and it does not ionise completely. Thus, due to presence of common ion NH4+ in NH4Cl, it suppresses the ionisation of weak base NH4OH in order to decrease the OH- concentration so that higher group cations will not get precipitated.
    Thus the pair NH4​OH+NH4​Cl shows a common ion effect. 
    Ammonium chloride suppresses the ionization of ammonium hydroxide
    .

     We eat a variety of foods still pH of our blood does not change every time. The reason is
    • a)
      Strong bases in the blood donot let pH change
    • b)
      Stomach wall is resistant
    • c)
      There are buffers in the blood which resist pH change
    • d)
      Strong acids in the blood donot let pH change
    Correct answer is option 'C'. Can you explain this answer?

    Pooja Shah answered
    The correct answer is option C
    pH of blood remains constant because of the buffer system present in the blood. Acid-base buffers confer resistance to a change in the pH of a solution when hydrogen ions (protons) or hydroxide ions are added or removed. An acid-base buffer typically consists of a weak acid, and its conjugate base (salt).It used to neutralized the extra added protons or OH- in blood.The buffer for maintaining acid-base balance in the blood is the carbonic-acid-bicarbonate buffer.So pH of blood remains even after eating spicy food.

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