All Exams  >   ACT  >   Chemistry for ACT  >   All Questions

All questions of Equilibrium for ACT 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

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?

Raghav Bansal answered
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

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?

Preeti Khanna answered
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.
 

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?

Naina Sharma 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.

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

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 

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.
 

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%

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?

Krishna Iyer answered
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

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?

Preeti Iyer 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

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. The equilibrium which is not affected by volume change at constant temperature is
  • a)
  • b)
  • c)
    N2O4(g)  2NO2(g)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Preeti Iyer answered
As the number of moles of reactants and products is the same. Hence, they both will occupy equal volume at the same temperature.
Therefore, the reaction A i.e., H2 (g) +I2 (g) ⇌ 2HI (g) is not affected by change in pressure and volume.

For the reaction, 
if Kp = Kc (RT)X, when the symbols have usual meaning, the value of x is (assuming ideality)
[jee Main 2014]
  • a)
    -1
  • b)
    -1/2
  • c)
    +1/2
  • d)
    +1
Correct answer is option 'B'. Can you explain this answer?

Gaurav Kumar answered
The correct answer is Option B.
SO2(g) + 1/2O2(g) ⇌ SO3(g)
KP = KC(RT)Δn
Δn= no. of gaseous moles of product minus no. of gaseous moles of reactant
Δn = 1−1−1/2
∴Δn = −1/2
 

At 273 K and 1 atm, 1 L of N2O4 (g) decomposes to NO2(g)a s given,
At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N2O4 (g) is thus, 
  • a)
    25%
  • b)
    50%
  • c)
    66.66%
  • d)
    33.33%
Correct answer is option 'D'. Can you explain this answer?

Suresh Reddy answered
Let the initial volume of N2O4 be x and initial volume of NO2 is 0
If the degree of dissociation is a, then the final volume of N2O4 is x(1−a) and NO2 is 2ax.
Initial
It equilibrium
N2O4            ⟶              2NO2
x                                       0
x(1−a)                               2ax
Total initial volume =x+0=x
Final volume =x(1−a)+2ax=x+ax=x(1+a)
It is given that the initial volume is 25% less than the final volume
x=0.75×(1+a)
1+a=1.33
a=0.33
So %age dissociation = 33.33%

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?

Naina Sharma 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

The concentration of the oxides of nitrogen are monitored in air-pollution reports. At 25°C, the equilibrium constant for the reaction,

 is 1.3 x 106 and that for 

is 6.5 x 10-16 (when each species is expressed in terms of partial pressure).
For the reaction,

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

Raghav Bansal answered
Given equations are 
NO (g) + ½ O2 (g) ⇌ NO2 (g) ----------(i) k1 = 1.3×106
And ½ N2 (g) + ½ O2 (g)     ⇌     NO(g) ----------(ii) k2 = 6.5×10-16
To get the reaction, N2(g) + 2O2(g) ⇌ 2NO2(g) ----------(iii) k3
We multiply eqn (i) and eqn (ii) by 2 and adding both reaction, we get eqn (iii)
k3 = k12×k22
    = (1.3×106)2×(6.5×10-16)2
    = 1.69×1012×42.25×10-32
    = 7.14×10-19

When hydrochloric acid is added to cobalt (II) nitrate solution at room temperature, the following reaction takes place
Q. The solution is blue at room temperature. However, it turns pink when cooled in a freezing mixture. Based upon this information, which of the following expression is correct for the forward reaction?
  • a)
    ΔH > 0
  • b)
    ΔH < 0
  • c)
    ΔH = 0
  • d)
    The sign of ΔH can’t be predicted based on the given information
Correct answer is 'A'. Can you explain this answer?

Preeti Iyer answered

According to the given information, the solution turns pink when cooled in freezing mixture. So, the equilibrium has shifted in backward direction. So we can say that the reaction is endothermic and ∆H for the reaction is greater than zero. This is because, endothermic reactions have ∆H > 0 and they proceed in reverse direction. 

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.)

Combustion of CO(g)can be increased in the following reaction by
  • a)
    decreasing volume at constant temperature
  • b)
    adding argon gas
  • c)
    adding O2 gas
  • d)
    decreasing pressure at constant temperature
Correct answer is 'A,B,C'. Can you explain this answer?

Gaurav Kumar answered

For option a; with decrease in volume or with increase in pressure, reaction shifts towards less no. of moles. So, here combustion of CO will increase.
   For option b; adding argon at constant volume doesn’t make any effect on equilibrium. Also, if we add argon at constant pressure, the reaction will shift towards more no. of moles and so the combustion of CO will decrease.
   For option c; adding O2 will shift the reaction in forward direction (according to Le Chatelier principle), so combustion of CO will increase.
   For option d; with decrease in pressure or increases in volume, reaction shifts towards more no. of moles and so combustion of CO will decrease.
 

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) 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

    H2S (g) initially at a pressure of 10.0 atm and a temperature of 800 K, dissociates as
    At equilibrium, the partial pressure of S2 vapour is 0.020 atm . Thus, Kp is 
    • a)
      3.23x 10-7
    • b)
      6.45x 10-7
    • c)
      1.55x 106
    • d)
      6.20x 107
    Correct answer is option 'A'. Can you explain this answer?

    Geetika Shah answered
    The correct answer is Option A.
        
                     2H2S(g) ⇌ 2H2(g) + S2(g)

    Pressure
    at t=0           Pi                −           −
    at eqm       Pi−P            2P          P
    as P=0.02    thus Pi−P=10−0.02
         Pi=10                     2P=0.04

    Kp = 3.23×10−7 atm.

    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

    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

    Chapter doubts & questions for Equilibrium - Chemistry for ACT 2025 is part of ACT exam preparation. The chapters have been prepared according to the ACT exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for ACT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

    Chapter doubts & questions of Equilibrium - Chemistry for ACT in English & Hindi are available as part of ACT exam. Download more important topics, notes, lectures and mock test series for ACT Exam by signing up for free.

    Chemistry for ACT

    110 videos|124 docs|114 tests

    Top Courses ACT

    Signup to see your scores go up within 7 days!

    Study with 1000+ FREE Docs, Videos & Tests
    10M+ students study on EduRev