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Equilibrium Test 12

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Equilibrium Test 12
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  • Question 1
    1 / -0

    For the reaction equilibrium :

    N2O4 (g)  2NO2(g) ; the concentration of N2O4 and NO2 at equilibrium are 4.8 × 10-2 and 1.2 × 10-2 mol/L respectively. The value of Kc for the reaction is :

    Solution


    conc. of N2O4 = 4.8 × 10–2 mol/lit
    NO2 = 1.2 × 10–2 mol/lit

  • Question 2
    1 / -0

    What is the equilibrium constant for the reaction P4(s) + 5O2(g)  P4O10(s) :

    Solution


  • Question 3
    1 / -0

    The equilibrium constant for the reaction :

    N2(g) + O2(g)  2NO(g) at temperature

    T is 4 × 10-4. The value of Kc for the reaction.

    NO(g)  N2(g) + O2(g) at the same temperature is :

    Solution


  • Question 4
    1 / -0

    The equilibrium constant for the given reaction :

    SO3(g)  SO2(g) + O2(g); Kc = 4.9 × 10-2

    The value of Kc for the reaction :

    2SO2(g) + O2(g)  2SO3(g), will be

    Solution


  • Question 5
    1 / -0

    For the following three reactions 1, 2 and 3, equilibrium constants are given :

    (1) CO(g) + H2O(g)CO2(g)+H2(g) ; K1 

    (2) CH4(g)+H2O(g)CO(g)+3H2(g) ; K2

    (3) CH4(g)+2H2O(g)CO2(g)+4H2(g) ; K3

    Which of the following relations is correct ?

    Solution

    CO(g) + H2O(g) ⇋  CO2(g)+H2(g) ; K1  --------(i)
    CH4(g)+H2O(g)  ⇋ CO(g)+3H2(g) ; K2  --------(ii)
    CH4(g)+2H2O(g)  ⇋ CO2(g)+4H2(g) ; K3 ------(iii)
    We can see that by adding eqn i and ii, we get eqn iii. So for K, on addition of reaction, the k value gets multiplied.
    Therefore, k3 = k1k2

  • Question 6
    1 / -0

    Consider following reactions in equilbrium with equilibrium concentration 0.01M of every species

    (I) PCl5(g)  PCl3(g) + Cl2(g)

    (II) 2HI(g)  H2(g) + I2(g)

    (III) N2(g) + 3H2 (g)  2NH3(g)

    Extent of the reactions taking place is :

    Solution




  • Question 7
    1 / -0

    A definite amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. NH4HS decomposes to give NH3 and H2S and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is :

    Solution



  • Question 8
    1 / -0

    For the reaction 3A(g) + B(g)  2C(g) at a given temperature, Kc = 9.0. What must be the volume of the flask, if a mixture of 2.0 mol each of A, B and C exist in equilibrium ?

    Solution



  • Question 9
    1 / -0

    Sulfide ion in alkaline solution reacts with solid sulfur to form polysulfide ions having formulas S22-, S32-, S42- and so on. The equilibrium constant for the formation of S22- is 12 (K1) & for the formation of S32- is 132 (K2), both from S and S2-. What is the equilibrium constant for the formation of S32- from S22- and S

    Solution


  • Question 10
    1 / -0

    1 mole N2 and 3 mol H2 are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained.

    N2(g) + 3H2(g)  2NH3(g). The equilibrium constant Kp for dissociation of NH3 is :

    Solution





     

  • Question 11
    1 / -0

    One mole of N2O4(g) at 300 K is left in a closed container under one atm. It is heated to 600 K when 20% by mass of N2O4 (g) decomposes to NO2(g). The resultant pressure is :

    Solution


  • Question 12
    1 / -0

    For the following gases equilibrium. N2O4(g)  2NO2(g) Kp is found to be equal to Kc. This is attained when temperature is

    Solution


  • Question 13
    1 / -0

    For the reaction :

    CO(g) + O2(g)  CO2(g), Kp/Kc is :

    Solution



  • Question 14
    1 / -0

    For the reaction :

    2NO2(g)  2NO(g) + O2(g) Kc = 1.8 × 10-6 at 184°C and R = 0.083 JK-1mol-1. When Kp and Kc are compared at 184°C, it is found that :

    Solution



  • Question 15
    1 / -0

    PCl5 dissociation a closed container as :

    PCl5(g)  PCl3(g) + Cl2(g)

    If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is α, the partial pressure of PCl3 will be :

    Solution

    PCl5 dissociation a closed container

  • Question 16
    1 / -0

    For the reaction : 2HI (g)  H2(g) + I2(g), the degree of dissociated (α) of HI(g) is related to equilibrium constant Kp by the expression :

    Solution





  • Question 17
    1 / -0

    The equilibrium constant for the reaction

    A(g) + 2B(g)  C(g) is 0.25 dm6 mol-2. In a volume of 5 dm3, what amount of A must be mixed with 4 mol of B to yield 1 mol of C at equilibrium

    Solution



  • Question 18
    1 / -0

    For the reaction

    A(g) + 2B(g)  C(g) + D(g); Kc = 1012 If the initial moles of A, B, C and D are 0.5, 1, 0.5 and 3.5 moles respectively in a one litre vessel. What is the equilibrium concentration of B ?

    Solution





  • Question 19
    1 / -0

    The equilibrium constant Kc for the reaction,

    A(g) + 2B(g)  3C(g) is 2 × 10-3

    What would be the equilibrium partial pressure of gas C if initial pressure of gas A & B are 1 & 2 atm respectively.

    Solution





  • Question 20
    1 / -0

    A 20.0 litre vessel initially contains 0.50 mole each of H2 and I2 gases. These substances react and finally reach an equilibrium condition. Calculate the equilibrium concentration of HI if Keq = 49 for the reaction H2 + I2  2HI.

    Solution




  • Question 21
    1 / -0

    A vessel of 250 litre was filled with 0.01 mole of Sb2S3 and 0.01 mole of H2 to attain the equilibrium at 440°c as

    Sb2S3(s) + 3H2(g)  2Sb(s) + 3H2S(g).

    After equilibrium the H2S formed was analysed by dissolving it in water and treating with excess of Pb2+ to give. 1.195 g of PbS (Molecular weight = 239) precipitate.

    What is value of Kc of the reaction at 440°C ?

    Solution



  • Question 22
    1 / -0

    The equilibrium constant for the reaction

    CO(g) + H2O(g)  CO2(g) + H2(g) is 3 at 500 K. In a 2 litre vessel 60 gm of water gas [equimolar mixture of CO(g) and H2(g)] and 90 gm of steam is initially taken.

    What is the equilibrium concentration of H2(g) at equilibrium (mole/L) ?

    Solution







  • Question 23
    1 / -0

    At 87°C, the following equilibrium is established

    H2(g) + S(s)  H2S(g) Kp = 7 × 10-2

    If 0.50 mole of hydrogen and 1.0 mole of sulfur are heated to 87°C in 1.0 L vessel, what will be the partial pressure of H2S at equilibrium ?

    Solution





  • Question 24
    1 / -0

    At certain temperature (T) for the gas phase reaction

    2H2O(g)+2Cl2(g) 4HCl(g)+O2(g)       ; Kp =12×108 atm

    If Cl2, HCl & O2 are mixed in such a manner that the partial pressure of each is 2 atm and the mixture is brought into contact with excess of liquid water. What would be approximate partial pressure of Cl2 when equilibrium is attained at temperature (T) ?

    [Given : Vapour pressure of water is 380 mm Hg at temperature (T)]

    Solution




  • Question 25
    1 / -0

    At 675 K, H2(g) and CO2(g) react to form CO(g) and H2O(g), Kp for the reaction is 0.16. If a mixture of 0.25 mole of H2(g) and 0.25 mol of CO2 is heated at 675 K, mole% of CO(g) in equilibrium mixture is :

    Solution





  • Question 26
    1 / -0

    In which of the following reactions, increase in the pressure at constant temperature does not affect the moles at equilibrium.

    Solution

  • Question 27
    1 / -0

     Change in volume of the system does not alter the number of moles in which of the following equilibrium

    Solution

  • Question 28
    1 / -0

    The conditions favourable for the reaction :

    2SO2(g)+O2(g)  2SO3(g) ; ΔH° = -198 kJ  are :

    Solution


    Exothermic reaction favourable at low temperature Δng < 0
    ⇒ favourable at high pressure

  • Question 29
    1 / -0

    The exothermic formation of ClF3 is represented by the equation :

    Cl2(g) + 3F2(g)  2ClF3(g)  ;  ΔH = -329 kJ

    Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3 :

    Solution


    ⇒ Exothermic low temperature
    ∴ Addition of reactant Δng < 0
    ⇒ Increase in pressure lower the volume.

  • Question 30
    1 / -0

    Densities of diamond and graphite are 3.5 and 2.3 gm/mL. respectively and for

    C(diamond)  C (graphite) ; ΔH= -1.9 kJ/mole favourable conditions for formation of diamond are

    Solution


    It is exothermic reaction so it favours low temperature density of graphite is less than diamondlow pessure.

    ⇒ but for formation of diamond Reverse condition high temperature & high pressure.

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