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Chemical Equilibrium Test - 8

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Chemical Equilibrium Test - 8
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Weekly Quiz Competition
  • Question 1
    1 / -0

    The decomposition reaction, 4HNO3 (g) ⇌ 4NO2 (g) + 2H2O (g) + O2 (g) is started with pure HNO3 (g). If p is the tot al pressure at equilibrium, then

    Solution

     

  • Question 2
    1 / -0

    For the equilibrium, SO2 (g) + 1/2 O2 (g) = SO3 (g); ∆H = −99.0 kJ. The extent of forward reaction can be increased by

    Solution

    Equilibrium shifts in the forward direction by removing SO3 , decreasing the temperature and increasing the pressure. Catalyst does not disturb the equilibrium. Therefore, extent of forward reaction increases by removing SO3 .

     

  • Question 3
    1 / -0

    Consider the following reaction, and choose the correct option.

    Solution

    The increase in temperature will favour the backward reaction, as the reaction is exothermic. With the increase of pressure on physical equilibria, the equilibrium will shift in that direction where the density is more. As the density of diamond is greater than graphite, thus increase in pressure will also favour backward reaction.

     

  • Question 4
    1 / -0

    For the equilibrium, 

    which of the following statement is true?

    Solution

    For heterogeneous physical equilibrium, with the increase of pressure, equilibrium shifts in the direction of physical state having higher density. This means that for the equilibrium, H2O (s) ⇌ H2O (I), more ice would melt on increasing the pressure of the system as density of water is more than ice.

     

  • Question 5
    1 / -0

    Ammonia gas at 76 cm Hg pressure was connected to a manometer. After sparking in the flask, ammonia partially dissociated as follows 

    The level in the mercury column of the manometer was found to show the difference of 18 cm. The partial pressure of H2(g) at equilibrium is

    Solution

     

  • Question 6
    1 / -0

    For the equilibrium A(g) + 2B(g) ⇌ AB2 (g) the value of equilibrium constant at 300 k is 3 ×10-3 mole-2 L2 and value of equilibrium constant at 500 k is 5 ×10-5 mole-2 L2

    What is the sign of enthalpy change for this reaction

    Solution

    Where, K1 is equilibrium constant at temperature, T1 and K2 is equilibrium constant at temperature T2 .

    then, we can either substitute the values in equation and solve the problem or we can apply a simple mathematical logic and find the answer.

    Alternately, You can remember this fact and apply it in questions directly. The fact is that the value of Keq decreases on increase in temperature and vice-versa in case of exothermic reaction. The value of Keq increases on increase in temperature in case of endothermic reaction.

     

  • Question 7
    1 / -0

    Following graph is plotted for the reaction:

    2A (g) + B (g) ⇌ 2C (g)

    In the graph the equilibrium state is reached at

    Solution

    At equilibrium, rate of forward reaction is equal to rate of backward reaction and at this situation concentration of each reactant and product becomes constant. Therefore, in graph-I at t3 equilibrium is attained.

     

  • Question 8
    1 / -0

    At –50°C, the self-ionization constant (ion product) of NH3 is KNH3 = 10–30 M2. How many amide ions are present per mm3 of pure liquid ammonia?

    Solution

     

  • Question 9
    1 / -0

    The approach to the following equilibrium was observed kinetically from both directions.

    At 25°C, it was found that

    The value of Kc (equilibrium constant) for the complexation of the fourth Cl by Pt(II) is

    Solution

    At equilibrium, the rate of change of concentration of any species (reactant or product) is zero.

    The equilibrium reaction for the complexation of the fourth Cl by Pt(II) is

    Since, water is a solvent in the given reaction, so its concentration remains constant.

    In the problem, expression of rate of change of concentration of is given, which at equilibrium is zero.

     

  • Question 10
    1 / -0

    20 ml of O2 contracts to 17 ml when subjected to silent electric discharge in an ozoniser. What is the volume of O3 formed at equilibrium?

    Solution

    ∴ x = 3

    ∴ Volume of O3 formed = 2 × 3 = 6 ml

     

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