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Chemical Equilibrium Test - 10
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Chemical Equilibrium Test - 10
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  • Question 1/10
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    The equilibrium, is attained at 25oC in a closed container and an inert gas He is introduced. Choose the correct statement.

    Solutions

    Since formation of NH3 is carried out in a closed container. When Inert gas (He) is introduced to the system in equilibriumatconstant volume then no any change occurs in the partial pressures of N2, 3H2 and 2NH3.

     

  • Question 2/10
    1 / -0

    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 KP is:

    Solutions

    CO2(g) +  C(s) ⇌ 2CO(g)

    at t = 0        0.5 atm      – 0

    at eqb.        (0.5 x)atm  – 2x atm

    As graphite is a pure solid, its concentration is invariable (or constant) during the course of reaction, it will not be a part of Kp 's expression. At equilibrium,

    Total pressure, PT = 0.8 atm

    PT =  0.5 − x + 2x = 0.8atm

    0.5 + x = 0.8

    x = 0.3 atm

     

  • Question 3/10
    1 / -0

    40% of a mixture of 0.2 mol of N2 and 0.6 mol of H2 react to give NH3 according to the equation, N2(g) + 3H2(g) ⇌ 2NH3(g), at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

    Solutions

     

  • Question 4/10
    1 / -0

    One mole of ethanol is treated with one mole of ethanoic acid at 25°C. One-fourth of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be

    Solutions

    C2H5OH + CH3COOH ⇌ CH3COOC2H5 + H2O

     

  • Question 5/10
    1 / -0

    One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N2O4 (g) 23 decomposes to NO2(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N2O4 (g) is

    Solutions

    N2O4(g) ⇌ 2NO2(g)

    Total pressure = 2 – 2α + 4α

    = 2 + 2α

    2 + 2α = 2.4 (Given)

    α = 0.2

    ∴ Percentage dissociation = 20%

     

  • Question 6/10
    1 / -0

    K for the synthesis of HI(g) is 50. The degree of dissociation of HI is

    Solutions

     

  • Question 7/10
    1 / -0

    For the reaction PCl3 (g) + Cl2 (g) ⇌ PCl5 (g), the value of Kp at 250°C is 0.61 atm–1. The value of Kc at this temperature will be

    Solutions

     

  • Question 8/10
    1 / -0

    In the reaction A2(g) + 4B2(g) ⇌ 2AB4(g) , ∆H > 0. The decomposition of AB4 (g) will be favoured at

    Solutions

    2AB4 (g) ⇌ A2 (g) + 4B2 (g) ∆H = –ve

    It is an exothermic reaction and hence favoured at low temperature. ∆n for the reaction is +3. Therefore low pressure will favour the forward reaction

     

  • Question 9/10
    1 / -0

    In a system, A (s) ⇌ 2B (g) + 3C (g)

    If the conc. of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to

    Solutions

    A(s) ⇌ 2B(g) + 3C(g)

    Let, x and y be the concentrations of B and C at equilibrium respectively.

    ∴ Kc = x2y3                           …. (1)

    Now, the concentration of C is changed from y to y′ such that y′ = 2y.

    If x′ is the new concentration of B

    ∴ Kc = (x′)2 (y′)3 = (x′)2(2y)3     …. (2)

    From Eqs. (1) and (2)

    (x′) 2 (8y3 ) = x2y3

    ∴Equilibrium concentration of B changes to 1/2√2 times the original value.

     

  • Question 10/10
    1 / -0

    For the following three reactions (i), (ii) and (iii), 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

    What is the relation between equilibrium constants of the three reactions?

    Solutions

    Reaction (iii) can be obtained by adding reactions (i) and (ii) therefore K3 = K1. K2.

     

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