BotanyIn the process  (in aq medium), initially there are 2 mole I2 & 2 mole I- . But at equilibrium, due to addition of AgNO3(aq), 1.75 mole yellow ppt is obtained. KC for the process is (Vflask=1 dm3) nearly

In the process 

large {I_2} + ,{I^ - } rightleftharpoons ,I_3^ -

 (in aq medium), initially there are 2 mole I2 & 2 mole I- . But at equilibrium, due to addition of AgNO3(aq), 1.75 mole yellow ppt is obtained. KC for the process is (Vflask=1 dm3) nearly

  1. A

    0.08

  2. B

    0.02

  3. C

    0.16

  4. D

    0.12

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    Solution:

    large {I_2}left( {aq} right) + {I^ - }left( {aq} right) rightleftharpoons I_3^ - left( {aq} right)

    Initial moles of I2 = 2

    Initial moles of I - = 2

    Volume of the vessel = 1 litre

    Yellow ppt formed by the addition of AgNO3 is AgI

    Number moles of AgI precepitated at equilibrium = 1.75

    large therefore

    number of moles of I- at equilibrium = 1.75

    Number of moles of I- reacted = (2 -1.75) = 0.25

    According to stoichiometry

    1 mole of I2 reacts with 1 mole of I-

    x mole of I2 reacts with 0.25 moles of I-

    large therefore boxed{x = 0.25}

    Similarly moles of 

    large I_3^ -

    formed due to the reaction of 0.25 moles each of I2 & I- will 0.25

     
    large {I_2}left( {aq} right)
    large {I^-}left( {aq} right)
    large rightleftharpoons
    large I_3^ - left( {aq} right)
    Initial moles22 -
    Moles at equilibrium(2 - 0.25)(2 - 0.25) 0.25
    Equilibrium concentration
    large frac{{1.75}}{1}
    large frac{{1.75}}{1}
     
    large frac{{0.25}}{1}
    large {K_C} = frac{{left[ {I_3^ - } right]}}{{left[ {{I_2}} right]left[ {{I^ - }} right]}}
    large {K_C} = frac{{0.25}}{{1.75 times 1.75}}
    large {K_C} = frac{{left( {frac{1}{4}} right)}}{{left( {frac{7}{4}} right)left( {frac{7}{4}} right)}}
    large {K_C} = frac{4}{{49}}
    large boxed{{K_C} approx 0.08{M^{ - 1}}}
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