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In a reaction between hydrogen and iodin...

In a reaction between hydrogen and iodine `6.84 mol` of hydrogen and `4.02` mol of iodine are found to be in equilibrium with `42.85` mol of hydrogen iodide at `350^(@)C.` Calculate the equilibrium constant.

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The correct Answer is:
B, D
`{:(,H_(2),+,I_(2),hArr,2HI),("A equilibrium",0.34,,4.02,,42.85):}`
`K=([HI]^(2))/([H_(2)][I_(2)])=([42.85]^(2))/([6.34][4.02])=72.042`
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Knowledge Check

  • An equilibrium system for the reaction between hydrogen and iodine to give hydrogen iodide at 765 K in a 5 litre volume contains 0.4 mole of hydrogen, 0.4 mole of iodine and 2.4 moles of hydrogen iodide, The equilibrium constant for the reaction is : H_2+I_2 hArr 2HI , is :

    A
    `36.0`
    B
    `15.0`
    C
    `0.067`
    D
    `0.28`

  • An equilibrium system for the reaction between hydrogen and iodien to give hydrogen iodide at 765 K in a 5 litre volume contains 0.4 mole of hydrogen iodide. The equilibrium constant for the reaction H_(2) + I_(2)hArr 2HI is

    A
    `36.0`
    B
    `15.0`
    C
    `0.067`
    D
    `0.028`

  • 1.50 moles each of hydrogen and iodine were placed in a sealed 10 litre container maintained at 717 K. At equilibrium 1.25 moles each of hydrogen and iodine were left behind. The equilibrium constant, K_(c) for the reaction , H_(2)(g)+I_(2)(g) hArr 2Hl(g) at 717 K is

    A
    `0.4`
    B
    `0.16`
    C
    25
    D
    50
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