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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 iodine at `350^(@)C.` Calculate the equilibrium constant.

Text Solution

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The reaction is
`{:(,H_2+, I_2 hArr , 2HI),("Equi conc.", 0.92,0.78,6.0):}`
`K_c=[HI]^2/([H_2][I_2])`
`=(6.0)^2/((0.92)xx(0.78))`=50.17
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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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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 :

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  • 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

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