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Derive the values of K_(C) and K_(P) for the synthesis of HI. |
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Answer» Solution :`H_(2) (g) + I_(2) (g) HARR 2 HI (g)` Let US consider the formation of HI in which .a. moles of hydrogen , .b. moles of iodine GAS are allowed to react in an container of volume .V. . Let .x. moles of each of `H_(2)` and `I_(2)` react together to form 2X moles of HI.  Applying law of MASS action, `K_(C) = ([HI]^(2))/([H_(2)][I_(2)])` `K_(C) = (((2x)/(V))^(2))/(((a - x))/(V) ((b- x))/(V)) = (4x^(2))/((a - x) (b - x))` Calculation of `K_(p) , K_(P) = K_(C) . RT^(Deltan_(g))` Here`Deltan_(g) = n_(p) - n_(r) = 2 - 2 = 0` Hence , `"" K_(P) = K_(C)` `K_(P) = (4x^(2))/((a - x) (b - x))` |
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