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For the reaction , H(2)(g) + I(2) (g) hA...

For the reaction , `H_(2)(g) + I_(2) (g) hArr 2HI(g), K = 55.3 " at " 699 K.` In a mixture consisting of 0.70 atm of HI and 0.02 atm each of `H_(2)` and `I_(2)` at 699 K, will there be any net reaction ? If so will HI be consumed or formed ?

Text Solution

Verified by Experts

The reaction quotiont `(Q_(p))` can be calculated as :
`Q_(p) = (P_(HI)^(2))/(P_(H_2) xx P_(H_2))= ((0.70 " atm ")^(2))/((0.02 " atm " ) xx ( 0.02 " atm ")) = 1225`
Since the value of `Q_(P)` is more than that of `K_(c)` the speed of the reverse reaction must increase in order to attain equilibrium.
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