The equilibrium constant, Kc, for 2HI(g) H2(g) + I2(g)  is 1.80×10-2 at 698 K. find the equilibrium concentrations of reactant and products ?

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  • 1 month ago
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    2HI  < === > H2  +  I2

    0.370...........0..........0......Initial

    -2x.............+x.........+x....Change

    0.37-2x........x...........x....Equilibrium

    Kc = [H2][I2] / [HI]^2 = 1.80x10^-2

    1.80x10^-2 = (x)(x) / (0.37-2x)^2

    Solve for x and that will be the equilibrium concentrations of both H2 and I2

    Multiply x by 2 and subtract that value from 0.370 to get equilibrium conc. of HI

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