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# 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 agoFavorite Answer
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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