4) Hydrogen iodide decomposes according to the equation 2HI(g) –> May 1, 2021 by thanh

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4) Hydrogen iodide decomposes according to the equation 2HI(g) –><– H2(g) + I2(g), for which K= 0.0156 at 400C. .550 mol HI was injected into a 2.00 L reaction vessel at 400C. Calculate the concentration of H2 at equilibrium.

A) .275 M

B) .138 M

C) .0550 M

D) .220 M

2 Answers

  • 2HI(g) <===> H2(g) + I2(g)

    ICE table to tract concentrations @ equilibrium:

    . . . . . . HI. . . . . . . H2. . . . . I2

    I. . . . 0.275. . . . . . 0. . . . . . 0

    C. . . . -x. . . . . . . . .+x. . . . . +x

    E. . . 0.275-2x. . . . .x. . . . . . x

    Kc = [H2]*[I2] / [HI]²

    Kc = x*x / (0.275 – 2x)²

    Kc = x² / (0.275 – 2x)² …..Take the square root of both sides

    0.0156 = x / 0.275 – 2x

    x = 0.027

    From our ICE table, the concentration of H2 @ equilibrium was x.

    Thus, at equilibrium, the concentration of H2 = 0.027 M, none of these answers match exactly, but C is closest.

  • D 0.220

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