Question #b54e2

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Answer 1 · 2015-07-17T21:14:10

$K_{c} = \frac{{[H I]}^{2}}{[H_{2}] [I_{2}]}$ $K_c=([HI]^(2))/([H_2][I_2])$

Where [ ] represents concentrations at equilibrium.

In general for a reaction:

$a A + b B ⇌ c C + d D$ $aA+bBrightleftharpoonscC+dD$

$K_{c} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$ $K_c=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))$

Where [ ] represents concentration of the species at equilibrium.

If we apply the general rule to this specific reaction:

$H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}$ $H_(2(g))+I_(2(g))rightleftharpoons2HI_((g))$

we get:

$K_{c} = \frac{{[H I]}^{2}}{[H_{2}] [I_{2}]}$ $K_c=([HI]^(2))/([H_2][I_2])$

Note that whenever you quote a value for $K_{c}$ $K_c$ you must also state the temperature as well.