Question

H2 + I2 ⇄ 2HI
what is the equilibrium constant if .o862 moles/liter H2, .263 moles/liter I2, and 1.02 moles/liter HI are found to be concentrations that occur in equilibrium?

Answer 1

30485, Chemistry, Inorganic Chemistry

$\mathrm{H}_{2} + \mathrm{I}_{2} = 2 \mathrm{HI}$

what is the equilibrium constant if .0862 moles/liter H2, .263 moles/liter I2, and 1.02 moles/liter HI are found to be concentrations that occur in equilibrium?

Solution:

The equilibrium constant for this equation is:

$$K_{\mathrm{eq}} = \frac{[\mathrm{HI}]_{\mathrm{eq}}^{2}}{[\mathrm{H}_{2}]_{\mathrm{eq}} \cdot [\mathrm{I}_{2}]_{\mathrm{eq}}}.$$

In this equation $[\mathrm{HI}]$, $[\mathrm{H}_{2}]$ and $[\mathrm{I}_{2}]$ are concentrations of corresponding compounds that occur in equilibrium.

So, equilibrium constant for this equation is:

$$K_{\mathrm{eq}} = \frac{1.02^{2}}{0.0862 \cdot 0.263} = 45.89$$

Answer:

The equilibrium constant is 45.89.