Question

A sample of HI $(9.30\times {10}^{-3})$mol was placed in an empty $2.00L$ container at $1000K$.After equilibrium was reached, the concentration of $I_2$ was $6.29\times {10}^{-4}M$. Calculate the value of $K_C$ at $1000K$ for the reaction $H_2\left(g\right){+I}_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$.

Answer 1

$H_2+I_2\rightleftharpoons 2HI.$

$t=0$ $4.65\times {10}^{-3}$

$t=t$ $x$ $x$ $4.65\times {10}^{-3}-x$

$x=6.29\times {10}^{-4}$

$\therefore \left[HI\right]=4.02\times {10}^{-3}$

$\therefore K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

$\Rightarrow K_c=40.89$