Answer to Question #300200 in General Chemistry for Tay

Question

The equilibrium constant, K_c, for the following reaction is 7.00×10^-5 at 673 K.

NH₄I(s) NH₃(g) + HI(g)

Calculate the equilibrium concentration of HI when 0.210 moles of NH₄I(s) are introduced into a 1.00 L vessel at 673 K.

[HI] =   M

Answer 1

NH$_4I_{(s)}\to\>NH_{3(g)}+HI_{(g)}$

$k_c=\frac{[NH_3][HI]}{[NH_4I]}$

$7×10^{-5}=\frac{(x)(x)}{0.210-x}$

$\implies\>x^2=1.470×10^{-5}$

$x=3.834×10^{-3}$

$[HI]=3.834×10^{-3}M$