Answer to Question #140271 in Inorganic Chemistry for Neo

Question

3. A closed system initially containing 1.000 × 10-3 M H2 and 2.000 × 10-3 M I2 at 448℃ is allowed to reach equilibrium. At equilibrium, the HI concentration is 1.87 × 10-3 M. Given that the reaction equation is H2(g) + I2(g) ↔ 2 HI(g), find:
a. Equilibrium concentrations of the two reactants
b. Kc of the reaction taking place.

Answer 1

$H_2(g) + I_2(g) ↔ 2 HI(g)$

initial concentration

$H_2 =1mM\\I_2=2mM$

$HI=0$

at equilibrium let's assume

$H_2 =(1-x)mM\\I_2=(2-x)mM\\HI=2x$

we are given equilibrium concentration of HI is 1.87mM

$2x=1.87\\x=0.935$

at equilibrium,

$H_2$ =$1-0.935=0.065mM$

$I_2=2-0.935=1.065mM$

$HI=2\times0.935=1.87mM$

Kc of the reaction =$\frac{[HI]^2}{H_2I_2}=\frac{1.87\times1.87}{1.065\times0.065}=50.5$