Question

At a certain temperature, the equilibrium constant, K c , Kc, for this reaction is 53.3. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 53.3 H2(g)+I2(g)↽−−⇀2HI(g)Kc=53.3 At this temperature, 0.700 mol H 2 0.700 mol H2 and 0.700 mol I 2 0.700 mol I2 were placed in a 1.00 L container to react. What concentration of HI HI is present at equilibrium?

Answer 1

Answer:

1.099 M is the concentration of HI is present at equilibrium.

Explanation:

Initial concentration of hydrogen gas = $[H_2]=\frac{0.700 mol}{1.00 L}=0.700 M$

Initial concentration of iodine gas = $[I_2]=\frac{0.700 mol}{1.00 L}=0.700 M$

$H_2+I_2\rightleftharpoons 2HI$

Initially

0.700 M     0.700 M         0

(0.700-x)M   (0.700-x)      2x

Expression of an equilibrium constant is given by :

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

$53.3=\frac{4x^2}{(0.700-x)(0.700-x)}$

Solving for x:

x = 0.5495 M

Concentration of HI is present at equilibrium:

$2\times 0.5495 M=1.099 M$