Question

The following reaction was performed in a sealed vessel at 791 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.50M . The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Answer 1

Answer:

4.31 × 10²

Explanation:

Equation of the reaction;

$H_{2(g)} + I_{2(g)}$     ⇌     $2HI_{(g)$

The ICE Table is shown as follows:

                            $H_{2(g)}$         $+$         $I_{2(g)}$        ⇌     $2HI_{(g)$

Initial                    3.10                     2.50                  0      

Change                 - x                       -x                     + 2x      

Equilibrium        (3.10 - x)                0.0800              2x

From $I_{2(g)}$   ;

We can see that 2.50 - x = 0.0800

So; we can solve for x;

x = 2.50 - 0.0800

x = 2.42

$H_{2(g)}$  which = (3.10 -x) will be :

= 3.10 - 2.42

= 0.68

$2HI_{(g)$ = 2x

= 2 (2.42)

= 4.84

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

$K_c = \frac{(4.84)^2}{(0.68)(0.0800)}$

$K_c =\frac{23.4256}{0.0544}$

$K_c =$ 430.62

$K_c$ ≅ 431

$K_c$ = 4.31 × 10²