Question

Initially the concentration of H2 is 2.0M , I2 is 2.0M and HI is 8.0M. The vessel is heated up to 675 degrees at equilibrium the concentration of HI changes to become 6.0M. Calculate the value of k at this temperature.

Answer 1

Answer:

The value of K at this temperature = 2.25

Explanation:

According to question the equilibrium reaction is

                                           H₂(g) + I₂(g) ⇄ 2 HI(g)

       Initial concentration   2           2             8

         At equilibrium     2 + 2x     2 + 2x          8 - 2x

At equilibrium concentration of HI changes to become 6 Molar

                      So 8 - 2x = 6

                      or, x = 1

So concentration of H₂ = 2 + 2 x 1 = 4 molar

concentration of I₂ = 2 + 2 x 1 = 4 molar

                  Kc = $\frac{[HI]^{2} }{[H]_{2} I_{2} }$

     ⇒          Kc = $\frac{6^{2} }{4 X 4}$

     ⇒         Kc = $\frac{36}{16}$ = 2.25