Question

The rate of the reaction`H_2 +I_2 ⇌ 2HI` is  given by:
                       Rate = 1.7 × 10^-19 [H₂] [I₂] at 25°C.
The rate of decomposition of gaseous HI to H₂ and I₂ is given by:
                      Rate = 2.4 × 10^-21 [HI]² at 25°C.
Calculate the equilibrium constant for the formation of HI from H₂ and I₂ at 25°C.

Answer 1

For the reversible reaction,

H₂ + I₂ ⇌ 2HI 

Equilibrium constant,        `K_c = ([HI]^2)/([H_2][I_2])`

At equilibrium,                     r_f = r_b 

Now,                                    r_f = 1.7 × 10^-19 [H₂][I₂] 

                                            r_b = 2.4 × 10^-21 [HI]² 

∴ At  equilibrium,    

          1.7 × 10^-19 [H₂][I₂] = 2.4 × 10^-21 [HI]² 

 or             `(1.7 xx 10^(19))/(2.4 xx 10^(-21)) = ([HI]^2)/([H_2][I_2]) = K_c`

                                     K_c = 70.83 at 25°C.