The equilibrium constant Kc for the reaction of hydrogen with iodine i

1.	The equilibrium constant Kc for the reaction of hydrogen with iodine is 54.0 at 700 K.Kc = 54.0 at 700 KIf kf is the rate constant for the formation of HI and kr is the rate constant for the decomposition of HI, deduce whether kr is larger or smaller than krii. If the value of kr at 700 K is 1.16 × 10-3 , what is the value of kf ?
Answer» Given: i. K_c = 54.0 at 700 K ii. k_r = 1.16 × 10^-3 at 700 K To find: i. Whether k_f is larger or smaller than k_r ii. Value of k_f Formula : K_c= \(\frac{K_f}{K_r}\) Calculation: i. As K_c = \(\frac{K_f}{K_r}\) = 54.0 K_fis greater than k_rby a factor of 54.0. ii. K_f = K_c x K_r = 54.0 x (1.16 x 10^-3) = 62.64 x 10^-3 i. K_fis greater than K_r, ii. Value of K_r is 62.64 x 10^-3

The equilibrium constant Kc for the reaction of hydrogen with iodine is 54.0 at 700 K.Kc = 54.0 at 700 KIf kf is the rate constant for the formation of HI and kr is the rate constant for the decomposition of HI, deduce whether kr is larger or smaller than krii. If the value of kr at 700 K is 1.16 × 10-3 , what is the value of kf ?

Answer»

Given: i. K_c = 54.0 at 700 K

ii. k_r = 1.16 × 10^-3 at 700 K

To find: i. Whether k_f is larger or smaller than k_r

ii. Value of k_f

Formula : K_c= \(\frac{K_f}{K_r}\)

Calculation:

i. As K_c = \(\frac{K_f}{K_r}\) = 54.0

K_fis greater than k_rby a factor of 54.0.

ii. K_f = K_c x K_r = 54.0 x (1.16 x 10^-3)

= 62.64 x 10^-3

i. K_fis greater than K_r,

ii. Value of K_r is 62.64 x 10^-3