Answer to Question #128600 in General Chemistry for A

Question

Good day,

For a certain reaction, Kc = 1.73×10−9 and kf= 2.21×104 M−2⋅s−1
. Calculate the value of the reverse rate constant kr,given that the reverse reaction is of the same molecularity as the forward reaction.

Answer 1

The equilibrium constant of any reversible reaction system is the ratio of forward and reverse rate constants:

$K = K_f/K_r$

The reverse rate constant is:

$K_r = K_f/K$

so K_r =(2.21*10⁴)/(1.73*10^-9)

K_r = 1.27* 10¹³ M^-2s^-1