Question

What is the rate constant of a reaction if rate = 0.2 (mol/L)/s, [A] and [B] are each 3 M, m = 1, and n = 2?

Answer 1

Answer : The rate constant of a reaction is, $7.407\times 10^{-3}M^{-2}s^{-1}$

Solution :

The general rate law expression is,

$rate=k[A]^m[B]^n$

where,

k = rate constant

[A] and [B] are the concentrations

m is the order of reactant A and n is the order of reactant B

Now put all the given values in the above rate law expression, we get the value for rate constant.

$0.2Ms^{-1}=k\times (3M)^1\times (3M)^2$

$k=7.407\times 10^{-3}M^{-2}s^{-1}$

Therefore, the rate constant of a reaction is, $7.407\times 10^{-3}M^{-2}s^{-1}$

Answer 2

M= 1 and n = 2
( m+ n =  1 +2  = 3)

rate = K [A] [B]^2

0.2  = K *  3  * 3 ^2
0.2 =  K  *  3 * 9 
 K =    0.2   / 27 
K =   7.408 *  10 ^ -3  m^-2  s^-1