This is a repost.
The following reaction is second order in A and first order in B.
2A+1B --> 3C
what is the rate law equation for the reaction below?
(Assume that A, B and C are the same compounds for each reaction.)
4A + 2B --> 6C
To determine the rate law equation for the reaction, we'll need to examine the reaction rate and the concentrations of the reactants.
Given that the reaction is second order in A and first order in B, we can write the rate law equation in the form:
Rate = k[A]^m[B]^n
Here, k is the rate constant that we need to determine, [A] represents the concentration of A, and [B] represents the concentration of B. The exponents m and n represent the reaction orders for A and B, respectively.
In the first reaction:
2A + 1B → 3C
We know that the reaction is second order in A, so m = 2, and it is first order in B, so n = 1. So, the rate law equation for the first reaction is:
Rate = k[A]^2[B]^1
Now, let's examine the second reaction:
4A + 2B → 6C
Since the reaction is the same as the first reaction, we can assume that the rate law equation for the second reaction will have the same form. Therefore, the rate law equation for the second reaction is also:
Rate = k[A]^2[B]^1
Both reactions have the same rate law equation since they involve the same compounds with the same reaction orders.
In summary, the rate law equation for the reaction:
4A + 2B → 6C
is:
Rate = k[A]^2[B]^1