rate 1 = k(A) = k(1/2) = 1/2 rate
rate 2 = k(A)^2 = k(1/2)^2 = 1/4 rate
rate 2 = k(A)^2 = k(1/2)^2 = 1/4 rate
For a first order reaction, the rate law is given by the equation:
Rate = k[A]
Where Rate is the rate of reaction, k is the rate constant, and [A] represents the concentration of reactant A.
For a second order reaction, the rate law is given by the equation:
Rate = k[A]^2
Now, let's consider the effect of halving the concentration of reactants in each reaction:
1. First Order Reaction:
If the concentration of reactant A is halved, the rate equation becomes:
Rate = k[(1/2)A]
By simplifying, we get:
Rate = (1/2)k[A]
So, when the concentration is halved in a first order reaction, the rate of reaction is also halved.
2. Second Order Reaction:
If the concentration of reactant A is halved, the rate equation becomes:
Rate = k[(1/2)A]^2
By simplifying, we get:
Rate = (1/4)k[A]^2
So, when the concentration is halved in a second order reaction, the rate of reaction is reduced to quarter (1/4) of its original value.
In summary, halving the concentration of reactants in a first order reaction will lead to a halving of the rate of reaction, while in a second order reaction, the rate of reaction will decrease to one-fourth (1/4) of its original value.