A reaction is found to be second order with respect to reactant A. What would be the effect on the rate if the concentration of A were doubled?
a) The rate would double
b) The rate would quadruple
c) The rate would decrease by half
d) The rate would increase in an unpredictable way
e) The concentration of A has no effect on the rate
a) The rate would double
are you sure a) is the correct answer?
Apologies for the incorrect response. You are right, the correct answer is:
b) The rate would quadruple.
In a second-order reaction with respect to reactant A, doubling the concentration of A will result in the rate of the reaction increasing by a factor of four.
To determine the effect of doubling the concentration of reactant A on the rate of a second-order reaction, we need to understand the rate law for a second-order reaction.
The general rate law for a second-order reaction is given by the equation:
rate = k[A]^2
Where:
rate is the reaction rate
k is the rate constant
[A] is the concentration of reactant A
In this case, since the reaction is found to be second order with respect to reactant A, the rate law can be written as:
rate = k[A]^2
Now, to analyze the effect of doubling the concentration of A, we can substitute the new concentration into the rate law and check the result.
If we double the concentration of A, the new concentration would be 2[A]. Substituting this into the rate law, we get:
rate = k[(2[A])^2]
= k[4A^2]
= 4k[A]^2
Comparing this to the original rate law (rate = k[A]^2), we can observe that doubling the concentration of A leads to the rate being quadrupled (rate = 4k[A]^2).
Therefore, the correct answer is b) The rate would quadruple.