A reaction between A and B is found to be first order in [A] and second order in [B]. If [A] is halved and [B] is doubled, the rate of the reaction will
(A) remain the same.
(B) be increased by a factor of 2.
(C) be increased by a factor of 4.
(D) be increased by a factor of 8.
I think the answer is A because I suspect that
The 1/2 and the double will cancel each other
Rate=k([A]/2).([B]exp2)x2
A is not right but your equation is correct. Just do the math.
[A]^1 x [B]^2.
A = 1/2 = 1/2 ^1 = 1/2
B = x2 and 2x2 = 4
so 1/2 x 4 = ?
To determine the effect of the changes in concentrations on the reaction rate, we can use the rate law expression for the given reaction.
Given that the reaction is first order in [A] and second order in [B], we can write the rate law expression as:
Rate = k[A]^1[B]^2
Now, let's analyze the changes in concentrations:
When [A] is halved, it becomes [A]/2.
When [B] is doubled, it becomes 2[B].
Substituting the new concentrations into the rate law expression, we get:
New Rate = k([A]/2)^1(2[B])^2
Simplifying this expression, we have:
New Rate = k([A]/2)(4[B]^2)
= 4k[A][B]^2
Comparing this with the original rate expression, we can see that the new rate is 4 times larger than the original rate. Therefore, the answer is (C) be increased by a factor of 4.