Question

Which of the following is true for the following rate law if [A] is doubled and [B] is decreased by 1/2 ?

Rate =k[A]¹[B]²
a.Rate stays the same
b.Rate =4k
c.Rate =2k
d.Rate = 1/2k

Answer 1

When the concentration of A is doubled and the concentration of B is halved, the new rate of the reaction is half the original rate, which is Rate = 1/2k.

Step-by-step explanation:

The given chemical reaction rate law is Rate = k[A]¹[B]². When the concentration of A ([A]) is doubled, the rate would also double (become 2k[A][B]²) because the reaction is first order with respect to A (power of 1). However, when the concentration of B ([B]) is halved, the rate would decrease by a factor of 2² (or 4) since the rate is second order with respect to B (power of 2). Thus, doubling [A] increases the rate 2 times and halving [B] decreases the rate 4 times, overall affecting the rate by a factor of 2/4 or 1/2.

The correct answer to what happens to the rate under these conditions is d. Rate = 1/2k