Given the rate equation, when the concentration of A is doubled while that of B is halved, the rate will Given the rate equation, when the concentration of A is doubled while that of B is halved, the rate will blank the original rate. the original rate.

I believe the question is incomplete as the rate equation was not included in the question, but no need to be worried I can be of help. Here is the complete question:

Given: A + 3B  2C + D

This reaction is first order with respect to reactant A and second order with respect to reactant B. If the concentration of A is doubled and the concentration of B is halved, the rate of the reaction would by a factor of .

(a) increase, 2

(b) decrease, 2

(c) increase, 4

(d) decrease, 4

(e) not change

Explanation:

Solving the above question, the rate of the reaction will decrease and by factor of 2. How?

K1 = [A] [B]^2 i.e, reactant A is first order, reactant B is second order.

k2 = 2[A] [B/2]^2 that is, concentration of reactant A is doubled, concentration of reactant B is halved.

Comparing both rate of reactions:

k1/k2 = [A] [B/2] / 2[A] [B/2]^2

k1/k2 = 4 [A] [B]^2 /2 [A] [B]^2

k1 /k2 = 4 [B]^2 / 2[B]^2

k1/k2 = 2

The rate of the reaction is therefore decreased and by 2.

With this ideas of how to solve this kind of question, you can use it to solve for the actual question if the question is solved is not the actual one.

it is stable as it is in group 0 which have full outer shells of electrons