Question

Suppose you wish to determine the order of the reaction. You initially measure a rate, r0, given some arbitrary concentration of reactants. You then proceed to double the amount of A in the reaction (keeping B the same) and find that the rate is now 2r0. In a reciprocal experiment, you double the amount of B (A same as initial) in the reaction and measure a rate of 8r0. What is the order of this reaction

Answer 1

Answer:

See explanation

Explanation:

Now taking A, we can see that if we double the amount of A while keeping the amount of B constant, the rate of reaction doubles, hence we can write;

2^n = 2^1

hence n =1

For B, when the amount of B is doubled while keeping the amount of A constant, the rate of reaction increases eight times. Hence we can write;

2^n = 8

2^n = 2^3

n=3

Hence this reaction has an overall order of 1 + 3 = 4

So we can write;

rate =k[A] [B]^3