The reactio

It is given that the reaction is of the first order with respect to A and of zero order with respect to B.

So, we can write the rate of the reaction as:

Rate = k [A]¹[B]⁰

⇒ Rate = k [A] …(1) [Anything to the power of 0 is 1]

From experiment I, it is given that rate = 2.0 x 10^-2mol L^-1min^-1 and the concentration of A,[A] =0.1 [M]. So, we put these values in equation (1), and we get:

2.0 x 10^-2 = k(0.1)

⇒ k= 0.2 min^-1

So, we get the value of rate constant, k.

Similarly, from II, we put the values of k and Rate to get the concentration of A.

From experiment II, we obtain Rate = 4.0 x 10^-2mol L^-1min^-1 and we already know the value of k. We put these values in equation (1), to get:

4.0 x 10^-2= 0.2 × [A]

⇒ [A] = 0.2 [M]

So, concentration of A in II is 0.2[M].

From experiment III, we have the concentration of A, [A] = 0.4 [M].Also, we know the value of rate constant k, and we put their values in equation (1) and we get rate as:

Rate = 0.2 × 0.4

⇒Rate = 0.08 mol L^-1min^-1 = 8.0×10^–2mol L^-1min^-1

From experiment IV, we have Rate as 2.0 × 10^-2mol L^-1min^-1, and we already know the value of rate constant k. Putting their values in equation (1), we get the concentration of A as:

2.0 × 10^-2 = 0.2 × [A]

⇒[A] = 0.1 [M]

So, concentration of A in II is 0.1[M].

Thus, the complete table looks like :