The reaction between A and B is first order with respect to A and zero order with respect to B. For this reaction, fill in the blanks in the following table.

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 :