Q. 155.0( 1 Vote )

# 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 :

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