For a general reaction a A + bB → products. The following initial rates are determined experimentally with the initial amounts of A and B

S.No.

A(M)

B(M)

Initial rate (M)

1.

1.00

1.00

1.2 x 10^–2

2.

1.00

2.00

4.8 x 10^–2

3.

1.00

4.00

1.9 x 10^–1

4.

4.00

1.00

4.9 x 10^–2

Assuming that rate law can be written as

$Rate = k {[A]}^{α} {[B]}^{β}$
Determine the value of $k, α and β .$

Solution

We have to assume that law canbe written as $Rate = k {[A]}^{α} {[B]}^{β}$

Thus

$\frac{r_{2}}{r_{1}} = \frac{k (A_{2})^{α} (B_{2})^{β}}{k {[A_{1}]}^{α} {[B_{1}]}^{β}}$

or $\frac{4.8 \times 10^{- 2}}{1.2 \times 10^{- 2}} = \frac{(1)^{α} (2.0)^{β}}{(1)^{α} {[B_{1}]}^{β}} or \frac{4}{1} = {(\frac{2}{1})}^{β} or β = 2 .$

From reaction no. (iv) and (i) $\frac{r_{4}}{r_{1}} = \frac{k (A_{4})^{α} (B_{4})^{β}}{k {[A_{1}]}^{α} {[B_{1}]}^{β}}$

or $\frac{4.9 \times 10^{- 2}}{1.2 \times 10^{- 2}} = \frac{(4)^{α}}{(1)^{α}} \times \frac{(1)^{β}}{(1)^{β}} or α = 1 .$

The order of the reaction is first order with respect to A and second order with respect to B. Overall order of the reaction is 3.

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Chemical Kinetics

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The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

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S.No.	A(M)	B(M)	Initial rate (M)
1.	1.00	1.00	1.2 x 10^–2
2.	1.00	2.00	4.8 x 10^–2
3.	1.00	4.00	1.9 x 10^–1
4.	4.00	1.00	4.9 x 10^–2