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Question: For the reaction \(A + B \rightarrow\)products, what will be the order of reaction with respect to A...

For the reaction A+B→A + B \rightarrowproducts, what will be the order of reaction with respect to A and B?

Exp.[A](molL−1)[B](mol−1L−1)

Initial rate

(molL−1s−1)

1.2.5 × 10−43 × 10−55 × 10−4
2.5 × 10−46 × 10−54 × 10−3
3.1 × 10−36 × 10−51.6 × 10−5
A

1 with respect to A and 2 with respect to B

B

2 with respect to A and 1 with respect to B

C

1 with respect to A and 1 with respect to B

D

2 with respect to A and 2 with respect to B

Answer

2 with respect to A and 1 with respect to B

Explanation

Solution

Rate =k[A]x[B]y= k\lbrack A\rbrack^{x}\lbrack B\rbrack^{y}

From exp. (1), 5×10−4=k[2.5×10−4]x[3×10−5]y5 \times 10^{- 4} = k\lbrack 2.5 \times 10^{- 4}\rbrack^{x}\lbrack 3 \times 10^{- 5}\rbrack^{y} …… (i)

From exp. (2), 4×10−3=k[5×10−4]x[6×10−5]y4 \times 10^{- 3} = k\lbrack 5 \times 10^{- 4}\rbrack^{x}\lbrack 6 \times 10^{- 5}\rbrack^{y} …… (ii)

Dividing (ii) by (i), 4×10−35×10−4=2x.2y=8\frac{4 \times 10^{- 3}}{5 \times 10^{- 4}} = 2^{x}.2^{y} = 8

From exp. (3), 1.6×10−2=k[1×10−3]x[6×10−5]y1.6 \times 10^{- 2} = k\lbrack 1 \times 10^{- 3}\rbrack^{x}\lbrack 6 \times 10^{- 5}\rbrack^{y} ……. (iii)

Dividing (iii) by (ii), 1.6×10−24×10−3=2x=4\frac{1.6 \times 10^{- 2}}{4 \times 10^{- 3}} = 2^{x} = 4

Or x = 2, y = 1

Hence, order with respect to A is 2 and with respect to b is 1.