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Answer the following in brief. For the reaction 2A + B → products, find the rate law from the following data. [A]/M [B]/M rate/M s-1 0.3 0.05 0.15 0.6 0.05 0.30 0.6 0.2 1.20

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Question

Answer the following in brief.

For the reaction 2A + B → products, find the rate law from the following data.

[A]/M [B]/M rate/M s-1
0.3 0.05 0.15
0.6 0.05 0.30
0.6 0.2 1.20
Answer in Brief
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Solution

From above observations (i) and (ii)

0.15 = (0.3)x (0.05)y  ......(i)

0.30 = (0.6)x (0.05) .......(ii)

Dividing (2) by (1)

`0.30/0.15 = 2 = ((0.6)^x (0.05)^y)/((0.3)^x (0.05)^y) = ((0.6)/(0.3))^x = 2^x`

Hence, x = 1

From observation (i) and (iii) separately in the rate law gives

0.15 = (0.3)x (0.05)y   ....(iii)

1.20 = (0.6)x (0.2)y  .......(iv)

Dividing (4) by (3)

`1.20/0.15 = 0.6/0.3 (0.2/0.05)^y`   (∵ x = 1)

∴ `8 = 2(0.2/0.05)^y = 2 xx 4^"y"`

∴ 4 = 4y

Therefore, y = 1

The rate law is rate = k [A][B].

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Chapter 6: Chemical Kinetics - Exercises [Page 137]

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Balbharati Chemistry [English] Standard 12 Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercises | Q 3. xvii. | Page 137

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