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

**chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution**

### 3. Answer the following in brief.

**chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution page 137**

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

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

**Answer:- **

From the above observations (i) and (ii):

0.15 = (0.3)^{x} (0.05)^{y} ......(i)

0.30 = (0.6)^{x} (0.05)^{y} .......(ii)

Dividing (ii) by (i):

\(0.30/0.15 = 2 = \frac{(0.6)^x (0.05)^y}{(0.3)^x (0.05)^y} = \left(\frac{0.6}{0.3}\right)^x = 2^x\)

Hence, \(x = 1\)

From observations (i) and (iii) separately in the rate law:

0.15 = (0.3)^{x} (0.05)^{y} ....(iii)

1.20 = (0.6)^{x} (0.2)^{y} .......(iv)

Dividing (iv) by (iii):

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

\(8 = 2(0.2/0.05)^y = 2 \times 4^y\)

Therefore, \(y = 1\)

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