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].