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For the reaction 2A + B rarrC, the value...

For the reaction `2A + B rarrC`, the values of ini concentrations are given in the table below. The rate law for the reaction is:

A

Rat `=k[A]^(2) [B]`

B

Rate `=k [A]^(2) [B]^(2)`

C

Rate =k [A] [B]

D

Rate `=k [A] [B]^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
From given data
Rate `=k [A]^(a) [B]^(b)`
Now, `0.045 =k [0.05]^(a+b)" ….(i) ", 0.090 =k[0.05]^(a+b) 2^(a)" ….(ii)"`
`0.72=k [0.5]^(a+b)2^(2a+b)" ….(iii)"`
From equation (i) and (ii) we get,
`rArr (1)/(2) =(1)/(2^(a)) rArr a=1`
And from equation (i) and (iii) we get
`rArr (0.045)/(0.72) = (1)/(2^(2+b)) rArr (45)/(720) =(1)/(2^(2+b)) rArr (720)/(45) =2^((2+b)) rArr Z^(4) =Z^(2+b) rArr b=2`
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