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4.10   In a reaction between A and B, the initial rate of reaction (r0) was measure for different initial concentrations of A and B as given below:

          

          What is the order of the reaction with respect to A and B?

              

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we know that 
rate law (r_{0}) = k[A]^{x}[B]^{y}
As per data 

\\5.07\times 10^{-5} =k[0.2]^{x}[0.3]^{y}\\ 5.07\times 10^{-5}=k[0.2]^{x}[0.1]^{y}\\ 1.43\times 10^{-4}=k[0.4]^{x}[0.05]^{y} these are the equation 1, 2 and 3 respectively

Now, divide eq.1 by equation2, we get
1= (0.3/0.1)^{y}
from here we calculate that y = 0

Again, divide eq. 2 by Eq. 3, we get
Since y =0 also substitute the value of y 
So, 
 =(\frac{1.43}{0.507})= (\frac{0.4}{0.2})^{x}
 = 2.821=2^{x}

taking log both side we get,

x = \frac{\log2.821}{\log2}
    = 1.496 
   = approx 1.5
Hence the order of reaction w.r.t A is 1.5 and w.r.t B is 0(zero)

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manish

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