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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 10 of 39
Marks: +1, -0
In a reaction between A and B, the initial rate of reaction (r0r_0) was measured for different initial concentrations of A and B as given below :
 A/mol L−1^{-1}  B/mol L−1^{-1}   r0r_0/mol L−1^{-1} s −1^{-1}
 0.20  0.30   5.07 × 10−5^{-5}
 0.20  0.10  5.07 × 10−5^{-5}
 0.40  0.05  1.43 × 10−4^{-4}
What is the order of the reaction with respect to A and B ?
Solution:  
dxdt=k[A]x[B]y\frac{dx}{dt}=k[A]^{x}[B]^{y}
5.07×10−5=k(0.2)x(0.3)y                  5.07 \times 10^{-5}=k(0.2)^{x}(0.3)^{y}\;\;\;\;\;\;\;\;\;...(i)
5.07×10−5=k(0.2)x(0.10)y              5.07 \times 10^{-5}=k(0.2)^{x}(0.10)^{y}\;\;\;\;\;\;\;...(ii)
Dividing (i) by (ii), we get
    1=3y⇒30=3y⇒y=0\;\;1=3^{y} \Rightarrow3^{0}=3^{y} \Rightarrow y=0
5.07×10−5=k(0.2)x(0.1)0                  5.07 \times 10^{-5}=k(0.2)^{x}(0.1)^{0}\;\;\;\;\;\;\;\;\;...(iii)
1.43×10−4=k(0.4)x(0.05)0              1.43 \times 10^{-4}=k(0.4)^{x}(0.05)^{0}\;\;\;\;\;\;\;...(iv)
Dividing (iv) by (iii), we get 2.8 = 2x^{x}
            log⁡2.8=xlog⁡2\;\;\;\;\;\;\log 2.8=x \log 2
or     x=log⁡2.8log⁡2.0\;\; x=\frac{\log 2.8}{\log 2.0}
            =0.44720.3010=1.48≈1.5\;\;\;\;\;\;= \frac{0.4472}{0.3010}=1.48 \approx 1.5
The order of reaction is 1.5 with respect to ‘A’ and zero with respect to ‘B’.
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