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InterviewSolution
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The experimental data for the reaction `2A+B_(2)rarr2AB,` is : The rate equation for the above data is :A. `rate =k [B_(2)]`B. `rate =k [B_(2)]_(2)`C. `rate =k[A]^(2)[B]^(2)`D. `rate=k [A]^(2)[B]` |
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Answer» Correct Answer - A Consider the following rate law equation . `(dx)/(dt)=K[A]^(m)[B_(2)]^(n)` `1.6xx10^(-4)=K[0.50]^(m)[0.50]^(n)` . ..(i) `3.2xx10^(-4)=K[0.50]^(m)[10]^(n)` . . .(ii) `3.2xx10^(-4)=K[100]^(m)[1.0]^(n)`. . .(iii) by dividing Eq.(iii) by (ii) we get , `(3.2xx10^(-4))/(3.2xx10^(-4))=(k[1.00]^(m)[1.0]^(n))/(k[0.50]^(m)[1.0]^(n))` `1=2 ^(m) or 2^(0)=2^(m)` ` therefore m=0` By dividing Eq(ii) by (i) `(3.2xx10^(-4))/(1.6xx10^(-4))=([0.50]^(m)[1.0]^(n))/([0.50]^(m)[0.50]^(n))` `1=2^(m) or 2^(0)=2^(n)` ` therefore m=0` by dividing Eq.s (ii) by (i) `(3.2xx10^(-4))/(1.6xx10^(-4))=([0.50]^(m)[1.0]^(n))/([0.50]^(m)[0.50]^(n))` `2=2^(n)` `or 2^(1)=2^(n)` `therefore n=1 ` hence rate . `((dx)/(dt))=K[A]^(0)[B_(2)]^(1)` `=K[b_(2)]` |
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