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  •  Consider a reaction  having rate low expres

 Consider a reaction \mathrm{A+B \rightarrow C}  having rate low expression is \mathrm{Rate=k(A)^{x}(B)^{y}}.on doubling the concentration of \mathrm{A} & having the concentration of \mathrm{B} Calculate the \mathrm{\left [ Rate \right ]'}to \mathrm{\left [ Rate \right ]} of the reaction.

Option: 1

\frac{1}{2(x+y)}


Option: 2

x+y


Option: 3

(y-x)


Option: 4

2^{y-x}


Answers (1)

best_answer

Given that \mathrm{[Rate]= =k(A)^{x}(B)^{y}}

Now if \mathrm{A \rightarrow 2 A }
           \mathrm{B \rightarrow B / 2 }

\mathrm{\text { [Rate] } \left.{ }^{\prime}=k(2 A)^{x}(B) 2\right)^{y}=2^{x}\left(\frac{1}{2}\right)^{y} \cdot k[A]^{x}[B]^{y} }
\mathrm{\frac{\text { [Rate }]^{\prime}}{\text { [Rate] }}=\frac{K(A)^{x}(B)^{y}}{2^{n}\left(\frac{1}{2}\right)^{y}(A)^{x}(B)^{y} x}=\frac{2^{y}}{2^{n}}=2^{y-x} }

Posted by

sudhir kumar

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