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In a reaction, A + B \rightarrow Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled, rate law for the reaction can be written as :

  • Option 1)

    Rate = k[A][B]2

  • Option 2)

    Rate = k[A]2[B]2

  • Option 3)

    Rate=k[A][B]

  • Option 4)

    Rate=k[A]2[B]

 

Answers (1)

\left [ A \right ]                         \left [ B \right ]                                 Rate 

x                             Y                                       R

x                             2Y                                      2R

2x                           2Y                                      8R

Let the rate law; rate=K\left [ A \right ]^{a}\left [ B^{} \right ]^{b}               

       From date R=Kx^{a}.y^{b}\rightarrow 1

and 2R=Kx^{a}.(2Y)^{b}\rightarrow 2

Dividing equation 2 and 1

2=2b or B=1

Also, 

K(2x)^{a}.(2y)^{b}=8R\rightarrow 3\\ From\: 1\: and\: 3 2^{a}.2^{b}=8 or 2^{a}=4\\ a=2\\ Rate law=K\left [ A^ \right ]^{2}\left [ B \right ]                 


Option 1)

Rate = k[A][B]2

This option is incorrect 

Option 2)

Rate = k[A]2[B]2

This option is incorrect 

Option 3)

Rate=k[A][B]

This option is incorrect 

Option 4)

Rate=k[A]2[B]

This option is correct 

Posted by

Vakul

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