A) \[{{s}^{-1}}\]
B) \[{{s}^{-1}}\,mol\,\,{{L}^{-1}}\]
C) \[{{s}^{-1}}\,mo{{l}^{-1}}\,\,L\]
D) \[s\,\,{{s}^{-1}}\,mo{{l}^{-2}}\,\,{{L}^{2}}\]
Correct Answer: C
Solution :
Let the initial rate is R and order with respect to A is x and B is y. Thus, rate law is, rate,. \[R={{[A]}^{x}}{{[B]}^{y}}\] ?...(i) On doubling the concentration of A, rate becomes 4R, \[4R={{[2A]}^{x}}{{[B]}^{y}}\] ?...(ii) On doubling concentration of B, rate remains R, \[R={{[A]}^{x}}{{[2B]}^{y}}\] ??.(iii) From Eq. (i) and (ii), we get \[\frac{R}{4R}={{\left( \frac{1}{2} \right)}^{x}}\] \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{x}}\] \[\therefore \] \[x=2\] From Eq. (i) and (iii), we get \[\frac{R}{R}={{\left[ \frac{1}{2} \right]}^{y}}\] \[1={{\left( \frac{1}{2} \right)}^{y}}\] \[\therefore \] \[y=0\] Thus, actual rate law is, rate, \[R={{[A]}^{2}}{{[B]}^{0}}\] The order of this reaction is 2 and for second order reaction units of rate constant are \[\mathbf{mo}{{\mathbf{l}}^{\mathbf{-1}}}\,\,\mathbf{L}\,\,{{\mathbf{s}}^{\mathbf{-1}}}\].
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