A) \[\text{Rate}\,\text{=}\,\text{k}\,\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\,{{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}^{\text{2}}}\]
B) \[\text{Rate}\,\text{=}\,\text{k}\,{{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }}^{\text{2}}}\,{{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}^{\text{2}}}\]
C) \[\text{Rate}\,\text{=}\,\text{k}\,\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\,\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }\]
D) \[\text{Rate}\,\text{=}\,\text{k }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }\]
Correct Answer: D
Solution :
Let the order of reaction with respect to A and B is x and y respectively. So, the rate law can be given as \[R=k{{[A]}^{x}}{{[B]}^{y}}\] ...(i) When the concentration of only B is doubled, the rate is doubled, so \[{{R}_{1}}=k{{[A]}^{x}}{{[2B]}^{y}}=2R\] ...(ii) If concentrations of both the reactants A and B are doubled, the rate increases by a factor of 8, so \[R''=k{{[2A]}^{x}}{{[2B]}^{y}}=8R\] ... (iii) \[\Rightarrow \] \[k{{2}^{x}}{{2}^{y}}{{[A]}^{x}}{{[B]}^{y}}=8R\] ...(iv) From Eq. (i) and (ii), we get \[\Rightarrow \] \[\frac{2R}{R}=\frac{{{[A]}^{x}}{{[2B]}^{y}}}{{{[A]}^{x}}{{[B]}^{y}}}\] \[2={{2}^{y}}\] \[\therefore \] \[y=1\] From Eq. (i) and (iv), we get \[\Rightarrow \]\[\frac{8R}{R}=\frac{{{2}^{x}}{{2}^{y}}{{[A]}^{x}}{{[B]}^{y}}}{{{[A]}^{x}}{{[B]}^{y}}}\]or\[8={{2}^{x}}{{2}^{y}}\] Substitution of the value of y gives, \[8={{2}^{x}}\,{{2}^{y}}\] \[4={{2}^{x}}\] \[{{(2)}^{2}}={{(2)}^{x}}\] \[\therefore \] \[x=2\] Substitution of the value of x and y in Eq. (i) gives, \[R=k{{[A]}^{2}}[B]\]
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