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 :
| [d] 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, \[R=k{{[A]}^{2}}[B]\] |
You need to login to perform this action.
You will be redirected in
3 sec
