| Mechanism of a hypothetical reaction\[{{X}_{2}}+{{Y}_{2}}\to 2XY\] is given below: [NEET-2017] |
| (i) \[{{X}_{2}}\to X+X(fast)\] |
| (ii) \[X+{{Y}_{2}}XY+Y(slow)\] |
| (iii)\[X+Y\to XY(fast)\] |
| The overall order of the reaction will be |
A) 1.5
B) 1
C) 2
D) 0
Correct Answer: A
Solution :
| [a] The solution of this question is given by assuming step (i) to be reversible which is not given in question |
| Overall rate = Rate of slowest step (ii) |
| \[=k[X][{{Y}_{2}}]\] (1) |
| K=rate constant of step (ii) |
| Assuming step (i) to be reversible, its equilibrium constant, |
| \[{{k}_{eq}}=\frac{{{[X]}^{2}}}{[{{X}_{2}}]}\Rightarrow [X]={{k}_{eq}}^{\frac{1}{2}}{{[{{X}_{2}}]}^{\frac{1}{2}}}\] (2) |
| Put (2) in (1) |
| \[Rate=k{{k}_{eq}}^{\frac{1}{2}}{{[{{X}_{2}}]}^{\frac{1}{2}}}[{{Y}_{1}}]\] |
| Overall order \[=\frac{1}{2}+1=\frac{3}{2}\] |
You need to login to perform this action.
You will be redirected in
3 sec
