Answer:
(slow)xy2+z→xy2z (fast) c step1:step2:xy2+z2→xy2z2
Explanation:
Step1: xy2+z2→xy2z2 (slow)
Step2: xy2z2→xy2z+z (fast)
2XY 2 + Z 2 → 2XY 2 Z
Rate= k[xy2][z2]
When the two elementary steps are summed up, the result is equivalent to the stoichiometric equation. Hence, this mechanism is acceptable. The order of both elementary steps is 2, which is ‘≤3’; this also makes this mechanism acceptable. Furthermore, the rate equation aligns with the experimentally determined rate equation, and this also makes this mechanism acceptable. Therefore, since all the three rules have been observed, this mechanism is possible.
Explanation:
The given reaction is as follows.
Value of equilibrium constant is given as 
= 4.3 \times 10^{6}[/tex].
Concentration of given species is ![[SO_2]](https://tex.z-dn.net/?f=%5BSO_2%5D)
= 0.010 M; ![[SO_3]](https://tex.z-dn.net/?f=%5BSO_3%5D)
= 10.M; ![[O_2]](https://tex.z-dn.net/?f=%5BO_2%5D)
= 0.010 M.
Formula for experimental value of equilibrium constant (Q) is as follows.
Q = ![\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%5D%5E%7B2%7D%7D%7B%5BSO_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D)
Putting the given concentration as follows.
Q =
Q = 
Q = 
It is known that when Q > 
, then reaction moves in the backward direction.
When Q < , then reaction moves in the forward direction.
When Q = , then reaction is at equilibrium.
As, for the given reaction Q > then it means reaction moves in the backward direction.
Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.
Answer:try to make a good pic please I have ideas about the topic
the answer to your question is
volume
Answer:
It is used as a catalyst.
