Answer:
Yes
2Ag + Cu(No3)2 -----(heat) ----- Cu + 2Agno3
When heat energy is passed through Cu(No3)2 then the reverse reaction will occur
If the system is cooled, more of the products will be formed as the equilibrium shifts to the right.
Explanation:
The reaction is an endothermic reaction which absorbs heat from the environment. This is why we see energy on the reactant side.
This reaction is in equilibrium when energy is absorbed in this manner.
According to Le Chatelier's principle "if any of the conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".
- A change in temperature of an equilibrium system shifts the system to a new equilibrium point.
- The backward reaction is favored by an increase in temperature because heat is absorbed in that step.
- This will eventually lead to the production of more of the product.
- Cooling the system will decrease the temperature and will favor the forward reaction.
- If the system is cooled, more of the products will be formed as the equilibrium shifts to the right.
- More NO gas will be produced in the process.
learn more:
Chemical equilibrium brainly.com/question/5877801
#learnwithBrainly
Explanation:
For the given reaction 
Now, expression for half-life of a second order reaction is as follows.
![t_{1/2} = \frac{1}{[A_{0}]k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5Dk%7D)
....... (1)
Second half life of this reaction will be 
. So, expression for this will be as follows.
= ![\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5B%5Cfrac%7B1%7D%7B%5BA%5D_%7Bf%7D%7D%20-%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%7D%5D)
...(2)
where ![[A]_{f}](https://tex.z-dn.net/?f=%5BA%5D_%7Bf%7D)
is the final concentration that is, ![\frac{[A]_{0}}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5D_%7B0%7D%7D%7B4%7D)
here and ![[A]_{i}](https://tex.z-dn.net/?f=%5BA%5D_%7Bi%7D)
is the initial concentration.
Hence, putting these values into equation (2) formula as follows.
= ![\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%7D%20%5B%5Cfrac%7B4%7D%7B%5BA%5D_%7B0%7D%7D%20-%20%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%7D%5D)
= ![\frac{3}{[A_{0}]k}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%5BA_%7B0%7D%5Dk%7D)
...... (3)
Now, dividing equation (3) by equation (1) as follows.
= ![\frac{3}{[A_{0}]k} \times [A_{0}]k](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%5BA_%7B0%7D%5Dk%7D%20%5Ctimes%20%5BA_%7B0%7D%5Dk)
= 3
or, = 3 
Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.
Answer:
dont know
Explanation:
what
ajjajaahhahahaahaahahqhqjqhwjahhahhqhqqhqhhqhahqah
