its properties I know it is
hope it help ;)
Answer : The partial pressure of
at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of
= 
The partial pressure of
= 
The partial pressure of
= 

The balanced equilibrium reaction is,

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴
At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)
The expression of equilibrium constant
for the reaction will be:

Now put all the values in this expression, we get :


The partial pressure of
at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of
at equilibrium is, 1.0 × 10⁻⁶
Group 1 elements since they have one outermost electron which they can give to chlorine which has 7 outermost electrons in order to form a stable compound.
Example
Pottasium (K) + Chlorine (Cl) = Potassium Chloride (KCL)
Answer:
Water to move into the cell
Answer :
(A) The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
(B) The average rate of the reaction during this time interval is, 0.00176 M/s
(C) The amount of Br₂ (in moles) formed is, 0.0396 mol
Explanation :
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The given rate of reaction is,

The expression for rate of reaction :
![\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DHBr%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DH_2%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DBr_2%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part A:</u>
The rate expression will be:
![Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
<u>Part B:</u>
![\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D)


The average rate of the reaction during this time interval is, 0.00176 M/s
<u>Part C:</u>
As we are given that the volume of the reaction vessel is 1.50 L.
![\frac{d[Br_2]}{dt}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D%3D0.00176M%2Fs)
![\frac{d[Br_2]}{15.0s}=0.00176M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BBr_2%5D%7D%7B15.0s%7D%3D0.00176M%2Fs)
![[Br_2]=0.00176M/s\times 15.0s](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.00176M%2Fs%5Ctimes%2015.0s)
![[Br_2]=0.0264M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D0.0264M)
Now we have to determine the amount of Br₂ (in moles).



The amount of Br₂ (in moles) formed is, 0.0396 mol