Le Chatelier's principle simply explains how equilibria change as you change the conditions of a reaction. If you have a reaction that is at equilibrium lets say (A + 2B <--> C + D) by removing C or D we can drive the reaction forward and products more products. I can provide a more in-depth description if needed.
Answer:
Explanation:
Cheese, Meat, dough, Sauce
The new volume at standard pressure of 1 atm is 21294 liters.
Explanation:
Data given:
Initial volume of the gas V1 = 338 liters
initial pressure on the gas P1 = 63 atm
standard pressure as P2 = 1 atm
Final volume at standard pressure V2 =?
The data given shows that Boyle's law equation is to used:
P1V1 = P2V2
rearranging the equation to calculate V2,
V2 = 
Putting the values in the equation:
V2 = 
= 21294 L
as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.
Answer:
Equilibrium constant for
is 0.5
Equilibrium constant for decomposition of
is 
Explanation:
dissociates as follows:

initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for
is 0.5
Now calculate the equilibrium constant for decomposition of 
It is given that
is decomposed.
decomposes as follows:

initial 1.0 M 0 0
at eq. concentration of
is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of
is 