The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.
The given chemical reaction at a stable equilibrium is,
2H₂O(g)+O₂(g) = 2H₂O₂(g)
According to the ideal gas equation,
PV = nRT
P is pressure,
V is volume,
n is moles
R is gas constant,
T is temperature.
Assuming the temperature is constant.
If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,
P₁V₁ = P₂V₂
Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.
If V₂ = 2V₁,
P₂ = P₁/2
So, the final total pressure will be half of the initial pressure.
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Answer:
The answer to your question is 2 molecules, or B(edg 2021).
Explanation:
edg 2021
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
Cu+ p=29 e=28 n=34
S2- p=16 e=18 n=16
Pb4+ p=82 e=78 p=125
I hope i did it right :))
