<u>Answer:</u> The value of
for the reaction at 690 K is 0.05
<u>Explanation:</u>
We are given:
Initial pressure of
= 1.0 atm
Total pressure at equilibrium = 1.2 atm
The chemical equation for the decomposition of phosgene follows:

Initial: 1 - -
At eqllm: 1-x x x
We are given:
Total pressure at equilibrium = [(1 - x) + x+ x]
So, the equation becomes:
![[(1 - x) + x+ x]=1.2\\\\x=0.2atm](https://tex.z-dn.net/?f=%5B%281%20-%20x%29%20%2B%20x%2B%20x%5D%3D1.2%5C%5C%5C%5Cx%3D0.2atm)
The expression for
for above equation follows:


Putting values in above equation, we get:

Hence, the value of
for the reaction at 690 K is 0.05
Answer:
Sodium is extracted from it's ore by electrolysis of fused sodium chloride.
Explanation:
he process is usually carried out ia a special electrochemical cell called the downs cell. While molten sodium metal is collected at the cathode and also sent to tanks for cooling and storage.
What best describes the result is a mixture
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)

x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer:
Eventually, these individual laws were combined into a single equation—the ideal gas ... We find that temperature and pressure are linearly related, and if the ... then P and T are directly proportional (again, when volume and moles of gas are ... of the variables, and they are more difficult to use in fitting theoretical equations ...
Explanation: