Answer: Option (B) is the correct answer.
Explanation:
As the given reaction is as follows.
Equilibrium constant for this reaction will be as follows.
![K_{c} = \frac{[CO_{2}]}{[CO]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCO_%7B2%7D%5D%7D%7B%5BCO%5D%5E%7B2%7D%7D)
According to Le Chatelier's principle, when we increase the temperature then the equilibrium will shift towards the right hand side.
As a result, concentration of carbon dioxide will decrease whereas concentration of carbon monoxide will increase.
Thus, we can conclude that in the given reaction equilibrium constant for this reaction will decrease with increasing temperature.
Answer:
for anapproximate result , divide the pressure value by 7.501
In each mole of carbon dioxide there will be one mole of O₂.
Let us calculate the moles of carbon dioxide gas present first
The conditions are NTP it means , Temperature = 293 K and P = 1 atm
We will use ideal gas equation
PV= nRT
Where
P = Pressure of gas = 1 atm
V= 112mL=0.112L
R= gas constant =0.0821 L atm /mol K
n = moles = ?
Putting values
moles = 0.00466
Thus moles of carbon dioxide will be 0.00466
The moles of O₂ = 0.00466
Carbon is special and unique because it is able to form different compounds with a lot of elements, including itself. When it bonds with itself, this is possible because of the concept of hybridization. It is the mixing of atomic orbitals into a new hybrid orbital. In this case, methane is formed through the sp³ hybridization.
