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COCl(2) gas decomposes as: COCl(2)(g) ...

`COCl_(2)` gas decomposes as:
`COCl_(2)(g) hArr CO(g)+Cl_(2)(g)`
If one mole of He gas is added in the vessel at equilibrium at constant pressure then

A

`[COCl_(2)]` increases.

B

"moles" of `CO` will increases.

C

The reaction goes in forward goes in forward direction.

D

`K_(c )=1`

Text Solution

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The correct Answer is:
To solve the problem regarding the decomposition of \( \text{COCl}_2 \) gas and the effect of adding helium gas at constant pressure, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Reaction**: The decomposition reaction is given as: \[ \text{COCl}_2 (g) \rightleftharpoons \text{CO} (g) + \text{Cl}_2 (g) \] Here, 1 mole of \( \text{COCl}_2 \) decomposes into 1 mole of \( \text{CO} \) and 1 mole of \( \text{Cl}_2 \), resulting in an increase in the total number of moles from 1 to 2. 2. **Identify the Effect of Adding Helium**: Helium is an inert gas, meaning it does not react with the components of the reaction. When an inert gas is added to a system at constant pressure, the total volume of the system increases, which leads to a decrease in the partial pressures of the reacting gases. 3. **Le Chatelier's Principle**: According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in conditions (such as pressure or concentration), the equilibrium will shift in a direction that counteracts the change. In this case, since we are increasing the total number of moles of gas by adding helium, the equilibrium will shift towards the side with more moles of gas. 4. **Determine the Direction of Shift**: In our reaction, the forward direction (producing \( \text{CO} \) and \( \text{Cl}_2 \)) has more moles of gas (2 moles) compared to the reactant side (1 mole). Therefore, the equilibrium will shift to the right (forward direction). 5. **Conclusion**: As a result of the shift in equilibrium towards the products, the concentrations of \( \text{CO} \) and \( \text{Cl}_2 \) will increase, while the concentration of \( \text{COCl}_2 \) will decrease. ### Final Answer: - The moles of \( \text{CO} \) will increase when 1 mole of helium gas is added at constant pressure.
To solve the problem regarding the decomposition of \( \text{COCl}_2 \) gas and the effect of adding helium gas at constant pressure, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Reaction**: The decomposition reaction is given as: \[ \text{COCl}_2 (g) \rightleftharpoons \text{CO} (g) + \text{Cl}_2 (g) \] ...
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