One of the following equilibria is not affected by change in volume of the flask (a) PCl5(g) PCl

Q: Why does volume change affect equilibrium?

Changing the volume changes the partial pressures (or concentrations) of the gases. The system shifts to counteract this change, moving towards more moles if volume increases, and fewer moles if volume decreases.

Q: What if inert gas is added at constant volume?

Adding an inert gas at constant volume increases total pressure but does not change the partial pressures or concentrations of the reacting gases, so the equilibrium is unaffected.

Q: Does $\Delta n_g$ include solid or liquid moles?

No, $\Delta n_g$ only considers the stoichiometric coefficients of gaseous species, because the volume of solids and liquids is negligible and their active mass is taken as unity.

Q: What happens if we increase the volume for option (a)?

For option (a), $\Delta n_g = 1$. Increasing the volume decreases pressure, so the system will shift forward to produce more moles and increase the pressure back.

Q: Is the equilibrium constant $K_c$ affected by volume change?

No, the equilibrium constant $K_c$ (or $K_p$) depends only on temperature. Volume changes only shift the reaction quotient $Q_c$ temporarily until equilibrium is re-established.

Question

One of the following equilibria is not affected by change in volume of the flask

Options:

(A) $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$

(B) $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

(C) $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$

(D) $SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g)$

Sigma Percentile · Accepted Answer

### The Core Concept: Le Chatelier's Principle

Imagine a chemical reaction happily sitting in a state of equilibrium inside a closed flask. Suddenly, you push a piston down, decreasing the volume of the flask. What happens?

According to **Le Chatelier's Principle**, the system will try to undo this change. Decreasing the volume increases the pressure of the gases inside. To reduce this pressure, the equilibrium will shift towards the side that has **fewer moles of gas**.

Conversely, if you increase the volume, the pressure drops, and the system shifts towards the side with **more moles of gas** to build the pressure back up.

### The Role of Gaseous Moles

The key to solving this problem lies in a simple value: $\Delta n_g$. This represents the change in the number of gaseous moles during the reaction.

Mathematically, it is calculated as:
$$ \Delta n_g = (	ext{Moles of gaseous products}) - (	ext{Moles of gaseous reactants}) $$

If $\Delta n_g$ is positive, the forward reaction produces more gas. If it's negative, the forward reaction consumes gas.

But what if $\Delta n_g = 0$? This means the number of gaseous moles is exactly the same on both sides of the equation. In this special case, changing the volume of the flask has **absolutely no effect** on the equilibrium state!

### Analyzing the Options

Let's put our options to the test by calculating $\Delta n_g$ for each one.

**Option (a):** 
$$ PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g) $$
Here, we have $1 + 1 = 2$ moles of products and $1$ mole of reactant. 
$$ \Delta n_g = 2 - 1 = 1 $$
Since $\Delta n_g 
eq 0$, this equilibrium will be affected by a volume change.

**Option (b):** 
$$ N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g) $$
We have $2$ moles of products and $1 + 3 = 4$ moles of reactants. 
$$ \Delta n_g = 2 - 4 = -2 $$
Again, $\Delta n_g 
eq 0$, so volume changes will shift this equilibrium.

**Option (c):** 
$$ N_2(g) + O_2(g) ightleftharpoons 2NO(g) $$
Look closely at this one. We have $2$ moles of product ($2NO$) and $1 + 1 = 2$ moles of reactants ($N_2$ and $O_2$). 
$$ \Delta n_g = 2 - 2 = 0 $$
Bingo! The number of moles is perfectly balanced.

**Option (d):** 
$$ SO_2Cl_2(g) ightleftharpoons SO_2(g) + Cl_2(g) $$
Here, we have $1 + 1 = 2$ moles of products and $1$ mole of reactant. 
$$ \Delta n_g = 2 - 1 = 1 $$
This one is also affected by volume changes.

### The Final Verdict

Because the reaction in option (c) has an equal number of gaseous moles on both the reactant and product sides ($\Delta n_g = 0$), altering the volume of the flask will not shift the equilibrium in either direction.

The system is perfectly immune to volume changes!

Animated Solution for Chemistry - Chemical Equilibrium: One of the following equilibria is not affected by change in volume of the flask

Select Answer:

Visualized Solution

Problem Statement

Le Chatelier's Principle

Checking Option (a)

Checking Option (b)

Checking Option (c)

Checking Option (d)

Final Conclusion

The Sigma Insight: Le-Chatelier's Principle

The Core Concept

The Role of Gaseous Moles

Analyzing the Options

The Final Verdict

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