The Question is from neet-chemistry chemical-equilibrium , and it was created by the author P SRAVAN KUMAR

Question

A system at equilibrium consisting of $N_2(g)$, $O_2(g)$ and $NO(g)$ in a sealed vessel at constant temperature is subjected to a sudden increase in pressure.  According to Le Chatelier's principle, which of the following is most likely to occur?

P SRAVAN KUMAR · Accepted Answer

## CORE INFORMATION
✓ **Answer**: 1
## CONCEPT FRAMEWORK
 **Primary Concept**: Le Chatelier's Principle
 **Related Topics**: Chemical Equilibrium, Gas Laws, Pressure and Volume
 **Prerequisites**: Understanding of equilibrium constants and the effect of changes in conditions on equilibrium positions.
## SOLUTION PATHWAY
**Given Information**:
  A system at equilibrium: $N_2(g) + O_2(g) ightleftharpoons 2NO(g)$
  Sudden increase in pressure at constant temperature.
  Sealed vessel.
**Method & Steps**:
 1. **Applying Le Chatelier's Principle:** Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.
  Reasoning: Increasing the pressure on a gaseous system at constant temperature will favor the reaction that produces fewer gas molecules to reduce the pressure.
  Formula/Rule: None directly applicable, but understanding the stoichiometry of the reaction is crucial.
 2. **Analyzing the Reaction Stoichiometry:** The balanced reaction is $N_2(g) + O_2(g) ightleftharpoons 2NO(g)$.  There are 2 moles of gas on the reactant side and 2 moles of gas on the product side.
  Working: Since the number of gas moles is equal on both sides, a change in pressure will not shift the equilibrium significantly.
  Key Point:  The key here is the equal number of moles of gas on both sides of the equation.  If the number of moles were different, the equilibrium would shift to favor the side with fewer moles of gas.
 3. **Determining the Equilibrium Shift:** Because the number of moles of gas is the same on both sides, the increase in pressure will not significantly affect the equilibrium position.
  Result: No significant shift in equilibrium occurs.
  Verification:  The system will adjust minimally to accommodate the pressure increase, maintaining the equilibrium constant.
 ## OPTION ANALYSIS
A :x: The equilibrium does not shift towards the reactants because the number of gas moles is equal on both sides of the equation.
B :x: The equilibrium does not shift towards the products because the number of gas moles is equal on both sides of the equation.
C ✓: This is correct. Since the number of moles of gas is equal on both sides of the equation, a pressure increase does not favor either side.
D :x: The concentration of NO will not increase significantly because there is no shift in equilibrium.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying Le Chatelier's principle without considering the stoichiometry of the reaction:  Always examine the number of moles of gaseous reactants and products.
  2.Assuming a pressure change always shifts the equilibrium: Pressure changes only affect the equilibrium position when the number of moles of gas is different on the reactant and product sides. 
 **Quick Tips**:
  1.Count the moles of gas on both sides of the equation before applying Le Chatelier's principle.
  2.Remember that increasing pressure favors the side with fewer moles of gas, while decreasing pressure favors the side with more moles of gas (at constant temperature).

P SRAVAN KUMAR · Answer

The equilibrium shifts towards the reactants.

P SRAVAN KUMAR · Answer

The equilibrium shifts towards the product.

P SRAVAN KUMAR · Answer

The correct Option is "No significant shift in equilibrium occurs."

P SRAVAN KUMAR · Answer

The concentration of NO increases significantly.

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