Concept:Pressure changes only affect the equilibrium position of gaseous reactions if there is a difference in the number of gaseous moles between products and reactants.
Formula:$$ \text{If } \Delta n = 0, \text{ Pressure has no effect} $$
Solution:- Analyze \( \Delta n \) for each option.
- Option A: \( \Delta n = 2 - 3 = -1 \). (Affected)
- Option C: \( \Delta n = 2 - 3 = -1 \). (Affected)
- Option D: \( \Delta n = 2 - 1 = +1 \). (Affected)
- Option B: \( \text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO} \). Reactant moles = \( 1 + 1 = 2 \). Product moles = 2.
- \( \Delta n = 2 - 2 = 0 \). Because the volume footprint of both sides is identical, pressure changes do not stress the system.
Why other options are incorrect:All other options have a non-zero \( \Delta n \), meaning Le Chatelier's principle would dictate a shift to compensate for pressure changes.
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