JEE Main · 2023 · Shift-IIeasyCEQ-001

The effect of addition of helium gas to the following reaction at constant volume in equilibrium state is: PCl5(g)…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The effect of addition of helium gas to the following reaction at constant volume in equilibrium state is:

PCl5(g)PCl3(g)+Cl2(g)\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}

Options
  1. a

    The equilibrium will shift in the forward direction and more of Cl2\mathrm{Cl_2} and PCl3\mathrm{PCl_3} will be produced

  2. b

    The equilibrium will go backward due to suppression of dissociation of PCl5\mathrm{PCl_5}

  3. c

    Helium will deactivate PCl5\mathrm{PCl_5} and reaction will stop

  4. d

    Addition of helium will not affect the equilibrium

Correct Answerd

Addition of helium will not affect the equilibrium

Detailed Solution

Key Concept: Adding an inert gas at constant volume does not change the partial pressures (or concentrations) of any reacting species.

Step 1 — Effect on partial pressures: At constant volume, adding He increases total pressure but the partial pressures of PCl5\mathrm{PCl_5}, PCl3\mathrm{PCl_3}, and Cl2\mathrm{Cl_2} remain unchanged.

Step 2 — Effect on equilibrium: Qp=PPCl3PCl2PPCl5Q_p = \frac{P_{\mathrm{PCl_3}} \cdot P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}} Since no partial pressure changes, Qp=KpQ_p = K_p \rightarrow equilibrium is unaffected.

Note: If He were added at constant pressure, the equilibrium would shift forward (mole fraction of reactants increases).

Answer: Option (4)

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