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Chemical Equilibrium question

2002 · Shift 0 · Q8
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Chemical Equilibrium question

2002 · Shift 0 · Q8

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
Change in volume of the system does not alter which of the following equilibria?
  1. A
    N2N_2N2​(g) + O2O_2O2​(g) ⇋\leftrightharpoons⇋ 2NONONO (g)
  2. B
    PCl5PCl_5PCl5​(g) ⇋\leftrightharpoons⇋ PCl3PCl_3PCl3​ (g) + Cl2Cl_2Cl2​ (g)
  3. C
    N2N_2N2​(g) + 3H2H_2H2​(g) ⇋\leftrightharpoons⇋ 2NH3NH_3NH3​ (g)
  4. D
    SO2Cl2SO_2Cl_2SO2​Cl2​ (g) ⇋\leftrightharpoons⇋ SO2SO_2SO2​ (g) + Cl2Cl_2Cl2​ (g)
View written solutionFree

Correct answer: A

  1. Principle involved: effect of change in volume on gaseous equilibrium

For a gaseous equilibrium, changing volume affects the equilibrium position only when the total number of moles of gaseous reactants and products are different.

  • If Δng=0\Delta n_g = 0Δng​=0, then change in volume/pressure does not shift equilibrium.
  • If Δng≠0\Delta n_g \neq 0Δng​=0, then equilibrium shifts on changing volume.

Here,

Δng=(total gaseous moles of products)−(total gaseous moles of reactants)\Delta n_g = (\text{total gaseous moles of products}) - (\text{total gaseous moles of reactants})Δng​=(total gaseous moles of products)−(total gaseous moles of reactants)
  1. Check each option

Option A

N2(g)+O2(g)⇋2NO(g)N_2(g) + O_2(g) \leftrightharpoons 2NO(g)N2​(g)+O2​(g)⇋2NO(g)

Reactant gaseous moles =1+1=2= 1+1 = 2=1+1=2

Product gaseous moles =2= 2=2

So,

Δng=2−2=0\Delta n_g = 2-2 = 0Δng​=2−2=0

Hence, change in volume does not affect this equilibrium.

Option B

PCl5(g)⇋PCl3(g)+Cl2(g)PCl_5(g) \leftrightharpoons PCl_3(g) + Cl_2(g)PCl5​(g)⇋PCl3​(g)+Cl2​(g)

Reactant gaseous moles =1= 1=1

Product gaseous moles =1+1=2= 1+1 = 2=1+1=2

So,

Δng=2−1=1\Delta n_g = 2-1 = 1Δng​=2−1=1

Since Δng≠0\Delta n_g \neq 0Δng​=0, volume change will affect the equilibrium.

Option C

N2(g)+3H2(g)⇋2NH3(g)N_2(g) + 3H_2(g) \leftrightharpoons 2NH_3(g)N2​(g)+3H2​(g)⇋2NH3​(g)

Reactant gaseous moles =1+3=4= 1+3 = 4=1+3=4

Product gaseous moles =2= 2=2

So,

Δng=2−4=−2\Delta n_g = 2-4 = -2Δng​=2−4=−2

Since Δng≠0\Delta n_g \neq 0Δng​=0, volume change will affect the equilibrium.

Option D

SO2Cl2(g)⇋SO2(g)+Cl2(g)SO_2Cl_2(g) \leftrightharpoons SO_2(g) + Cl_2(g)SO2​Cl2​(g)⇋SO2​(g)+Cl2​(g)

Reactant gaseous moles =1= 1=1

Product gaseous moles =1+1=2= 1+1 = 2=1+1=2

So,

Δng=2−1=1\Delta n_g = 2-1 = 1Δng​=2−1=1

Since Δng≠0\Delta n_g \neq 0Δng​=0, volume change will affect the equilibrium.

  1. Conclusion

Only Option A has equal total moles of gaseous reactants and products, so change in volume does not alter the equilibrium.

A\boxed{A}A​
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