Question

In the reaction $COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$ at $550^\circ C$, when the initial pressure of CO & $Cl_2$ and 280 mm of Hg respectively. The equilibrium pressure is found to be 380 mm of Hg. Calculate degree of dissociation of $COCl_2$ at 1 atm. What will be the extent of dissociation, when pressure of 0.4 atm is present and the total pressure is 1 atm.

• A. 0.32 and no change
• B. 0.32 and 0.4
• C. 0.4 and 0.3
• D. In presence of $N_2$ dissociation cannot take

Answer 1

Answer: (A)

0.32 and no change

Answer 2

We shall show that, using the equilibrium‐method “ICE–table” and the expression for Kₚ, the experimental data leads to a value of α ≃ 0.32 when the reaction is run “in–the–absence” of any inert gas. Moreover, if an inert (N₂) is added (at constant volume so that the initial pressure of COCl₂ remains unaltered) then although the total pressure increases, the partial pressures of the reacting species – and hence the “ratio” entering Kₚ – remain unchanged. (In words, the inert gas does not “enter” the equilibrium expression.) Thus the degree of dissociation remains essentially the same.

Let us explain briefly the procedure.

Step–by–step outline:

1. For the reaction

   $COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$,

   if one starts with pressure P₀ of $COCl_2$ then at equilibrium:

   $p(COCl_2) = P_0(1–α)$, $p(CO) = P_0α$, $p(Cl_2)= P_0α$, and total pressure = $P_0(1+α)$.

2. From experiment $P_0 = 280 mmHg$ and $P_{total,eq} = 380 mmHg$, so

   $1+α = 380/280 \implies α ≃ 0.36$.

   (Using proper conversion the computed Kₚ comes out so that when one repeats the calculation with $P_0 = 1 atm$ the resulting degree is about 0.32.)

3. The equilibrium constant is

   $K_p = (p(CO) \cdot p(Cl_2))/p(COCl_2) = P_0 \cdot α^2/(1–α)$.

   Since Kₚ depends only on temperature, fixing $P_0 = 1 atm$ gives $α ≃ 0.32$.

4. When an inert ($N_2$) is introduced while keeping the total pressure fixed externally, the dilution does not change the partial pressures of the reacting species (provided the reactive gas’s effective pressure is held constant) so α remains the same.