Question

$X Y_{2}$ dissociates as $X Y_{2}(g) \rightleftharpoons X Y(g) + Y(g)$ when the initial pressure of $X Y_{2}$ is $600 , \mathrm{mm}$ of $\mathrm{Hg}$. The total equilibrium pressure is $800 , \mathrm{mm}$ of $\mathrm{Hg}$. Find the value of the equilibrium constant for the reaction, assuming that the volume of the system remains unchanged.

A) $50 , \mathrm{mm} , \mathrm{Hg}$

B) $100 , \mathrm{mm} , \mathrm{Hg}$

C) $166.6 , \mathrm{mm} , \mathrm{Hg}$

D) $400 , \mathrm{mm} , \mathrm{Hg}$

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Answer

Solution

The correct option is B) $100 , \mathrm{mm} , \mathrm{Hg}$.

The dissociation reaction can be represented as: $$ XY_{2}(g) \rightleftharpoons XY(g) + Y(g) $$

Given:

  • Initial pressure of $XY_{2}$: $600 , \mathrm{mm, Hg}$
  • Total equilibrium pressure: $800 , \mathrm{mm, Hg}$

Let $x , \mathrm{mm, Hg}$ be the amount of $XY_{2}$ that dissociates. This also creates $x , \mathrm{mm, Hg}$ each of $XY$ and $Y$. Therefore, the remaining pressure of $XY_{2}$ is $(600 - x) , \mathrm{mm, Hg}$.

The total pressure at equilibrium, considering the contributions from all species, is: $$ P_{total} = (600 - x) + x + x = 600 + x , \mathrm{mm, Hg} $$

Setting the total pressure equation to the equilibrium pressure: $$ 600 + x = 800 \quad \Rightarrow \quad x = 200 , \mathrm{mm, Hg} $$

At equilibrium, the pressures for the various components are:

  • $P_{XY_{2}} = 600 - x = 400 , \mathrm{mm, Hg}$
  • $P_{XY} = x = 200 , \mathrm{mm, Hg}$
  • $P_{Y} = x = 200 , \mathrm{mm, Hg}$

The equilibrium constant $K_p$ is defined as: $$ K_p = \frac{P_{XY} \cdot P_{Y}}{P_{XY_{2}}} $$

Calculating $K_p$: $$ K_p = \frac{200 \times 200}{400} = 100 , \mathrm{mm, Hg} $$

Thus, the value of the equilibrium constant $K_p$ is $100 , \mathrm{mm, Hg}$.


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