Question

$\sf PCl_5$ dissociates as follows in a closed reaction $\sf PCl_5\rightleftharpoons{PCl_3 + Cl_2}$. If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of $\sf PCl_5$ is x, the partial pressure of $\sf PCl_3$ will be

Answer 1

Answer:-

Partial pressure of PCl₃ is (x/1+x)P

Explanation:-

According to the problem, we have :-

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Initial moles:-

→ PCl₅ = 1

→ PCl₃ = 0

→ Cl₂ = 0

Moles at equilibrium:-

→ PCl₅ = a - ax

→ PCl₃ = ax

→ Cl₂ = ax

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Total moles at equilibrium :-

= a - ax + ax + ax

= a + ax

= a(1 + x)

Mole fraction of PCl₃ :-

= ax/a(1 + x)

= x/(1 + x)

Thus, partial pressure of PCl₃ :-

= (x/1+x) × P

= (x/1+x)P

Answer 2

Answer:

Given :-

• PCl₅ dissociated as follows in a closed reaction PCl₅ $\rightleftharpoons$ PCl₃ + Cl₂.
• If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl₅ is x.

To Find :-

• What is the partial pressure of PCl₃.

Solution :-

$\longmapsto$ $\sf PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$

➲ At initial moles :

⇢ $\sf PCl_5 =\: 1$

⇢ $\sf PCl_3 =\: 0$

⇢ $\sf Cl_2 =\: 0$

And,

➲ At equilibrium :

⇢ $\sf PCl_5 =\: (1 - x)$

⇢ $\sf PCl_3 =\: x$

⇢ $\sf Cl_2 =\: x$

Again, total number of moles :

↦ $\sf 1 - x + x + x =\: 1 + x$

↦ $\sf\bold{\green{x =\: 1 + x}}$

Now, we have to find the partial pressure of PCl₃,

↦ $\sf PCl_3 =\: \bigg \lgroup \dfrac{x}{1 + x} \bigg \rgroup \times P$

➠ $\sf\bold{\red{PCl_3 =\: \bigg \lgroup \dfrac{x}{1 + x} \bigg \rgroup P}}$

$\therefore$ The partial pressure of PCl₃ is $\sf\bold{PCl_3 =\: \bigg \lgroup \dfrac{x}{1 + x} \bigg \rgroup P}$.