Question

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Answer 1

Explanation:

It is given that x−x1=8, by squaring both sides we get:

(x−x1)2=82⇒(x)2+(x1)2−(2×x×x1)=64(∵(a−b)2=a2+b2−2ab)⇒x2+x21−2=64⇒x2+x21=64+2⇒x2+x21=66....(1)

Now, consider (x+x1)2 then we have:

(x+x1)2=(x)

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 :-

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

➲ At initial moles :

$→ \sf PCl_5 =\: 1PCl$

$→\sf PCl_3 =\: 0PCl$

$→ \sf Cl_2 =\: 0Cl$

And,

➲ At equilibrium :

$→\sf PCl_5 =\: (1 - x)PCl$

$→\sf PCl_3 =\: xPCl$

$→ \sf Cl_2 =\: xCl$

Again, total number of moles :

$↦ \sf 1 - x + x + x =\:$

$↦ \sf\bold{\green{x =\: 1 + x}}$

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

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

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