Chemical and Ionic Equilibrium - Result Question 62

####6. Two solids dissociate as follows:

$$ \begin{aligned} & A(s) \rightleftharpoons B(g)+C(g) ; K _{p _1}=x atm^{2} \ & D(s) \rightleftharpoons C(g)+E(g) ; K _{p _2}=y atm^{2} \end{aligned} $$

The total pressure when both the solids dissociate simultaneously is

(2019 Main, 12 Jan I)

(a) $\sqrt{x+y}$ atm

(b) $x^{2}+y^{2}$ atm

(c) $(x+y) atm$

(d) $2(\sqrt{x+y}) atm$

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Answer:

Correct Answer: 6. (c)

Solution:

  1. The equilibrium reaction for the dissociation of two solids is given as:

At equilibrium

$$ A(s) \rightleftharpoons B(g)+C(g) $$

$$ K _{p _1}=x=p _B \cdot p _C=p _1\left(p _1+p _2\right) $$

Similarly, $D(s) \rightleftharpoons C(g)+E(g)$

At equilibrium

$$ Im _{p _2}=y=p _C \cdot p _E=\left(p _1+p _2 \cdot p _2\right) p _2 $$

On adding Eq. (i) and (ii), we get.

$K _{p _1}+K _{p _2}=x+y=p _1\left(p _1+p _2\right)+p _2\left(p _1+p _2\right)$

$$ =\left(p _1+p _2\right)^{2} $$

or $\sqrt{x+y}=p _1+p _2$

Now, total pressure is given as

$$ \begin{aligned} p _T & =p _B+p _C+p _E \ & =p _1+\left(p _1+p _2\right)+p _2 \ & =2\left(p _1+p _2\right) \end{aligned} $$

On substituting the value of $p _1+p _2$ from Eq. (iii) to Eq. (iv), we get

$$ p _T=2 \sqrt{x+y} $$



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