Chemical and Ionic Equilibrium - Result Question 6
####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:
- 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
$$ p _T=2 \sqrt{x+y} $$
