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. (d)
Solution:
- The equilibrium reaction for the dissociation of two solids is given as:
$ A(s) \rightleftharpoons \underset{p_1}B(g)+\underset{p_1+p_2}{C(g)}$
At equilibrium
$ K _{p _1}=x=p _B \cdot p _C=p _1\left(p _1+p _2\right) \hspace{20mm}…(i) $
Similarly, $D(s) \rightleftharpoons C(g)+E(g)$
At equilibrium $\hspace{10mm} p_1 + p_2 + p_3$
$ K_{p_2} =y=p _C \cdot p _E=\left(p _1+p _2 \cdot p _2\right) p _2\hspace{20mm}…(ii) $
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 $\sqrt{x+y}=p _1+p _2 \hspace{20mm}…(iii)$
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)\hspace{20mm}…(iv) \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} $
BETA
