Chemical and Ionic Equilibrium 1 Question 6
6. Two solids dissociate as follows:
$$ \begin{aligned} & A(s) \rightleftharpoons B(g)+\mathrm{C}(g) ; K_{p_{1}}=x \mathrm{~atm}^{2} \ & D(s) \rightleftharpoons \mathrm{C}(g)+E(g) ; K_{p_{2}}=y \mathrm{~atm}^{2} \end{aligned} $$
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) $(x+y) \mathrm{atm}$
(d) $2(\sqrt{x+y}) \mathrm{atm}$
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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
$$ \mathrm{Im}{p{2}}=y=p_{C} \cdot p_{E}=\left(p_{1}+p_{2} \cdot p_{2}\right) p_{2} $$
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
$$ p_{T}=2 \sqrt{x+y} $$
