Chemical and Ionic Equilibrium - Result Question 112
####56. The incorrect statement among the following for this reaction, is
(2016 Adv.)
(a) Decrease in the total pressure will result in the formation of more moles of gaseous $X$
(b) At the start of the reaction, dissociation of gaseous $X _2$ takes place spontaneously
(c) $\beta _{\text {equilibrium }}=0.7$
(d) $K _C<1$
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Solution:
- (a)
$$ K _p=\frac{4 p x^{2}}{\left(4-x^{2}\right)}=p x^{2} $$
$$ \therefore \quad x \propto \sqrt{\frac{1}{p}} $$
$(\because 4»>x)$
If $p$ decreases, $x$ increases. Equilibrium is shifted in the forward side. Thus, statement (a) is correct.
(b) At the start of the reaction, $Q=0$ where, $Q$ is the reaction quotient $\Delta G=\Delta G^{\circ}+2.303 R T \log Q$
Since, $\Delta G^{\circ}>0$, thus $\Delta G$ is $-ve$.
Hence, dissociation takes place spontaneously.
Thus, (b) is correct.
(c) If we use $x=0.7$ and $p=2$ bar then $K _p=\frac{4 \times 2(0.7)^{2}}{\left[4-(0.7)^{2}\right]}$
Thus, (c) is incorrect.
$$ =1.16>1 $$
(d) At equilibrium, $\Delta G=0$
$$ \begin{array}{ll} \therefore & \Delta G^{\circ}=-2.303 R T \log K _p \ \text { Since, } & \Delta G^{\circ}=+ \text { ve } \end{array} $$
Hence, $K _p<1$
$$ K _C=\frac{K _p}{(R T)} $$
Then $K _C<1$. Thus, (d) is correct.
Topic 2 lonic Equilibrium
| $Al(OH) _3 \rightleftharpoons$ | $Al^{3+}$ | $+3 OH^{-}$ | |
|---|---|---|---|
| Initiall | 1 | 0 | 0 |
| At equilibrium | $1-S$ | $S$ | $3 S+0.2$ |
