Chemical and Ionic Equilibrium - Result Question 56
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Thermal decomposition of gaseous $X_2$ to gaseous $X$ at $298$ K takes place according to the following equation:
$ X_2(g) \rightleftharpoons 2 X(g) $
The standard reaction Gibbs energy, $\Delta_r G^{\circ}$, of this reaction is positive. At the start of the reaction, there is one mole of $X_2$ and no $X$. As the reaction proceeds, the number of moles of $X$ formed is given by $\beta$. Thus, $\beta_{\text {equilibrium }}$ is the number of moles of $X$ formed at equilibrium. The reaction is carried out at a constant total pressure of $2$ bar. Consider the gases to behave ideally. (Given, $R=0.083 \mathrm{~L}$ $ \mathrm{bar} $ $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
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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Answer:
Correct Answer: 56. (c)
Solution:
- (a) $ K_p=\dfrac{4 p x^2}{\left(4-x^2\right)}=p x^2 \quad$ $ (\because 4 > > > x) $
$\therefore x \propto \sqrt{\dfrac{1}{p}}$
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 $K_p=\dfrac{4 \times 2(0.7)^2}{\left[4-(0.7)^2\right]}$
$ =1.16>1 $
Thus, (c) is incorrect.
(d) At equilibrium, $\Delta G=0$
$ \therefore \quad \Delta G^{\circ}=-2.303 $ $R T $ $\log$ $ K_p $
Since, $ \Delta G^{\circ}=+$ ve
Hence, $K_p < 1$
$ K_C=\dfrac{K_p}{(R T)} $
Then $K_C < 1$.
Thus, (d) is correct.
BETA
