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:

  1. (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 $1-S$ $S$ $3 S+0.2$


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