Chemical and Ionic Equilibrium - Result Question 55
Passage
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} $ $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
55. The equilibrium constant $K _p$ for this reaction at $298 K$, in terms of $\beta _{\text {equilibrium }}$ is
(2016 Adv.)
(a) $\dfrac{8 \beta^{2}{ } _{\text {equilibrium }}}{2-\beta _{\text {equilibrium }}}$
(b) $\dfrac{8 \beta^{2}{ } _{\text {equilibrium }}}{4-\beta _{\text {equilibrium }}^{2}}$
(c) $\dfrac{4 \beta^{2}{ } _{\text {equilibrium }}}{2-\beta _{\text {equilibrium }}}$
(d) $\dfrac{4 \beta^{2}{ } _{\text {equilibrium }}}{4-\beta^{2}{ } _{\text {equilibrium }}}$
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Answer:
Correct Answer: 55. (b)
Solution:
- $ \quad\quad\quad\quad\quad X_2(g) \rightleftharpoons 2 X(g) $
At $t=0$ $\quad\quad\quad\quad\quad$ 1 $\quad\quad$ 0
At equilibrium $\quad\left(1-\dfrac{x}{2}\right) \quad x \quad$ (where, $x=\beta_{\text {eq }}$ )
Total moles $=\left(1+\dfrac{x}{2}\right)$ and Mole fraction, $X_2(g)=\dfrac{\left(1-\dfrac{x}{2}\right)}{\left(1+\dfrac{x}{2}\right)}$
$X(g)=\left(\dfrac{x}{1+\dfrac{x}{2}}\right)$ and $p=2$ bar
$ \text { Partial pressure, } p_{X 2}=\left(\dfrac{1-\dfrac{x}{2}}{1+\dfrac{x}{2}}\right) \cdot p \text { and } p_X=\dfrac{p \cdot x}{\left(1+\dfrac{x}{2}\right)} $
$ \begin{aligned} & \therefore \quad K_p=p_X^2 / p_{X_2}=\dfrac{\left[p x /\left(1+\dfrac{x}{2}\right)\right]^2}{p \dfrac{(1-x / 2)}{\left(1+\dfrac{x}{2}\right)}} \\ & \quad=\dfrac{4 p x^2}{\left(4-x^2\right)}=\dfrac{8 \beta_{\mathrm{eq}}^2}{\left(4-\beta_{\mathrm{eq}}^2\right)} \end{aligned} $
BETA
