Chemical and Ionic Equilibrium - Result Question 111
####55. The equilibrium constant $K _p$ for this reaction at $298 K$, in terms of $\beta _{\text {equilibrium }}$ is
(2016 Adv.)
(a) $\frac{8 \beta^{2}{ } _{\text {equilibrium }}}{2-\beta _{\text {equilibrium }}}$
(b) $\frac{8 \beta^{2}{ } _{\text {equilibrium }}}{4-\beta _{\text {equilibrium }}^{2}}$
(c) $\frac{4 \beta^{2}{ } _{\text {equilibrium }}}{2-\beta _{\text {equilibrium }}}$
(d) $\frac{4 \beta^{2}{ } _{\text {equilibrium }}}{4-\beta^{2}{ } _{\text {equilibrium }}}$
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Solution:
$$ \begin{aligned} & X _2(g) \rightleftharpoons 2 X(g) \ & \text { At } \begin{array}{lll} t=0 & 1 & 0 \end{array} \ & \text { At equilibrium } \quad\left(1-\frac{x}{2}\right) \quad x \quad\left(\text { where, } x=\beta _{eq}\right) \ & \text { Total moles }=\left(1+\frac{x}{2}\right) \text { and Mole fraction, } X _2(g)=\frac{\left(1-\frac{x}{2}\right)}{\left(1+\frac{x}{2}\right)} \ & X(g)=\left(\frac{x}{1+\frac{x}{2}}\right) \text { and } p=2 \text { bar } \ & \text { Partial pressure, } p _{X 2}=\left(\frac{1-\frac{x}{2}}{1+\frac{x}{2}}\right) \cdot p \text { and } p _X=\frac{p \cdot x}{\left(1+\frac{x}{2}\right)} \ & \begin{array}{l} \therefore \quad K _p=p _X^{2} / p _{X _2}=\frac{\left[p x /\left(1+\frac{x}{2}\right)\right]^{2}}{p \frac{(1-x / 2)}{\left(1+\frac{x}{2}\right)}} \ \quad=\frac{4 p x^{2}}{\left(4-x^{2}\right)}=\frac{8 \beta _{eq}^{2}}{\left(4-\beta _{eq}^{2}\right)} \end{array} \end{aligned} $$
