$2A{B_2}(g) \rightleftharpoons 2AB(g) + {B_2}(g)$
$1$ $0$ $0$
Initially
$(1-x)$ $x$ $\frac {x}{2}$
At equilibrium
Total moles at equilibrium $=$
$1-x+x+\frac{1}{2}=1+\frac{x}{2}=1$
$[\because \,{\text{x}}$ is small incomparision to unity $]$
${P_{{\text{A}}{{\text{B}}_2}}} = (1 - {\text{x}}){\text{P}}\quad {P_{{\text{AB}}}} = {\text{xP}}\quad {{\text{p}}_{{{\text{B}}_2}}} = \frac{{{\text{xP}}}}{2}$
$\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{x}^{3} \mathrm{P}^{3}}{2(1-\mathrm{x})^{2} \mathrm{P}^{2}}=\frac{\mathrm{x}^{3} \mathrm{P}}{2} \quad[\because(1-\mathrm{x}) \approx 1]$
$K_{p}=\frac{x^{3} P}{2}$
$x^{3}=\frac{2 K_{p}}{P}$
${x}=\sqrt{\frac{2 K_{p}}{P}}$
$(2)$ is correct answer.
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$\Delta_f G^0[\mathrm{C}(\text { graphite })]=0 \mathrm{kJmol}^{-1}$
$\Delta_f G^0[\mathrm{C}(\text { diamond })]=2.9 \mathrm{kJmol}^{-1}$
The standard state means that the pressure should be $1$ bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mol}^{-1}$. If $\mathrm{C}$ (graphite) is converted to $\mathrm{C}$ (diamond) isothermally at $\mathrm{T}=298 \mathrm{~K}$, the pressure at which $C$ (graphite) is in equilibrium with C(diamond), is
[Useful information: $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^5 \mathrm{~Pa}$ ]