Thermal decomposition of gaseous $X_{2}$ to gaseous X at 298 K takes place according to the following equation: $X_{2} (g) \overset{}{⇌} 2 X (g)$
The standard reaction Gibbs energy, $Δ_{r} G^{0}$ , 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 $β$ . Thus, $β_{e q u i l i b r i u m}$ 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 L bar $K^{- 1} m o l^{- 1}$ )

The equilibrium constant $K_{P}$ for this reaction at 298 K, in terms of $β_{e q u i l i b r i u m}$ , is

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$\frac{8 {β^{2}}_{e q u i l i b r i u m}}{2 - β_{e q u i l i b r i u m}}$

$\frac{4 {β^{2}}_{e q u i l i b r i u m}}{2 - β_{e q u i l i b r i u m}}$

$\frac{8 {β^{2}}_{e q u i l i b r i u m}}{4 - {β^{2}}_{e q u i l i b r i u m}}$

$\frac{4 {β^{2}}_{e q u i l i b r i u m}}{4 - {β^{2}}_{e q u i l i b r i u m}}$

answer is B.

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Detailed Solution

$\begin{array}{l} X_{2} ⇌ 2 x, Δ = v e \\ 1 \\ 1 - \frac{β}{2} β \\ P_{x_{2}} = \frac{1 - \frac{β}{2}}{1 + \frac{β}{2}} x^{2} = \frac{2 - β}{2 + β} \times 2 \\ P_{x} = \frac{β}{1 + \frac{β}{2}} \cdot 2 = \frac{4 β}{2 + β} \\ k_{p} = \frac{{(\frac{4 β}{2 + β})}^{2}}{(\frac{2 - β}{2 + β}) \cdot 2} = \frac{16 β^{2}}{(2 + β)^{2}} \frac{(2 + β)}{2 (2 - β)} \\ = \frac{8 β^{2}}{4 - β^{2}} \end{array}$