The equilibrium constant K P for this reaction at 298 K , in terms of β e q u | TestHub

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Paragraph: Thermal decomposition of gaseous $X_{2}$ to gaseous $X$ at $298 K$ takes place according to the following equation: $X_{2} (g) ⇌ 2 X (g)$ The standard reaction Gibbs energy, $Δ_{r} G^{\circ}$ , 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 $β$ . Thus, $β_{equilibrium}$ 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 : $R = 0.083 L$ bar $K^{- 1} mol^{- 1}$ )

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

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Solution:

$\underset{at equilibrium}{X_{2} (g)} ⇌ \underset{β_{equilibrium}}{2 X (g)}$
$n_{T} = \frac{1 - β_{equilibrium}}{2} + β_{equilibrium}$
$= 1 + \frac{β_{equilibrium}}{2}$
$1 - \frac{β_{equilibrium}}{2}$
$K_{P} = \frac{P_{x}^{2}}{P_{x_{2}}} = \frac{{(\frac{β_{equilibrium}}{1 + \frac{β_{equilibrium}}{2}} P_{T})}^{2}}{(\frac{1 - \frac{β_{equilibrium}}{2}}{1 + \frac{β_{equilibrium}}{2}} P_{T})}$
$K_{P} = \frac{β_{e q u i l i b r i u m}^{2}}{1 - \frac{β_{e q u i l i b r i u m}^{2}}{4}} P_{T} = \frac{2 β_{e q u i l i b r i u m}^{2}}{1 - \frac{β_{e q u i l i b r i u m}}{4}}$
$= \frac{8 β_{e q u i l i b r i u m}^{2}}{4 - β_{e q u i l i b r i u m}^{2}}$

Chemistry - Thermodynamics - II Question with Solution | TestHub

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