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Arrhenius Equation

The equilibri...

Question

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:

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

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\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}}

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

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\frac{4 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}}

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Solution

The correct option is B $\frac{8 β_{e q u i l i b r i u m}^{2}}{4 - β_{e q u i l i b r i u m}^{2}}$
$X_{2} (g) ⇌ 2 X (g)$
$\begin{matrix} Initial mole & 1 & 0 t = t_{e q} & (1 - α) & 2 α \end{matrix}$

Given $2 α = β_{e q u i l i b r i u m}$

$α = \frac{β_{e q u i l i b r i u m}}{2}$

Total mole at equilibrium $= (1 + α) = (1 + (\frac{β e q}{2})$

$P_{x_{2}} = ⎡ ⎢ ⎢ ⎢ ⎣ \frac{1 - \frac{β e q}{2}}{1 + \frac{β e q .}{2}} P_{t o t a l} ⎤ ⎥ ⎥ ⎥ ⎦ = [\frac{2 - β e q}{2 + β e q} P_{t o t a l}] = [\frac{2 - β e q}{2 + β e q} P_{t o t a l}]$

$P_{x (g)} = ⎡ ⎢ ⎢ ⎢ ⎣ \frac{β e q}{1 + \frac{β e q}{2}} P_{t o t a l} ⎤ ⎥ ⎥ ⎥ ⎦ = [\frac{2 β e q}{2 + β e q}] P_{t o t a l}$

So, $K_{p} = \frac{(P x)^{2}}{(P x_{2})} = \frac{{[\frac{2 β e q}{2 + β e q} \times P_{t o t a l}]}_{t o t a l}^{2}}{[\frac{2 - β e q}{(2 + β e q)} \times P_{t o t a l}]}$

$K_{p} = \frac{4 β^{2} e q .}{4 - β^{2} e q} \times P_{T o t a l} = (\frac{8 β^{2} e q}{4 - β^{2} e q})$

So, answer is $= B$ .

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The equilibrium constant Kp for this reaction at 298K, in terms of βequilibrium, is:

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: