Paragraph: Thermal decomposition of gaseous X_ 2 to gaseous X at 298 ~K takes place according to the following equation: X_ 2 (~g) leftharpoons 2 X(g) The standard reaction Gibbs energy, Δ_ r G^ , of this reaction is

Correct answer: B. 8 β _ equilibrium ^ 2 4 - β _ equilibrium ^ 2 X _ 2 ( g ) _ atequilibrium ⇌ 2X( g ) _ β _ equilibrium n _ T = 1 - β _ equilibrium 2 + β _ equilibrium = 1 + β _ equilibrium 2 1 - β _ equilibrium 2 K _ P = P _ x ^ 2 P _ x _ 2 = ( β _ equilibrium 1 + β _ equilibrium 2 P _ T ) ^ 2 ( 1 - β _ equilibrium 2 1 + β _ equilibrium 2 P _ T ) K _ P = β _ equilibrium ^ 2 1 - β _ equilibrium ^ 2 4 P _ T = 2 β _ equilibrium ^ 2 1 - β _ equilibrium 4 = 8 β _ equilibrium ^ 2 4 - β _ equilibrium ^ 2

JEE Advanced 2016Paper 2

Exam: JEE Advanced
Year: 2016
Paper: Paper 2
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Advanced 2016 · Paper 2 · Chemistry · Chemical Equilibrium

Paragraph: Thermal decomposition of gaseous X_ 2 to gaseous X at...

Paragraph: Thermal decomposition of gaseous

\mathrm{X}_{2}

to gaseous

\mathrm{X}

298 \mathrm{~K}

takes place according to the following equation:

\mathrm{X}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{X}(\mathrm{g})

The standard reaction Gibbs energy,

\Delta_{r} G^{\circ}

, of this reaction is positive. At the start of the reaction, there is one mole of

\mathrm{X}_{2}

and no

\mathrm{X}

. As the reaction proceeds, the number of moles of

\mathrm{X}

formed is given by

\beta

. Thus,

\beta_{\text{equilibrium }}

is the number of moles of

\mathrm{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 \mathrm{~L}

bar

\mathrm{K}^{-1} \mathrm{~mol}^{-1}

)
Question: The equilibrium constant

K_{P}

for this reaction at

298 \mathrm{~K},

in terms of

\beta_{equilibrium}

Options

A $\dfrac{8{\beta}_{equilibrium}^{2}}{2 - {\beta}_{equilibrium}}$
B $\dfrac{8{\beta}_{equilibrium}^{2}}{4 - {\beta}_{equilibrium}^{2}}$
C $\dfrac{4{\beta}_{equilibrium}^{2}}{2 - {\beta}_{equilibrium}}$
D $\dfrac{4{\beta}_{equilibrium}^{2}}{4 - {\beta}_{equilibrium}^{2}}$

Correct answer

B. $\dfrac{8{\beta}_{equilibrium}^{2}}{4 - {\beta}_{equilibrium}^{2}}$

Step-by-step solution

\mathop{{\text{X}}_{\text{2}} ( \text{g} )}_{\text{at}\text{equilibrium}} ⇌ \mathop{2\text{X}( \text{g} )}_{{\text{\beta}}_{\text{equilibrium}}}

{\text{n}}_{\text{T}} = \dfrac{1 - {\text{\beta}}_{\text{equilibrium}}}{2} + {\text{\beta}}_{\text{equilibrium}}

= 1 + \dfrac{{\text{\beta}}_{\text{equilibrium}}}{2}

1 - \dfrac{{\text{\beta}}_{\text{equilibrium}}}{2}

{\text{K}}_{\text{P}} = {\text{P}}_{\text{x}}^{2}{\text{P}}_{{\text{x}}_{2}} = \dfrac{{\left(\dfrac{{\text{\beta}}_{\text{equilibrium}}}{1 + \dfrac{{\text{\beta}}_{\text{equilibrium}}}{2}}{\text{P}}_{\text{T}}\right)}^{2}}{\left(\dfrac{1 - \dfrac{{\text{\beta}}_{\text{equilibrium}}}{2}}{1 + \dfrac{{\text{\beta}}_{\text{equilibrium}}}{2}}{\text{P}}_{\text{T}}\right)}

{K}_{P} = \dfrac{{\beta}_{equilibrium}^{2}}{1 - \dfrac{{\beta}_{equilibrium}^{2}}{4}}{P}_{T} = \dfrac{2{\beta}_{equilibrium}^{2}}{1 - \dfrac{{\beta}_{equilibrium}}{4}}

= \dfrac{8{\beta}_{equilibrium}^{2}}{4 - {\beta}_{equilibrium}^{2}}

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