JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Equilibrium

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_{equilibrium}

, is

(A) $\frac{8 β _{equilibrium}^{2}}{2 - β _{equilibrium}}$
(B) $\frac{8 β _{equilibrium}^{2}}{4 - β _{equilibrium}^{2}}$ ✓ Correct
(C) $\frac{4 β _{equilibrium}^{2}}{2 - β _{equilibrium}}$
(D) $\frac{4 β _{equilibrium}^{2}}{4 - β _{equilibrium}^{2}}$

Answer: (B)

Solution

At equilibrium: moles of

X_{2} = 1 - \frac{β _{e}}{2}

, moles of

X = β_{e}

. Total moles

= 1 + \frac{β _{e}}{2}

. Partial pressures:

p_{X} = \frac{β _{e}}{1 + \frac{β _{e}}{2}} \times 2 = \frac{2 β _{e}}{2 + β _{e}}

and

p_{X_{2}} = \frac{1 - \frac{β _{e}}{2}}{1 + \frac{β _{e}}{2}} \times 2 = \frac{2 - β _{e}}{2 + β _{e}}

. Then:

K_{P} = \frac{p _{X}^{2}}{p _{X_{2}}} = \frac{( \frac{2 β _{e}}{2 + β _{e}} ) ^{2}}{\frac{2 - β _{e}}{2 + β _{e}}} = \frac{4 β _{e}^{2}}{( 2 + β _{e} ) ( 2 - β _{e} )} = \frac{4 β _{e}^{2}}{4 - β _{e}^{2}} \times 2 = \frac{8 β _{e}^{2}}{4 - β _{e}^{2}}

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option

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JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Equilibrium

Solution

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