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 INCORRECT statement among the following, for this reaction, is

(A)Decrease in the total pressure will result in formation of more moles of gaseous X
(B)At the start of the reaction, dissociation of gaseous $X_{2}$ takes place spontaneously
(C) $β_{equilibrium} = 0.7$ ✓ Correct
(D) $K_{C} < 1$

Answer: (C)

Solution

Since

Δ_{r} G^{\circ} > 0

, we have

K < 1

. From

Δ G^{\circ} = - 2.303 R T lo g K_{c}

, if

Δ G^{\circ} > 0

then

K_{c} < 1

. Decreasing pressure favors the side with more moles (forward direction), so (A) is correct. At the start,

Q = 0 < K

, so the reaction proceeds forward spontaneously despite

Δ G^{\circ} > 0

, making (B) correct. Statement (D) is correct. Statement (C) claims a specific value for

β_{equilibrium}

but no data is given to calculate this exact value, making it incorrect.

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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