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: C. β _ equilibrium = 0.7 A] On decreasing P _ T [Q = n _ x ^ 2 P _ T n _ x _ 2 n _ T ] Q will be less than K _ p reaction will move in forward direction B] At the start of the reaction ∆ G = ∆ G ^ o + RT ln Q t = 0, Q = 0 ⇒ ∆ _ rxn G = – ve ( spontaneous ) C] If β _ equilibrium = 0.7 K _ p = (8 × 0.49)/(4 - 0.49) = (3.92)/(3.51) K _ p > 1 Since it is given that ∆ G ^ o > 0 ⇒ K _ p ∴ This is incorrect D] K _ p = K _ C × (RT) ^ ∆ n _ g K _ C = K _ p (R × 298) ^ 1 K _ C < 1

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 INCORRECT statement among the following, for this reaction, is

Options

ADecrease in the total pressure will result in formation of more moles of gaseous X
BAt the start of the reaction, dissociation of gaseous ${X}_{2}$ takes place spontaneously
C ${\beta}_{equilibrium} = 0.7$
D ${K}_{C} < 1$

Correct answer

C. ${\beta}_{equilibrium} = 0.7$

Step-by-step solution

A] On decreasing

{P}_{T}

\left[Q = \dfrac{{n}_{{x}^{2}}{P}_{T}}{{n}_{{x}_{2}}{n}_{T}}\right]

Q will be less than

{K}_{p}

reaction will move in forward direction<br />B] At the start of the reaction

∆ G = ∆ {G}^{o} + RT

ln Q<br />

t = 0, Q = 0 ⇒ {∆}_{rxn}G = – ve ( spontaneous )

<br />C] If

{\beta}_{equilibrium} = 0.7

<br />

{K}_{p} = \dfrac{8 \times 0.49}{4 - 0.49} = \dfrac{3.92}{3.51}

<br />

{K}_{p} > 1

<br />Since it is given that<br />

∆ {G}^{o} > 0 ⇒ {K}_{p} < 1

<br />

∴

This is incorrect<br />D]

{K}_{p} = {K}_{C} \times {\left(RT\right)}^{∆ {n}_{g}}

<br />

{K}_{C} = \dfrac{{K}_{p}}{{\left(R \times 298\right)}^{1}}

<br />

{K}_{C} < 1

Practice Chemical Equilibrium on Quantrex Academy →

More from Chemical Equilibrium

Consider the following reactions in which all the reactants and products are present in… JEE Main 2026 Solid carbon, CaO and CaCO _3 are mixed and allowed to attain equilibrium at T K.… JEE Main 2026 In a closed flask at 600 K, one mole of X _2 Y _4 (g) attains equilibrium as given below… JEE Main 2026 One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to… JEE Main 2026 The reaction A(g) leftharpoons B(g) + C(g) was initiated with the amount ' a ' of A(g) .… JEE Main 2026 The values of pressure equilibrium constant recorded at different temperatures for the… JEE Main 2026 For the following reaction at 50° C and at 2 atm pressure, 2N_2O_5(g) leftharpoons… JEE Main 2026 At T(K) , the equilibrium constant of A_2(g) + B_2(g) leftharpoons C(g) is 2.7 × 10^ -5 .… JEE Main 2026 Full Chemical Equilibrium list All JEE Advanced PYQs