JEE Advanced 2018 · Paper 2 · Chemistry Q6

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ChemistryMultiple correct+4 / -2Question 24 of 54

For a reaction, A ⇌ P, the plots of [A] and [P] with time at temperatures $T_1$ and $T_2$ are given below.

If $T_2 > T_1$ , the correct statement(s) is (are)

(Assume $\Delta H^\ominus$ and $\Delta S^\ominus$ are independent of temperature and ratio of $\ln K$ at $T_1$ to $\ln K$ at $T_2$ is greater than $T_2/T_1$ . Here $H$ , $S$ , $G$ and $K$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Two panels. Left: [A] (mol L⁻¹) against time, starting at 10 and falling below the dashed line at 5; the T₂ curve falls faster at first and levels off higher (about 3.5), while the T₁ curve levels off lower (about 2.5). Right: [P] against time rising from 0 past 5; the T₂ curve rises faster and levels off lower (about 6.5), while the T₁ curve levels off higher (about 7.5).

\Delta H^\ominus < 0

\Delta S^\ominus < 0

\Delta G^\ominus < 0

\Delta H^\ominus > 0

\Delta G^\ominus < 0

\Delta S^\ominus < 0

\Delta G^\ominus < 0

\Delta S^\ominus > 0

Answer and worked solution

Official answer: A, C

Equilibrium constants. At both temperatures the plateaus have $[\ce{P}]>[\ce{A}]$ , so $K>1$ and $\Delta G^\circ<0$ .

Sign of ΔH°. At the higher $T_2$ , $[\ce{P}]_{eq}$ is lower, so $K$ falls with temperature. The reaction is exothermic: $\Delta H^\circ<0$ .

Sign of ΔS°. From $\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}R$ :
$T_1\ln K_1-T_2\ln K_2=\frac{\Delta S^\circ}R(T_1-T_2).$
The given condition $\dfrac{\ln K_1}{\ln K_2}>\dfrac{T_2}{T_1}$ (both logs positive) means $T_1\ln K_1>T_2\ln K_2$ . So $\Delta S^\circ(T_1-T_2)>0$ . Since $T_1<T_2$ , $\Delta S^\circ<0$ .

So $\Delta H^\circ<0$ , $\Delta S^\circ<0$ and $\Delta G^\circ<0$ .

(A), (C)

Topics:Equilibrium and Gibbs energy Gibbs energy and spontaneity Le Chatelier's principle

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JEE Advanced 2018 · Paper 2, Chemistry question 6