For a reaction- A- P- the plots of -A- and -P- with time at temperature T1 and T2 are given below-If T2 - T1- the correct statements is-are-Assume H- and S- are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T2T1- Here H- S- G and K are enthalpy- entropy- Gibbs energy and equilibrium constant- respectively-

The correct options are A H^θ C G^θ A⇌ P given T_2 gt; T_1 lnK_1lnK_2 gt; T_2T_1 ⇒ T_1 lnK_1 gt; T_2 lnK_2 ⇒ G_1^∘ gt; ...

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Byju's Answer

Standard XII

Physics

Banking of Roads

For a reactio...

Question

For a reaction, $A ⇌ P$ , the plots of $[A]$ and $[P]$ with time at temperature $T_{1}$ and $T_{2}$ are given below.
If $T_{2} > T_{1}$ , the correct statement(s) is/are:
(Assume $△ H^{θ}$ and $△ S^{θ}$ are independent of temperature and ratio of $l n K$ at $T_{1}$ to $l n K$ at $T_{2}$ is greater than $\frac{T_{2}}{T_{1}}$ . Here $H, S, G$ and $K$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

△ H^{θ} < 0, △ S^{θ} < 0

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△ G^{θ} < 0, △ H^{θ} > 0

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△ G^{θ} < 0, △ S^{θ} < 0

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△ G^{θ} < 0, △ S^{θ} > 0

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Solution

The correct options are
A $△ H^{θ} < 0, △ S^{θ} < 0$
C $△ G^{θ} < 0, △ S^{θ} < 0$
$A ⇌ P$

given $T_{2} > T_{1}$

$\frac{l n K_{1}}{l n K_{2}} > \frac{T_{2}}{T_{1}}$

$\Rightarrow T_{1} l n K_{1} > T_{2} l n K_{2}$

$\Rightarrow - △ G_{1}^{\circ} > - △ G_{2}^{\circ}$

$\Rightarrow (- △ H^{\circ} + T_{1} △ S^{\circ}) > (- △ H^{\circ} + T_{2} △ S^{\circ})$

$\Rightarrow T_{1} △ S^{\circ} > T_{2} △ S^{\circ}$
$\Rightarrow △ S^{\circ} < 0$ .

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The correct options are A △Hθ<0,△Sθ<0 C △Gθ<0,△Sθ<0A⇌Pgiven T2>T1lnK1lnK2>T2T1⇒T1 lnK1>T2 lnK2⇒−△G∘1>−△G∘2⇒(−△H∘+T1△S∘)>(−△H∘+T2△S∘)⇒T1△S∘>T2△S∘⇒△S∘<0.