Chemistry

Equilibrium

Relation between equilibrium constants and total pressures of two reaction (Equilibrium) (Chemistry) (Class 11)

For a reaction, A P, the plots of [A] and [P] with time at te

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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 $Δ H^{⊖}$ and $Δ S^{⊖}$ 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.)

Name: Video answer #2 for For a reaction, A⇌P, the plots of [A] and [P] with time at temperatures T1 and T2 are given below.If T2>T1, the correct statement(s) is are (Assume ΔH⊖ and ΔS⊖ are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T2/T1. Here H,S,G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)
Uploaded: 2024-01-01T03:06:45.000Z
Duration: 4 min
Description: For a reaction, A⇌P, the plots of [A] and [P] with time at temperatures T1 and T2 are given below.If T2>T1, the correct statement(s) is are (Assume ΔH⊖ and ΔS⊖ are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T2/T1. Here H,S,G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

SELECT MULTIPLE OPTION

Δ H^{⊖} < 0, Δ S^{⊖} < 0

Δ G^{⊖} < 0, Δ H^{⊖} > 0

Δ G^{⊖} < 0, Δ S^{⊖} < 0

Δ G^{⊖} < 0, Δ S^{⊖} > 0

Views: 5,154 students

Updated on: Jan 1, 2024

Exam:

JEE Advanced 2018

Found 3 tutors discussing this question

Charlotte Discussed

For a reaction,

A ⇌ P

, the plots of

[A]

and

[P]

with time at temperatures

T_{1}

and

T_{2}

are given below.

T_{2} > T_{1}

, the correct statement(s) is are (Assume

Δ H^{⊖}

and

Δ S^{⊖}

are independent of temperature and ratio of

ln K

T_{1}

ln K

T_{2}

is greater than

T_{2} / T_{1}

. Here

H, S, G

and

K

are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

8 mins ago

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8 mins ago

Text solutionVerified

For the reaction,

A ⇌ P

Given,

T_{1} < T_{2} \frac{l n K _{1}}{l n K _{2}} > \frac{T _{2}}{T _{1}}

It shows,

On

increasing the temperature,

K

decreases so reaction is exothermic i.e.,

Δ H^{\circ} < 0

Besides, the graph shows

K > 1

Δ G^{\circ} < 0

Now from equation (i)

T_{1} ln K_{1} > T_{2} ln K_{2} - Δ G_{1}^{\circ} > - Δ G^{\circ}_{2}

Likewise

(- Δ H^{0} + T_{1} Δ S^{0}) > (- Δ H^{\circ} + T_{2} Δ S^{\circ})

or simply

T_{1} Δ S^{\circ} > T_{2} Δ S^{\circ}

So,

(T_{2} - T_{1}) Δ S^{\circ} < 0

∴ Δ S^{0} < 0

In other words, increase of

Δ G

with increase in temperature is possible only when

Δ S^{\circ} < 0

. Hence, options (a) and (c) are correct.

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Question Text	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 $Δ H^{⊖}$ and $Δ S^{⊖}$ 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.)
Updated On	Jan 1, 2024
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	238
Avg. Video Duration	3 min

Text solutionVerified

Video solutions (2)

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