Question

Which of the following expression is correct for a reversible process at equilibrium?

Options are above in image...
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Answer 1

Solution:-

=>This Reaction Reversible process

$\to\tt A + B \leftrightharpoons C+D$

=> The equation of Gibb's Free Energy ( ΔG )

$\to \tt \: \Delta{G} = \Delta{G}^{o} + 2.303RT logk$

We know that for Reversible Process ΔG = 0

So Put the value

$\to \tt \: 0= \Delta{G}^{o} + 2.303RT logk$

$\to \tt \: \Delta{G}^{o} = - 2.303RT logk$

We Got Relationship Between ΔG⁰ and Equilibrium Constant

Where k = equilibrium Constant

Answer:-

=> Option B is correct

More Information About Gibbs Free Energy

It is defined as the energy available in the system for Conversion into useful work

it is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the useful work done by the system

G = H - TS

Where H is the heat content. T is the absolute temperature and S is the entropy of the system. As before for the isothermal process,

Answer 2

Answer:

$\huge\bf\red{\underline{\underline{aNswER}}}$

=>This Reaction Reversible process

\to\tt A + B \leftrightharpoons C+D→A+B⇋C+D

=> The equation of Gibb's Free Energy ( ΔG )

\to \tt \: \Delta{G} = \Delta{G}^{o} + 2.303RT logk→ΔG=ΔG

o

+2.303RTlogk

We know that for Reversible Process ΔG = 0

So Put the value

\to \tt \: 0= \Delta{G}^{o} + 2.303RT logk→0=ΔG

o

+2.303RTlogk

\to \tt \: \Delta{G}^{o} = - 2.303RT logk→ΔG

o

=−2.303RTlogk

We Got Relationship Between ΔG⁰ and Equilibrium Constant

Where k = equilibrium Constant

Answer:-

=> Option B is correct

More Information About Gibbs Free Energy

It is defined as the energy available in the system for Conversion into useful work

it is that thermodynamic quantity of a system, the decrease in whose value during a process is equal to the useful work done by the system

G = H - TS

Where H is the heat content. T is the absolute temperature and S is the entropy of the system. As before for the isothermal process,