For a system in equilibrium- - G - 0 under conditions of constant-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Conservative and Non Conservative Forces

For a system ...

Question

For a system in equilibrium, $Δ G = 0$ under conditions of constant:

temperature and pressure.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

temperature and volume.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

energy and volume.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

pressure and volume.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B temperature and pressure.
When a system is at equilibrium under constant temperature and pressure, its free energy change is zero $(Δ G = 0)$ .

Suggest Corrections

Similar questions

Q. For a system in equilibrium,

Δ G = 0

, under conditions of constant ________.

A change in the free energy of a system at constant temperature and pressure will be:

Δ_{s y s} G = Δ_{s y s} H - T Δ_{s y s} S

At constant temperature and pressure,

Δ_{s y s} G < 0 (s p o n t a n e o u s)

Δ_{s y s} G = 0 (e q u i l i b r i u m)

Δ_{s y s} G > 0 (n o n - s p o n t a n e o u s)

For a system in equilibrium,

Δ G = 0

, under conditions of constant_____

Q. In an equilibrium reaction for which

Δ G^{\circ}

= 0. Determine the value of equilibrium constant K.

δ G

Q. Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.

Δ G = Δ G^{\circ} + R T l n Q_{P}

the condition for a system to be at equilibrium is that

Δ G = 0 and Q_{p} = K_{P}

Thus at equilibrium

Δ G^{\circ} = - R T l n K_{P}

Note: In the reaction, where all gaseous reactants and products;

K represents K_{P}

, we may conclude that for standard reactions, i.e., at

1 M or 1 atm

When

Δ G^{\circ} = - v e or K > 1 : forward reaction is feasible

Δ G^{\circ} = + v e or K < 1 : reverse reaction is feasible

Δ G^{\circ} = 0 or K = 1 : reaction is at equilibrium (very rare)

For the equilibrium

N i O (s) + C O (g) ⇌ N i (s) + C O_{2} (g), Δ G^{\circ} ({J Mol}^{- 1}) = - 20700 - 11.97 T .

. Calculate the temperature at which the product gases at equilibrium at

1 atm

will conatin

400 ppm

of carbon monoxide.

Q. From application of thermodynamics on chemical reaction,
we get

Δ G = Δ G^{0} + R T

ln Q
Also

Δ G = Δ H - T Δ S

.
If

Δ G = 0

, reaction is at equilibrium.
If

Δ G > 0

reaction is non-spontaneous under given condition.

I f Δ G < 0

, reaction is spontaneous under given condition.

Consider the reaction given below.

A (s) ⇌ 2 B (g) Δ H^{0} = 160

KJ/mol.

Δ S^{0} = 400

J/mol-K at

400

K
Which of the following is correct at

400

Join BYJU'S Learning Program

For a system in equilibrium, ΔG=0 under conditions of constant:

The correct option is B temperature and pressure.When a system is at equilibrium under constant temperature and pressure, its free energy change is zero (ΔG=0).

For a system in equilibrium, $Δ G = 0$ under conditions of constant:

The correct option is B temperature and pressure.
When a system is at equilibrium under constant temperature and pressure, its free energy change is zero $(Δ G = 0)$ .