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**Condition of equilibrium and criterion for a spontaneous process.**

The entropy of a system remains unchanged in a reversible change while it increases in an irreversible change i.e.

0 for reversible change

and 0 for irreversible change

For a system with its surrounding

d_{system }+ dS_{surrounding }= 0 for reversible change

and d_{system }+ dS_{surrounding} > 0 for irreversible change

then we have

d_{system }+ dS_{surrounding }≥ 0

The criteria for reversible and irreversible process in terms. Of entropy as well as in terms U, H, A & G.

By the equation,

We know that

q_{irr} < q_{rev} then

TdS > d_{qrev} (for irreversible process)

Thus we write, TdS = dU + PdV (for reversible process)

TdS > dU + PdV (for irreversible process)

Combining we get, TdS ≥ dU + PdV

(i) Criterion in terms of change of entropy: Let U & V remains constant.

T(dS)U, V ≥ 0 or (dS)U, V ≥ 0

The equal sign refers to reversible and greater-sign refers to an irreversible process.

(ii) Criterion in terms of change of internal energy: Let S & V constant then

(dS)U, V ≥ 0

(iii) Criterion in terms of change of enthalpy: Let S & P are constant then

TdS ≥ dU + PdV

⇒ dU + PdV ≤ 0

But dU + PdV = dH

⇒ (dH)S, P ≤ 0

(iv) Criterion in terms of change of work function:

We know that dA = dU – TdS & TdS ≥ dU + PdV

then we get dA ≤ –PdV

at constant volume and temperature (dA)V, T ≤ 0

(v) Criterion in terms of change in free energy:

We know that G = H – TS

⇒ G = (U + PV) – TS

⇒ dG = dU + PdV + VdP – TdS – SdT

& TdS ≥ dU + PdV, we get

⇒ dG ≤ VdP – SdT

Let P & T are constant then

(dG)P, T ≤ 0

Using thermodynamic magic square we can find these criterion

(dU)S, V ≤ 0,

(dH)S, P ≤ 0 (dG)P, T ≤ 0,

(dA)V, T ≤ 0 (dS)U, V ≥ 0

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