For a reversible process at equilibrium, the change in entropy may be expressed as:

Δ S = T q_{r e v}

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Δ S = \frac{q_{r e v}}{T}

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Δ S = \frac{- Δ H}{T}

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Δ S = Δ H

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Solution

The correct option is B $Δ S = \frac{q_{r e v}}{T}$
For a reversible process at equilibrium, $Δ S = \frac{q_{r e v}}{T}$ .

Option B is correct.

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Similar questions

The entropy change can be calculated by using the expression $Δ S = \frac{q_{r e v}}{T}$ .
When water freezes in a glass beaker, choose the correct statement amongst the following.
(a) $Δ$ S (system) decreases but $Δ$ S (surroundings) remains the same
(b) $Δ$ S (system) increases but $Δ$ S (surroundings) decreases
(c) $Δ$ S (system) decreases but $Δ$ S (surroundings) increases
(d) $Δ$ S (system) decreases but $Δ$ S (surroundings) also decreases

Q. The change in entropy,

Δ S

is positive for an endothermic reaction. If enthalpy change

Δ H

occurs at the same temperature T, then the reaction is feasible:

Q. For which of the following cases

Δ S = \frac{Δ H}{T}

Q. For which of the following cases,

Δ S = \frac{Δ H}{T} ?

Q. The correct statement for an exergonic gaseous reaction

A + B ⟶ 3 C

will be:

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For a reversible process at equilibrium, the change in entropy may be expressed as:

The correct option is B ΔS=qrevTFor a reversible process at equilibrium, ΔS=qrevT.Option B is correct.

The correct option is B $Δ S = \frac{q_{r e v}}{T}$
For a reversible process at equilibrium, $Δ S = \frac{q_{r e v}}{T}$ .

Option B is correct.