For a particular reversible reaction at temperature $\mathrm{T}, \Delta \mathrm{H}$ and $\Delta \mathrm{S}$ were found to be both +ve . If $\mathrm{T}_{\mathrm{e}}$ is the temperature at equilibrium, the reaction would be spontaneous when

(B) $\mathrm{T}>\mathrm{T}_{\mathrm{e}}$

(D) $\mathrm{T}=\mathrm{T}_{\mathrm{e}}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Chemical Thermodynamics

For a particular reversible reaction at temperature

T, Δ H

and

Δ S

were found to be both +ve . If

T_{e}

is the temperature at equilibrium, the reaction would be spontaneous when

Answer: (B)

Spontaneity requires

Δ G = Δ H - T Δ S < 0

. At the equilibrium temperature

Δ G = 0

, which fixes

T_{e} = \frac{Δ H}{Δ S}

. Since both

Δ H

and

Δ S

are positive, the

- T Δ S

term wins over the positive

Δ H

only once the temperature is high enough, that is when

T > \frac{Δ H}{Δ S} = T_{e}

. The reaction becomes spontaneous for

T > T_{e}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option