For a particular reaction, Δ H°=-40 kJ and Δ S° =-200 J K -1 mol -1. The reaction is

Question

For a particular reaction, Δ H°=-40 kJ and Δ S° =-200 J K -1 mol -1. The reaction is (A) spontaneous at all temperatures (B) nonspontaneous at all temperatures (C) spontaneous at temperature below -73° C (D) spontaneous at temperature above -73° C

Tardigrade Admin · Accepted Answer

Δ G°=Δ H°-T Δ S° When Δ G°=0, T=(Δ H°/Δ S°) =(-40 × 103/-200)=200 K

Q. For a particular reaction, $Δ H^{\circ} = - 40 k J$ and $Δ S^{\circ}$ $= - 200 J K^{- 1} m o l^{- 1}$ . The reaction is

spontaneous at all temperatures

nonspontaneous at all temperatures

spontaneous at temperature below $- 7 3^{\circ} C$

spontaneous at temperature above $- 7 3^{\circ} C$

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
When $Δ G^{\circ} = 0, T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}}$
$= \frac{- 40 \times 1 0 ^{3}}{- 200} = 200 K$

Q. For a particular reaction, ΔH∘=−40kJ and ΔS∘ =−200JK−1mol−1. The reaction is

spontaneous at all temperatures

nonspontaneous at all temperatures

spontaneous at temperature below −73∘C

spontaneous at temperature above −73∘C

ΔG∘=ΔH∘−TΔS∘ When ΔG∘=0,T=ΔS∘ΔH∘​ =−200−40×103​=200K

Q. For a particular reaction, $Δ H^{\circ} = - 40 k J$ and $Δ S^{\circ}$ $= - 200 J K^{- 1} m o l^{- 1}$ . The reaction is

spontaneous at temperature below $- 7 3^{\circ} C$

spontaneous at temperature above $- 7 3^{\circ} C$

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
When $Δ G^{\circ} = 0, T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}}$
$= \frac{- 40 \times 1 0 ^{3}}{- 200} = 200 K$