For a given reaction, Δ H =35.5 kJ mol -1 and Δ S =83.6 JK -1 mol -1 . The reaction is spontaneous at: (Assume that Δ H and Δ S do not vary with temperature)

Question

For a given reaction, Δ H =35.5 kJ mol -1 and Δ S =83.6 JK -1 mol -1 . The reaction is spontaneous at: (Assume that Δ H and Δ S do not vary with temperature) (A) T > 425 K (B) All temperatures (C) T > 298 K (D) T < 425 K

Tardigrade Admin · Accepted Answer

∵ Δ G =Δ H - T Δ S For a reaction to be spontaneous, Δ G =- ve i.e., Δ H < T Δ S T >(Δ H /Δ S )=(35.5 × 103 J /83.6 JK -1) i.e., T >425 K

Q. For a given reaction, $Δ H = 35.5 k J m o l^{- 1}$ and $Δ S = 83.6 J K^{- 1} m o l^{- 1} .$ The reaction is spontaneous at $:$ (Assume that $Δ H$ and $Δ S$ do not vary with temperature)

T > 425 K

All temperatures

T > 298 K

T < 425 K

$∵ Δ G = Δ H - T Δ S$

For a reaction to be spontaneous, $Δ G = - v e$

i.e., $Δ H < T Δ S$

$T > \frac{Δ H}{Δ S} = \frac{35.5 \times 1 0 ^{3} J}{83.6 J K ^{- 1}}$

i.e., $T > 425 K$

Q. For a given reaction, ΔH=35.5kJmol−1 and ΔS=83.6JK−1mol−1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with temperature)