For a certain process, and Δ H=178 kJ Δ S=160 J / K. What is the minimum temperature at which the process is spontaneous (assuming that Δ H and Δ S do not vary with temperature)

Question

For a certain process, and Δ H=178 kJ Δ S=160 J / K. What is the minimum temperature at which the process is spontaneous (assuming that Δ H and Δ S do not vary with temperature) (A) 2112.3 K (B) 136.7 K (C) 275.8 K (D) 1112.5 K

Tardigrade Admin · Accepted Answer

According to Gibbs-Helmholtz reaction, Δ G=Δ H-T Δ S For a spontaneous reaction, Δ G<0 i.e., Δ H-T Δ S<0 or T>(Δ H/Δ S) or T>(178 × 1000/160) T>1112.5 K ∴ For spontaneous process, minimum temperature must be 1112.5 K.

Q. For a certain process, and $Δ H = 178 k J Δ S = 160 J / K$ . What is the minimum temperature at which the process is spontaneous (assuming that $Δ H$ and $Δ S$ do not vary with temperature)

2112.3 K

136.7 K

275.8 K

1112.5 K

According to Gibbs-Helmholtz reaction,
$Δ G = Δ H - T Δ S$
For a spontaneous reaction, $Δ G < 0$
i.e., $Δ H - T Δ S < 0$
or $T > \frac{Δ H}{Δ S}$
or $T > \frac{178 \times 1000}{160}$
$T > 1112.5 K$
$∴$ For spontaneous process,
minimum temperature must be $1112.5 K$ .

Q. For a certain process, and ΔH=178kJΔS=160J/K. What is the minimum temperature at which the process is spontaneous (assuming that ΔH and ΔS do not vary with temperature)