For a particular reaction, ΔH=-38.3kJ and ΔS=-113JK- 1mol- 1 . This reaction is:

Question

For a particular reaction, ΔH=-38.3kJ and ΔS=-113JK- 1mol- 1 . This reaction is: (A) spontaneous at all temperatures (B) non-spontaneous at all temperatures (C) spontaneous at temperatures below 66° C (D) spontaneous at temperatures above 66° C

Tardigrade Admin · Accepted Answer

ÎG=ÎH-TÎS For ÎG=0 T=(ÎH/ÎS) =(- 38 . 3 × 103/113) =338.9K (65 . 78 ° C)

Q. For a particular reaction, $Δ H = - 38.3 k J$ and $Δ S = - 113 J K^{- 1} m o l^{- 1}$ . This reaction is:

spontaneous at all temperatures

non-spontaneous at all temperatures

spontaneous at temperatures below $6 6^{\circ} C$

spontaneous at temperatures above $6 6^{\circ} C$

$Δ G = Δ H - T Δ S$
For $Δ G = 0$
$T = \frac{Δ H}{Δ S}$
$= \frac{- 38.3 \times 1 0 ^{3}}{113}$
$= 338.9 K$
$(65.7 8^{\circ} C)$

Q. For a particular reaction, ΔH=−38.3kJ and ΔS=−113JK−1mol−1 . This reaction is:

spontaneous at all temperatures

non-spontaneous at all temperatures

spontaneous at temperatures below 66∘C

spontaneous at temperatures above 66∘C

ΔG=ΔH−TΔS For ΔG=0 T=ΔSΔH​ =113−38.3×103​ =338.9K (65.78∘C)

Q. For a particular reaction, $Δ H = - 38.3 k J$ and $Δ S = - 113 J K^{- 1} m o l^{- 1}$ . This reaction is:

spontaneous at temperatures below $6 6^{\circ} C$

spontaneous at temperatures above $6 6^{\circ} C$

$Δ G = Δ H - T Δ S$
For $Δ G = 0$
$T = \frac{Δ H}{Δ S}$
$= \frac{- 38.3 \times 1 0 ^{3}}{113}$
$= 338.9 K$
$(65.7 8^{\circ} C)$