What can be concluded about the reaction represented below in terms of spontaneity?

Question

#2A + B ->2C#
#DeltaH =89 (kJ)/(mol)#
#DeltaS = 0.070 (kJ)/(mol)#

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Answer 1 · 2016-07-16T17:13:58

The reaction is not spontaneous below 1000 °C, at equilibrium at 1000 °C, and spontaneous above 1000 °C.

#"2A + B" → "2C"#

#ΔH = "89 kJ·mol"^"-1"#; #ΔS = "0.070 kJ·mol"^"-1""K"^"-1"#

A reaction is

#ΔG = ΔH - TΔS#

#ΔH# is +, and #ΔS# is +.

At low temperatures, the #ΔH# term will predominate.

#ΔG# will be +, and the reaction will not be spontaneous.

At high temperatures, the #TΔS# term will predominate.

#ΔG# will be negative, and the reaction will be spontaneous.

At equilibrium,

#ΔG = ΔH -TΔS = 0#

#89 color(red)(cancel(color(black)("kJ·mol"^"-1"))) - T × 0.070 color(red)(cancel(color(black)("kJ·mol"^"-1")))"K"^"-1" = 0#

#T = 89/("0.070 K"^"-1") = "1271 K" = "1000 °C"#

The reaction is at equilibrium at 1000 °C.