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

Question

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

2853 views around the world

You can reuse this answer
Creative Commons License

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.