Question

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

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

Answer 1

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

Explanation:

`"2A + B" → "2C"`

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

A reaction is

• spontaneous if `ΔG < 0`
• at equilibrium if `ΔG = 0`
• not spontaneous if `ΔG > 0`

`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.