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

## $2 A + B \to 2 C$ $\Delta H = 89 \frac{k J}{m o l}$ $\Delta S = 0.070 \frac{k J}{m o l}$

Jul 16, 2016

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

#### Explanation:

$\text{2A + B" → "2C}$

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.