Question

For a particular reaction, Δ H ∘ = − 93.8 kJ and Δ S ∘ = − 156.1 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Answer 1

To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity

At constant temperature and pressure, the change in Gibbs free energy is defined as

$\Delta G = \Delta H - T\Delta S$

Where,

H = Entalpy

T = Temperature

S = Entropy

When the temperature is less than that number it is negative meaning it is a spontaneous reaction. $\Delta G$ is also always 0 when using single element reactions. In numerical that implies $\Delta G = 0$

At the equation then,

$\Delta G = \Delta H - T\Delta S$

$0 = \Delta H - T\Delta S$

$\Delta H = T\Delta S$

$T = \frac{\Delta H}{\Delta S}$

$T = \frac{-93.8kJ}{-156.1J/K}$

$T = \frac{-93.8*10^3J}{-156.1J/K}$

$T = 600.89K$}

Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K