Question

For a particular reaction, Δ H ∘ = − 46.7 kJ and Δ S ∘ = − 123.8 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Answer 1

Answer: 377 K

Step-by-step explanation:

According to Gibbs equation

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

`\Delta G` = Gibbs free energy

`\Delta H` = enthalpy change = -46.7 kJ= -46700 J

`\Delta S` = entropy change = -123.8 J/K

T = temperature in Kelvin = ?

`\Delta G`= +ve, reaction is non spontaneous

`\Delta G` = -ve, reaction is spontaneous

`\Delta G` = 0, reaction is in equilibrium

`\Delta H=T\Delta S`

`T=(\Delta H)/(\Delta S)=( -46700J)/(-123.8J/K)=377K`

Thus at 377 K the reaction change from nonspontaneous to spontaneous