Chemistry · General chemistry II · Worked example
Find the temperature where ΔG changes sign
A reaction has ΔH = −100. kJ and ΔS = −200. J/K, both nearly constant with temperature. Find ΔG at 298 K and the temperatures at which the reaction is spontaneous.
Read the signs
ΔH < 0 favors the reaction and ΔS < 0 opposes it, so the answer depends on temperature: the entropy term −TΔS grows as T rises.
ΔG at 298 K
With ΔS = −0.200 kJ/K:
Find where ΔG = 0
Set ΔH − TΔS = 0 and solve for T.
State the range
Below 500. K, ΔG < 0 and the reaction is spontaneous; above it, the reverse reaction is. At 600 K, for example, ΔG = +20 kJ.
Result
ΔG = −40.4 kJ at 298 K; the reaction is spontaneous below 500. K.
Your turn
A reaction has ΔH = +30.0 kJ and ΔS = +75.0 J/K. Above what temperature is it spontaneous?
Show the answer and explanation
Above 400. K.
Both signs are positive, so the reaction becomes spontaneous once TΔS exceeds ΔH: T = ΔH/ΔS = 30.0 kJ ÷ 0.0750 kJ/K = 400. K.
Keep exploring
In Hess law & thermodynamics, raise the temperature to 600 K. ΔG becomes +20 kJ, and the reaction is no longer spontaneous.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
Find it in Hess law & thermodynamics Check the crossover in a Chemistry box Open worked example on a board Chemistry formulas: energy and thermodynamicsYour existing work stays on this device. Examples open as editable copies.