Chemistry

Chemistry

Problem

At 1 bar, the temperature dependence of the standard entropy change for a reaction is given by ΔS∘(T)=150 J mol−1 K−1−0.050 J mol−1 K−2 T.\Delta S^\circ(T) = 150\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1} - 0.050\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-2}\,T. The standard enthalpy change is temperature-independent at ΔH∘=+50.0 kJ mol−1.\Delta H^\circ = +50.0\,\text{kJ}\,\text{mol}^{-1}. Determine the temperature (in K) at which the reaction is at equilibrium.
Intuitive explanation

At equilibrium, the driving force for a reaction vanishes—meaning ΔG∘=0\Delta G^\circ = 0. Use the fundamental relationship ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ to set up an equation where you plug in the given expressions for enthalpy and entropy. Since both are provided explicitly as functions of temperature (or constants), you can solve directly for the temperature where these two competing terms balance perfectly.

Intuitive explanation

At equilibrium, the driving force for a reaction vanishes—meaning ΔG∘=0\Delta G^\circ = 0. Use the fundamental relationship ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ to set up an equation where you plug in the given expressions for enthalpy and entropy. Since both are provided explicitly as functions of temperature (or constants), you can solve directly for the temperature where these two competing terms balance perfectly.

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