Question

If ΔH = -80.0 kJ and ΔS = -0.500 kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature

Express your answer numerically in kelvins.

Answer 1

Concepts and reason

• Gibbs free energy is explained by using the parameters entropy, enthalpy and temperature. Gibbs free energy is explained as the difference of enthalpy to the product of temperature and entropy. Gibbs energy is termed as a state function, since the function is depending only on initial and final states of the system.

• Temperature condition can be explained by using this Gibbs free energy equation. According to the nature of the reaction temperature can be calculated by using change in entropy and change in enthalpy values.

Fundamentals

Gibbs free energy: In Gibbs free energy calculation, three parameters are used; Temperature, entropy and enthalpy.

$\begin{array}{l}\\{\rm{G}}\,{\rm{ = }}\,{\rm{H - TS}}\\\\{\rm{where,}}\\\\{\rm{G}}\,{\rm{ = }}\,{\rm{Gibbs}}\,{\rm{free}}\,{\rm{energy}}\\\\{\rm{H}}\,{\rm{ = }}\,{\rm{enthalpy}}\,{\rm{of}}\,{\rm{the}}\,{\rm{system}}\\\\{\rm{T}}\,{\rm{ = }}\,{\rm{absolute}}\,{\rm{temperature}}\\\\{\rm{S}}\,{\rm{ = }}\,{\rm{entropy}}\,{\rm{of}}\,{\rm{the}}\,{\rm{system}}\\\end{array}$

Since Gibbs free energy is a state function, a change in Gibbs free energy exists as follows.

$\begin{array}{l}\\{\rm{\Delta G}}\,{\rm{ = }}\,{\rm{\Delta H}}\,{\rm{ - }}\,{\rm{T\Delta S}}\\\\{\rm{i) if}}\,{\rm{\Delta G}}\,{\rm{ < }}\,{\rm{0,}}\,{\rm{reaction}}\,{\rm{is}}\,{\rm{spontaneous}}{\rm{.}}\\\\{\rm{ii)}}\,{\rm{if}}\,{\rm{\Delta G > 0,}}\,{\rm{reaction}}\,{\rm{is}}\,{\rm{non}}\,{\rm{spontaneous}}{\rm{.}}\\\\{\rm{iii)}}\,{\rm{if}}\,{\rm{\Delta G = 0,}}\,{\rm{indicates}}\,{\rm{equilibrium}}\,{\rm{reaction}}{\rm{.}}\\\end{array}$

Given,

Change in enthalpy ${\rm{\Delta H}}\,{\rm{ = }}\,{\rm{ - 80}}{\rm{.0kJ}}$

Change in entropy ${\rm{\Delta S}}\,{\rm{ = }}\,{\rm{ - 0}}{\rm{.500kJ/K}}$

Given that the reaction is spontaneous at a certain temperature.

$\begin{array}{l}\\{\rm{\Delta G}}\,{\rm{ = }}\,{\rm{\Delta H}}\,{\rm{ - }}\,{\rm{T\Delta S}}\\\\{\rm{\Delta G}}\,{\rm{ < }}\,{\rm{0,}}\,{\rm{the reaction}}\,{\rm{is}}\,{\rm{spontaneous}}{\rm{.}}\\\end{array}$

Therefore,

$\begin{array}{l}\\{\rm{\Delta H - T\Delta S < 0}}\\\\{\rm{\Delta H < T\Delta S}}\\\\\frac{{{\rm{\Delta H}}}}{{{\rm{\Delta S}}}}\,{\rm{ < }}\,{\rm{T}}\\\end{array}$

Equation is given below for the change in enthalpy and change in entropy.

$\frac{{{\rm{\Delta H}}}}{{{\rm{\Delta S}}}}\,{\rm{ < }}\,{\rm{T}}$

Given that,

Change in enthalpy ${\rm{\Delta H}}\,{\rm{ = }}\,{\rm{ - 80}}{\rm{.0kJ}}$

Change in entropy ${\rm{\Delta S}}\,{\rm{ = }}\,{\rm{ - 0}}{\rm{.500kJ/K}}$

Now, substitute these values in the above equation.

$\begin{array}{l}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{\rm{\Delta H}}}}{{{\rm{\Delta S}}}}\,{\rm{ < }}\,{\rm{T}}\\\\\frac{{{\rm{ - 80}}{\rm{.0kJ}}}}{{{\rm{ - 0}}{\rm{.500kJ/K}}}}\,{\rm{ < }}\,{\rm{T}}\\\\{\rm{now,}}\\\\\frac{{{\rm{80}}{\rm{.0kJ}}}}{{{\rm{0}}{\rm{.500kJ/K}}}}\,{\rm{ > }}\,{\rm{T}}\\\\{\rm{ 160K > }}\,{\rm{T}}\\\end{array}$

Therefore, the reaction is spontaneous, where the temperature is below 160K.

Ans:

The reaction is spontaneous at below 160K temperature.