Question

If ΔH = -70.0 kJ and ΔS = -0.300 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. It is explained as the difference of enthalpy in the product of temperature and entropy. It is also a state function in which, the function depends only on the 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, the temperature can be calculated by using a change in the entropy and enthalpy values.

Fundamentals

Gibbs free energy: In Gibbs free energy calculation, three parameters are used such as, the temperature, entropy and enthalpy.

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

Since Gibbs free energy is a state function, a change in the gibbs free energy exists as follows:

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

Given,

Change in enthalpy ${\rm{\Delta H}}\,{\rm{ = }}\, - 7{\rm{0}}{\rm{.0kJ}}$

Change in entropy ${\rm{\Delta S}}\,{\rm{ = }}\, - {\rm{0}}{\rm{.300kJ/K}}$

Given that the reaction is spontaneous at a certain temperature.

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

Therefore,

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

An equation for the change in enthalpy and entropy is given below:

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

Given that,

Change in enthalpy ${\rm{\Delta H}}\,{\rm{ = }}\, - 7{\rm{0}}{\rm{.0kJ}}$

Change in entropy ${\rm{\Delta S}}\,{\rm{ = }}\, - {\rm{0}}{\rm{.300kJ/K}}$

Now, substitute these values in the above equation.

$\begin{array}{c}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{\rm{\Delta H}}}}{{{\rm{\Delta S}}}}\,{\rm{ < }}\,{\rm{T}}\\\\\frac{{ - 7{\rm{0}}{\rm{.0kJ}}}}{{ - {\rm{0}}{\rm{.300kJ/K}}}}\,{\rm{ < }}\,{\rm{T}}\\\\\frac{{{\rm{70}}{\rm{.0kJ}}}}{{{\rm{0}}{\rm{.300kJ/K}}}}\,{\rm{ > }}\,{\rm{T}}\\\\{\rm{ 233}}{\rm{.3K > }}\,{\rm{T}}\\\end{array}$

Therefore, the reaction is spontaneous when the temperature of the system is below 233.3K.

Ans:

The reaction is spontaneous at below 233.3K temperature.