Question

Consider a reaction with ΔH = 15 kJ and ΔS = 50 J · K⁻¹ . Is the reaction spontaneous (a) at 10°C, (b) at 80°C?

Short answer

When ΔH = 15 kJ and ΔS = 50 J · K⁻¹, the reaction is spontaneous at 10°C, and at 80°C.

Step-by-step solution

 Definition and Formula

Enthalpy reflects the capacity to do a non-mechanical work and the capacity to release heat. It is denoted by H and its unit is joules.

Entropy indicates the degree of randomness of a system and is abbreviated as S. Its unit is J . K^-1.

Free energy indicates the part of the total energy of a system which is available for useful work. A change in the free energy of a process is ΔG.

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

For a spontaneous press process ΔG < zero and such processes are said to be exergonic. Non spontaneous process has a positive ΔG value and are said to be endergonic.

Solution

According to the question ΔH = 15 kJ and ΔS = 50 J · K⁻¹

a) at 10°C

$\begin{array}{l}\Delta G=\Delta H-T\Delta S\\ \Delta G=15-(10\times 50)\\ \Delta G=-485\end{array}$

As ΔG is - 485 (negative), the reaction will be spontaneous.

b) at 80°C

$\begin{array}{l}\Delta G=\Delta H-T\Delta S\\ \Delta G=15-(80\times 50)\\ \Delta G=-3985\end{array}$

As ΔG is -3985 (negative), the reaction will be spontaneous.

Thus, the reaction will be spontaneous at 10°C and 80°C.