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Chemistry · Ch 4 — Chemical Thermodynamics

Second law of thermodynamics

4.11.4

Second law of thermodynamics

The second law of thermodynamics states that the total entropy of a system and its surroundings increases in a spontaneous process. For the process to be spontaneous,

ΔStotal=ΔSsys+ΔSsurr>0...(4.33)\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0 \qquad \text{...(4.33)}

Consider

2H2(g)+O2(g)⟶2H2O(l)2\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2\mathrm{H_2O}(l)

with ΔS=−327\Delta S = -327 J K−1^{-1} and ΔH=−572\Delta H = -572 kJ (both at 298 K).

To find ΔStotal\Delta S_{total}, we need to know ΔSsurr\Delta S_{surr}. ΔH\Delta H for the reaction is -572 kJ: when 2 moles of H2_2 and 1 mole of O2_2 gas combine to form 2 moles of liquid water, 572 kJ of heat is released, which is received by the surroundings at constant pressure (and 298 K). The entropy change of the surroundings is

ΔSsurr=QrevT=572×103 J298 K=1919 J K−1\Delta S_{surr} = \frac{Q_{rev}}{T} = \frac{572 \times 10^3\ \mathrm{J}}{298\ \mathrm{K}} = 1919\ \mathrm{J\,K^{-1}}

ΔStotal=ΔSsys+ΔSsurr=−327 J K−1+1919 J K−1=+1592 J K−1\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = -327\ \mathrm{J\,K^{-1}} + 1919\ \mathrm{J\,K^{-1}} = +1592\ \mathrm{J\,K^{-1}}

ΔStotal>0\Delta S_{total} > 0, and the reaction is spontaneous. It follows that, to decide the spontaneity of reactions, we need to consider the entropy of the system and of its surroundings.

Note

The book's prose at this step says "527 kJ of heat is released", while its own data line and calculation use ΔH=−572\Delta H = -572 kJ and 572×103572 \times 10^3 J — the "527" is the print's own slip; the value used consistently throughout the working is 572 kJ. …

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