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As we have seen above, the entropy change of the ammonia / hydrogen chloride reaction (‘the system’) is –284 J K^{-1} mol^{-1}. It is negative as we have calculated (and predicted from the reaction being two gases going to a solid). But how can we evaluate the entropy change caused by ‘dumping’ 176 kJ mol^{-1} of heat energy into the surroundings? (Notice that this is 176 000 J mol^{-1}.) It must be positive as more quanta of energy lead to more possible arrangements and it must be more than 284 J K^{-1} mol^{-1}.

The formal derivation is complex but leads to the expression ∆*S* of the surroundings (∆*S*_{surroundings}) = –∆*H* / *T*. This makes sense because:

- the negative sign means that an exothermic reaction (∆
*H*is negative, heat given out) produces an increase in the entropy of the surroundings - the more negative the value of ∆
*H*, the more positive the entropy increase of the surroundings - the same amount of energy dumped into the surroundings will make more difference at lower temperature – this rationalises the ‘division by
*T*’

For the ammonia / hydrogen chloride reaction at 298 K:

∆*S*_{surroundings} = –∆*H* / *T*= –(–176 000)/298

∆*S*_{surroundings} = 591 J K^{-1} mol^{-1}, more than enough to outweigh the value of ∆S_{system }of –284 J K^{‑1} mol^{-1}.

So the total entropy change (of the Universe, *ie* system + surroundings) brought about by the reaction is +307 J K^{‑1} mol^{-1}.

So if we want to predict the direction of a chemical reaction we must take account of the *total* entropy change of the system *and* the surroundings, and that includes the effect on entropy of any heat change from the system to the surroundings (or in the other direction, heat taken in from surroundings to system).

So

∆*S*_{total} = ∆*S*_{system} + ∆*S*_{surroundings}

If ∆*S*_{total} for a reaction is positive, the reaction will be feasible, if negative it will not be feasible.

Total entropy change |