Topic 12 · page 6 of 17

Surroundings and total entropy

Heat passed to the surroundings changes their entropy by −ΔH / T. Add that to the entropy change of the system to get the total, and a change is feasible when the total is positive.

14 min read12.412.612.8

Read first: Entropy change of the system

System and surroundings 12.6

The system is the chemicals that are reacting. The surroundings are everything else: the solvent, the flask, the air, the bench. Together they make the universe, and the entropy that counts is the entropy of the whole universe.

Two panels. Each shows a box labelled system inside a dashed box labelled surroundings. In the left panel, arrows point out from the system: an exothermic reaction, with delta H negative and the entropy change of the surroundings positive. In the right panel, arrows point into the system: an endothermic reaction, with delta H positive and the entropy change of the surroundings negative. — Heat flows between the system and its surroundings. Exothermic: heat flows out, and the surroundings gain energy quanta. Endothermic: heat flows in.

\Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surroundings}

Heat and the surroundings 12.8

An exothermic reaction gives out heat. That heat becomes extra energy quanta in the surroundings, spread over a huge number of molecules, so the entropy of the surroundings rises. An endothermic reaction takes energy quanta from the surroundings, so their entropy falls. This is how enthalpy enters the entropy rule. The size of the effect is:

\Delta S_\text{surroundings} = -\frac{\Delta H}{T}

The minus sign. An exothermic reaction has $\Delta H$ negative, so $\Delta S_\text{surroundings}$ is positive.
The division by T. The same heat matters more to cold surroundings than to hot ones. One shout is very noticeable in a silent library and barely noticed in a busy station. The same energy adds more disorder to cold surroundings.

MethodCalculating the entropy change of the surroundings

Convert $\Delta H$ from kJ mol⁻¹ to J mol⁻¹ by multiplying by 1000.
Convert the temperature to kelvin: K = °C + 273.
Calculate $-\Delta H / T$ . Keep the sign: exothermic gives a positive answer.
Add $\Delta S_\text{system}$ , which is already in J K⁻¹ mol⁻¹, to get $\Delta S_\text{total}$ .

Worked exampleAmmonia synthesis at 298 K

$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ , $\Delta H$ = −92 kJ mol⁻¹ and $\Delta S_\text{system}$ = −198.8 J K⁻¹ mol⁻¹. Is the reaction feasible at 298 K?

$\Delta H$ = −92 × 1000 = −92 000 J mol⁻¹
$\Delta S_\text{surroundings}$ = −(−92 000) ÷ 298 = +309 J K⁻¹ mol⁻¹
$\Delta S_\text{total}$ = −198.8 + 308.7 = +110 J K⁻¹ mol⁻¹
$\Delta S_\text{total}$ is positive, so the reaction is feasible at 298 K, even though $\Delta S_\text{system}$ is negative.

Worked exampleBurning magnesium at 298 K

$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$ , $\Delta H$ = −1204 kJ mol⁻¹ and $\Delta S_\text{system}$ = −216.6 J K⁻¹ mol⁻¹.

$\Delta S_\text{surroundings}$ = 1 204 000 ÷ 298 = +4040 J K⁻¹ mol⁻¹
$\Delta S_\text{total}$ = −216.6 + 4040.3 = +3824 J K⁻¹ mol⁻¹. The reaction is strongly feasible. Page 9 explains why magnesium ribbon still does not catch fire by itself.

Total entropy decides the direction 12.412.6

The system on its own can mislead. Ammonia synthesis has a negative $\Delta S_\text{system}$ and still goes at 298 K, because the heat given out raises the entropy of the surroundings by even more. The rule that never fails is:

Worked exampleDoes water freeze?

Water at 0 °C is the border between liquid and solid. $\ce{H2O(l) -> H2O(s)}$ has $\Delta H$ = −6.01 kJ mol⁻¹ (freezing gives out heat) and $\Delta S_\text{system}$ = 48 − 70 = −22 J K⁻¹ mol⁻¹. Find $\Delta S_\text{total}$ at 263 K (−10 °C) and at 283 K (+10 °C).

At 263 K: $\Delta S_\text{surroundings}$ = 6010 ÷ 263 = +22.85. $\Delta S_\text{total}$ = −22 + 22.85 = +0.85 J K⁻¹ mol⁻¹. Positive: water freezes.
At 283 K: $\Delta S_\text{surroundings}$ = 6010 ÷ 283 = +21.24. $\Delta S_\text{total}$ = −22 + 21.24 = −0.76 J K⁻¹ mol⁻¹. Negative: water does not freeze.
The total is zero at T = 6010 ÷ 22 = 273 K, which is 0 °C. The numbers reproduce the freezing point.

The same heat released is worth more entropy to colder surroundings. That is why temperature decides who wins, the subject of page 7.

Check yourself

Calculate $\Delta S_\text{surroundings}$ at 298 K for a reaction with $\Delta H$ = −46 kJ mol⁻¹.
Calculate $\Delta S_\text{surroundings}$ at 500 K for a reaction with $\Delta H$ = +178 kJ mol⁻¹. Why is the sign different from question 1?
For $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$ , $\Delta H$ = −890 kJ mol⁻¹ and $\Delta S_\text{system}$ = −242.7 J K⁻¹ mol⁻¹. Calculate $\Delta S_\text{total}$ at 298 K.
For $\ce{N2O4(g) -> 2NO2(g)}$ , $\Delta H$ = +57 kJ mol⁻¹ and $\Delta S_\text{system}$ = +176 J K⁻¹ mol⁻¹. Is it feasible at 298 K?
A student calculates $\Delta S_\text{surroundings}$ for $\Delta H$ = −92 kJ mol⁻¹ at 298 K as +0.309 J K⁻¹ mol⁻¹. What is wrong?
Why does the same amount of heat released produce a larger entropy change in cold surroundings than in hot ones?

Answers cover these until you have tried the questions

$\Delta S_\text{surroundings}$ = −(−46 000) ÷ 298 = +154 J K⁻¹ mol⁻¹
$\Delta S_\text{surroundings}$ = −178 000 ÷ 500 = −356 J K⁻¹ mol⁻¹. The reaction in question 2 is endothermic: it takes heat from the surroundings and lowers their entropy. Question 1 is exothermic and raises it.
$\Delta S_\text{surroundings}$ = 890 000 ÷ 298 = +2986.6. $\Delta S_\text{total}$ = −242.7 + 2986.6 = +2744 J K⁻¹ mol⁻¹ (feasible).
$\Delta S_\text{surroundings}$ = −57 000 ÷ 298 = −191.3. $\Delta S_\text{total}$ = +176 − 191.3 = −15 J K⁻¹ mol⁻¹. This is negative, so the reaction is not feasible at 298 K.
The student left $\Delta H$ in kJ. It must be −92 000 J mol⁻¹, giving −(−92 000) ÷ 298 = +309 J K⁻¹ mol⁻¹.
In cold surroundings there are few energy quanta already, so a given amount of extra energy increases the number of ways of sharing them by a large factor. In hot surroundings the same energy is a small addition to what is already there. This is the 1 ÷ T in $-\Delta H / T$ .