# Why does the reaction between Na (s) and Cl2 (g) occur spontaneously if the entropy of the system decreases?

1- The reaction is exothermic. 2- My book says that it is because the total entropy change is very positive. Here is my book's explanation and it is the highlighted part which i dont understand : "Despite the fact that there is a decrease in the entropy of the system, the energy leaving the system produces a substantial increase in the entropy of the surroundings, because there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone.

• The book is correct. Spontaneity is governed by the change in entropy of the universe which is the sum of the change in entropy of the system and the change in entropy of the surroundings. The change in entropy of the surroundings is derived from the enthalpy change of the system. – Zhe Jul 12 '17 at 15:30
• @Zhe okay what i didnt understand is what i wrote in bold. So what if there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone? – Max white Jul 12 '17 at 15:33
• That's the definition of entropy... – Zhe Jul 12 '17 at 15:34
• @Zhe i think i got it. It is because there will be a smaller chance of having the energy quanta divided equally over the molecules in the surroundings right ? – Max white Jul 12 '17 at 15:34
• "It is because there will be a smaller chance of having the energy quanta divided equally over the molecules in the surroundings" Not sure what that means... The number of microstates available in the surroundings per unit increase in energy is larger. That is the definition of entropy. – Zhe Jul 12 '17 at 15:38

As you point out there are two things to consider, the heat released in the reaction and the change in entropy. In forming NaCl from chlorine gas and solid sodium the entropy decreases because the gas is converted into solid NaCl which has a far far smaller total volume. The energy released has the effect of heating the solid NaCl just formed and then the surroundings as energy flows away to eventually reach equilibrium again but now at a higher temperature than that at equilibrium before reaction occurred.

When a quantity of heat $dq$ from a body at $T_2$ transfers to one at $T_1$ the overall entropy change is $dS = dS_1 + dS_2 = dq/T_1 - dq/T_2$. As $T_2 \gt T_1$ the entropy change is positive.

In the reaction the entropy change overall depends on comparing that decreased due to volume decrease vs. that increased due to the energy released on reaction.

Entropy is a measure of the number of ways that the particles can be placed into their energy levels, this is often called the number of configurations $\Omega$, and Boltzmann found that $S=k\ln(\Omega)$ so that increasing the distribution of energy into the energy levels increases $\Omega$ and hence S. As the temperature increases more vibrational and rotational levels are populated and the kinetic energy distribution widens, all this means that there is an increased number of configurations and hence entropy just because energy can be placed into energy levels and in more ways.

If there was just one level then all 'particles' would have to be in this level and so entropy would be zero, $\Omega=1$. When there are two levels some can be in the first and some in the second level and now the entropy is increased ($\Omega \gt 1$), and so on for many levels.

• So the released energy increases the entropy of the surroundings because there are more possible arrangements of energy quanta which means that there is a lesser chance of the particles being in the same energy level, right ? – Max white Jul 12 '17 at 20:30
• yes, just so, entropy interpreted at a molecular level is all about the number of ways a molecule's energy can be distributed among its energy levels. – porphyrin Jul 12 '17 at 20:36

"...there are more ways of arranging the energy quanta in the rest of the universe than there are of arranging them in the system alone."

This sentence is poetical gibberish if one does not first study the chemical reaction at hand by the fundamental equation.

### 1 Prelininary remarks

Because of his question about statistical mechanics, I suppose that the questioner has (at least superficial) knowledge of the fundamental equation of (chemical) thermodynamics $\ce{\Delta G} = \Delta H – {T\Delta S}$, otherwise it would be nearly impossible to explain the touched entanglement without gibberish.

In the formation reaction of $\ce{NaCl}$ from the elements, $\ce{\Delta G}$ is negative if the amount of (negative) $\Delta H$ (exothermic reaction) exceeds the amount of (positive) $\ce{–T\Delta S}$; this is the indication that this reaction may proceed spontaneously. If during the reaction, the punctual temperature near the piece of sodium in the porcelain boat is “relatively low” (perhaps 1000°C, when chlorine is channeled slowly over the sodium), the positive value of $\ce{–T\Delta S}$ does not exceed $\Delta H$.

### 2 The core argument

While it is otherwise a good practical jingle, this antagonistic view of $\Delta H$ and ${T\Delta S}$ is confusing if one tries to apply it (as in the question) from the point of view of the law of increasing entropy. Entropy is defined as $\Delta S$ = $\frac{\Delta q}{T}$ (where q is transferred “heat” $\Delta H$). The real meaning of $\ce{\Delta G}$ may only be revealed if it is transformed into its entropy representation, by dividing the fundamental equation by (-T) (note that $\Delta H$ and $\Delta S$ are by definition $\Delta H_{sys}$ and ${\Delta S}_{sys}$):

$$\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}$$

In this equation $\frac{\Delta H_{sys}}{T}$ is the entropy change of the system by “heat” exchange with the surroundings, hence $\frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}$. On this basis, the above equation may be identified as:

$$\Delta S_{total} = {\Delta S_{surroundings} + \Delta S}_{sys}$$

$\Delta S_{sys}$ and $\Delta S_{surroundings}$ strive mutually to a maximum of $\Delta S_{total}$.

Based on the comments, it looks like you are struggling with the base concept of spontaneity.

Spontaneity is characterized by having the total entropy change in a process be non-negative:

$$\Delta S_{\mathrm{universe}} \geq 0$$

This we can decompose into two pieces:

$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{surroundings}} + \Delta S_{\mathrm{system}}\geq 0$$

So, the system or the surroundings can experience a decrease in entropy as long as the other component more than compensates for the decrease with an increase in entropy that is at least as large as the decrease.

The definition of entropy is: $$S = k \log \Omega$$ where $\Omega$ is the density of states, that is the number of states available for some small increase in energy:

$$\Omega = \frac{\delta \Phi}{\delta E}$$

The bold statement is trying to say that because the "number of states" (even though they mean density of states) is decreasing in the system, the entropy change is negative. On the other hand, because of the high exothermicity of reaction, the density of states in the surroundings is increasing by an amount that more than compensates for this (i.e., the entropy of the surroundings goes up a lot). This maintains the overall relationship that $\Delta S_{\mathrm{universe}}\geq 0$ even though the system's entropy change is negative.

Specifically, in your reaction, the increase in entropy of the surroundings is derived from the massive quantity of heat that you're dissipating from the system.

• Okay i understood what you just said in this answer but what i dont understand is what i wrote in bold. Did i write something wrong in the last two comments ? – Max white Jul 12 '17 at 18:03
• @Maxwhite No, the comments were getting long, and I thought I understood your problem, but maybe not. I'm having trouble understanding what about the bold statement you don't understand... – Zhe Jul 12 '17 at 18:38
• The thing is that i dont see why they even mentioned the number of possible arrangements of quanta in the system. They are explaining why the energy leaving the system produces an increase in the entropy of the surroundings right ? If i had to explain why that energy increases the entropy of the surroundings i would say : After the energy has been transferred to the surroundings, there are more possible arrangements of quanta in the universe. If we thought of the "universe" as the gas molecules in the air, this transfer of energy to the surroundings means that now there is a smaller chance of – Max white Jul 12 '17 at 19:04
• (Continued) the molecules sharing their quanta equally since there are more possible arrangements and only one of these arrangements has all the quanta shared equally between the molecules. Now, is my explanation correct ? – Max white Jul 12 '17 at 19:07
• @Maxwhite Updated answer. Hope this helps. – Zhe Jul 12 '17 at 19:22