Interpreting entropy as "disorder", which comes from thermodynamics applications, is only one means to analyze entropy. Entropy deserve to likewise be described according to statistical mechanics as the statistical distribution of possible degenerate microstates:
wbelow $k_B$ is Boltzman"s consistent and $Omega$ is the variety of microstates. Therefore it is a measure of uncertainty. The even more equivalent/degenerate microclaims there are, the less likely we know specifically which among them the device is in, and thus, the better the entropy. This probabilistic interpretation provides a much more intuitive and also satisfactory expertise of the connection between variety of molecules and also entropy of the mechanism.
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A decrease in the number of molecules leads to a decrease in the complete number of degenerate microclaims, which leads to a decrease in entropy. Tright here are even more degenerate means to distribute N objects in M spaces than tbelow are degenerate means to distribute (N-1) objects in M spaces.
Let"s look at a really simple instance, in which we have 3 molecules $ce2H2 + O2$ in a container through room for 4 molecules. How many kind of means deserve to we fill four spaces through these three molecules? Tright here are four spaces in which we might put the $ceO2$ molecule:
$$eginpmatrixceO2 & ... \ ... & ... \ endpmatrixeginpmatrix... & ceO2 \ ... & ... \ endpmatrixeginpmatrix... & ... \ ceO2 & ... \ endpmatrixeginpmatrix... & ... \ ... & ceO2 \ endpmatrix$$
How many type of means can we put two $ceH2$ molecules in one of these arrangements? In order to answer this, we should temporarily distinguish one $ceH2$ from the various other $ceH2^circ$ (or to differentiate the one we place first from the one we location second) in order to geneprice all of the degeneprice says.
$$eginpmatrixceO2 & ceH2 \ ceH2^circ & ... \ endpmatrixeginpmatrixceO2 & ceH2 \ ... & ceH2^circ \ endpmatrixeginpmatrixceO2 & ceH2^circ \ ceH2 & ... \ endpmatrixeginpmatrixceO2 & ... \ ceH2 & ceH2^circ \ endpmatrixeginpmatrixceO2 & ceH2^circ \ ... & ceH2 \ endpmatrixeginpmatrixceO2 & ... \ ceH2^circ & ceH2 \ endpmatrix$$
Tright here are six arrangements of two $ceH2$ molecules for each area of $ceO2$ for a total of $Omega=24$ arrangements. The entropy $S=k_Bln(24)=4.39 imes10^-23) frac extJ extK$. Keep in mind that this occurred to be $Omega=4 imes3 imes2=12$, or 4 methods of placing the initially molecule, 3 methods of placing the second, and two ways of placing the third. We will certainly usage this shortreduced next time we do this.
Now, let us run the reaction, converting hydrogen and also oxygen to water.
$$ce2H2 + O2 -> 2H2O$$
And let us repeat the exercise. How many type of methods have the right to we fit 2 $ceH2O$ molecules in 4 spots? Tright here are 4 means to place the first one, and 3 means to place the second for each of those 4 placements of the initially. Hence, $Omega=4 imes3=12$ and the entropy is $S=k_Bln(12)=3.43 imes10^-23) frac extJ extK$.
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Keep in mind that tright here is a subtle flaw in my explacountry. The molecules themselves have an intrinsic entropy concealed in their structures as well. According to this site the conventional molar entropies of $ceO2, ceH2,$ and also $ceH2O$ are (in$frac extJ extKcdot extmol$) 205.1, 130.7, and also 188.8, respectively. Thus, we have an entropy adjust of reaction:
$$Delta_rS^circ=sum S^circ_products - sum S^circ_reactants$$ $$ Delta_rS^circ =left(2 imes188.8 - (205.1+2 imes130.7) ight)frac extJ extKcdot extmol=-40frac extJ extKcdot extmol$$
By comparing standard entropies, we additionally gain a decrease in entropy. Standard entropies are easier to occupational with bereason we are frequently working on the molar range. If we adjusted our streamlined example to be moles of $ceH2, ceO2,$ and $ceH2O$, in 4 moles of spaces we would certainly have actually the original number of microclaims be:
When you plug in $N_A=6.022 imes10^23$, all but the many staunch calculators and also computer system programs are going to provide up. Wolfram Alpha have the right to perform it bereason it does so symbolically, and also we obtain
$$S=8.314frac extJ extKcdot extmolln(1.398 imes10^73)=1.400 imes10^3frac extJ extKcdot extmol$$
(At the molar scale we replace $K_B$ via $R$, the appropriate as consistent.)
Another obvious weakness of the statistical strategy is that it assumes no phase readjust. As F"x stated, there is an entropy adjust associated through a phase adjust. Roughly, the number of accessible spaces increases or decreases.