<p>My chem text only shows how to calculate the entropy change in the surroundings, which is -dH/S with respect to the system. If I wanted to calculate, say, the entropy change in the system that is water being heated to a vapor, do I just flip-flop system and surroundings? So that the equation I use is just -dH/S with respect to the surroundings for the dS of the water?</p>
<p>So say 43 kJ/mol is absorbed by 1 mol of H2O. Then is dS = (43 kJ/mol)(10^3 J/KJ)/298K ?</p>
<p>No! dH is related to the change in entropy of the surroundings because when a system releases heat, the surrounding molecules increase in speed and therefore have more entropy. (And when the system absorbs heat, the cooled surroundings have less entropy.) This is independent of the change in entropy of the system.
Qualitative determination of the change in entropy of a reaction system is done by comparing reactants to products. Changing solids to liquids, or liquids to gases, or increasing the # of moles of gas increases entropy.
Quantitative determination of entropy requires reference data. Like dH, dS is a state function. That means you can do the "same type of math" with dS as you do with dH. (1) [Sum of products]-[sum of reactants] = dS. (2) Hess' Law type calculations</p>
<p>On Zumdahl's site with extra quizes, some of the questions seem to suggest that the sum entropy is zero. I am assuming that the questions assume that the heat flow is reversible process so that entropy change is zero.</p>
<p>Could you provide an example of a question? dS(universe) can be zero [in fact dS(univ) always equals zero at equilibrium], but dS(system) is rarely zero. When AP questions ask about dS, they always refer to dS(system) The Zumdahl text, in establishing the theoretical foundation of dG=dH-TdS, refers to dS(system), dS(surroundings), and dS(universe). I'm not sure which is being used in the quiz you are referring to.</p>
<p>In any case, when dS(of any variety) equals 0, it doesn't have anything to do with whether heat flow is reversible.
dS(surroundings) = 0 when dH=0
dS(system)=0 when there is zero change in positional probabilities between reactants and products (rare)
dS(universe) = 0 when (a) the enthalpy and entropy driving forces are in opposition to each other and (b) the temperature is such that they exactly balance out.</p>
<p>One mole of calcium chloride is vaporized at 1940?C. If the heat of vaporization for CaCl2 is 235.1 kJ mol-1, calculate the change in entropy for the process.</p>
<pre><code>A. -12.65 J/K
B. 140.7 J/K
C. 0.1186 J/K
D. -98.06 J/K
E. 106.2 J/K
</code></pre>
<p>Answer is E. This seems to be done by using dS<em>sys = -sS</em>surroundings = -H/T = -(235.1 x 10^3 J)/(1940 + 273 K)</p>
<p>When a pure substance is at the temperature of a phase change (melt point or boiling point) the system is at equilibrium, therefore the following substitutions can be made:</p>
<p>dG = dH - TdS
0 = dHvap - (BP)dS or 0=dHfus - (MP)dS
Therefore, dH = TdS or dS = dH/T</p>
<p>This is essentially the same thing as what you said - the change in entropy of surroundings is equal and opposite to the change in entropy of the system. (That's what creates the equilibrium - opposing, equally strong driving forces.) This <em>only</em> works, though, at an equilibrium temperature. For example, if they said "Reaction X occurs at 1940C and dHrxn=235.1kJmol-1", you couldn't calculate dS because you can't assume dG = 0. You would need further information.</p>
<p>Oh, I see. Thanks a lot. When you say equilibrium, you mean the states are in equilibrium right? </p>
<p>Because in chemical equilibrium the entropy of sys/surr are not necessarily the same as it is when Gibb's Free Energy is as low as possible and hence when universal entropy is as positive as possible.</p>
<p>At equilibrium, dG is always zero - for both phase equilibria and chemical equilibria. I think you are confusing using dG to predict the direction of a reaction versus dG once the reaction is at equilibrium. If a reaction has a large negative dG, that means it spontaneously proceeds to the right because the products have a lower free energy than the reactants. As the reaction proceeds, however, free energy of the reactants and products changes until, at the equilibrium point, G(reactants)=G(products). At that point, dG=0.
From a mathematical perspective,
dG = dG0 + RTlnQ (dG0 is standard free energy - pretend the 0 is a degree sign.)</p>
<p>Since dG0= -RTlnKeq, when Q=K (which by definition means you are at equilibrium), dG=0.</p>
<p>Oh yeah. I meant total G, not the change. dG is zero at equilibrium so that's why no net change occurs, but equilibrium point is the when the system reaches the lowest G value possible, or equivalently when the universe reaches the greatest value of entropy as -G/T at constant P and T = dS_univ.</p>
<p>Thanks for that explanation though, it made me double check the text. I feel I understand this now thanks.</p>