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Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

Chapter 18, Problem 15

Ammonium hydrogen sulfide, a stink bomb ingredient, decomposes to ammonia and hydrogen sulfide: Calculate the standard free-energy change for the rection at 25 °C if the total pressure resulting from the solid NH4S placed in an evacuated container is 0.658 atm at 25 °C. (a) -43.8 kJ (b) +1.04 kJ (c) -462 kJ (d) +5.51 kJ

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everyone in this video being total. When we're heating solid C A C 03 results in the formation of a solid ceo and carbon dioxide gas here are being asked to calculate for the delta gene at 500 kelvin's the total pressure from heating solid C A C. 03 and a vacuum container is at 1.86 18 for pressure. So the reaction that's going on here Is that we have again our solid CAC. yielding R C A O. Which they sell it as well. And our carbon dioxide gas. So we know that my delta G of reaction is equal to negative R. Times T. Multiplied by the natural log of our K constant. So my key constant here is equal to the pressure by the C. 02 which is given to us in this problem to be 1.86 A. T. M. Let's go ahead and plug in or utilize this formula to find our answer. So again we have delta G. Of the reaction equaling two are are here as our guest constant. So that's 8.314 joules per mole times kelvin. We're multiplying this by T. For temperature. And that's given to us at 500 Kelvin's. And then we have the natural log of K. And we established that to be 1.86. Alright, so once I put everything into a calculator, I get them, I delta G of reaction is equal to negative 2557. jewels per mole. When I convert these jewels into killer jewels so we can do a direct conversion. So for every one killer jewel we have, we have 1000 jewels. And once you put that into the calculator we get the my Delta G. of reaction is equal to negative 2.56 killer jewels per mole. So then this right here is going to be my final answer for this problem.
Related Practice
Textbook Question
Calculate ∆Stotal, and determine whether the reaction is spon-taneous or nonspontaneous under standard-state conditions. (a) -429 J/K; nonspontaneous (b) -123 J/K; spontaneous (c) +3,530 J/K; nonspontaneous (d) +184 J/K; nonspontaneous
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Textbook Question
Consider the following endothermic reaction of gaseous AB3 molecules with A2 molecules.

Identify the true statement about the spontaneity of the reaction. (a) The reaction is likely to be spontaneous at high temperatures. (b) The reaction is likely to be spontaneous at high temperatures. (c) The reaction is always spontaneous. (d) The reaction is always spontaneous.
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Textbook Question
Nitrogen reacts with fluorine to form nitrogen trifluoride: Calculate ∆G°, and determine whether the equilibrium composition should favor reactions or products at 25 °C (a) ∆G° = -6.7 kJ; the equilibrium composition should favor products. (b) ∆G° = -332 kJ; the equilibrium composition should favor reactants (c) ∆G° = -166 kJ; the equilibrium composition should favor products (d) ∆G° = +82.6 kJ; the equilbirum composiiton should favor reactants.
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Textbook Question
Consider the following graph of total free energy of reactants and products versus reaction progress for the general reaction, Reactants -> Products. At which of the four points (labeled a, b, c, and d) is Q < K?

(a) Point a (b) Point c and d (c) Point a, c, and d (d) Point b
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Textbook Question

Spinach contains a lot of iron but is not a good source of dietary iron because nearly all the iron is tied up in the oxalate complex [Fe(C2O4)3]3-.

(b) Under the acidic conditions in the stomach, the Fe3+ concentration should be greater because of the reaction

[Fe(C2O4)3]3-(aq) + 6 H3O+(aq) ⇌ Fe3+(aq) + 3 H2C2O4(aq) + 6 H2O(l)

Show, however, that this reaction is nonspontaneous under standard-state conditions. (For H2C2O4, Ka1 = 5.9 × 10-2 and Ka2 = 6.4 × 10-5.)

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Textbook Question

Formation constants for the ammonia and ethylenediamine complexes of nickel(II) indicate that Ni(en)32+ is much more

stable than Ni(NH3)62+:

(1) <REACTION>

(2) <REACTION>

The enthalpy changes for the two reactions, ΔH°1 and ΔH°2, should be about the same because both complexes have six Ni﹣N bonds. 

(c) Assuming that ΔH°2 - ΔH°1 is zero, calculate the value of ΔS°2 - ΔS°1.

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