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Ch.19 - Free Energy & Thermodynamics

Chapter 19, Problem 41c

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) c. ΔH°rxn = -115 kJ; ΔS°rxn = -263 J>K; T = 298 K

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Hello everyone in this video we're trying to calculate for our delta S. Of the universe as well as predicted. The reaction is going to be spontaneous or not given this following data right over here. All right. First thing we need to go ahead and realize is that the delta US of the universe is composed of the delta S. Of this system, plus our delta S of the surroundings. Alright, so first they were going to solve for is our delta S. Of the universe or the delta as our system. I apologize. So that is actually equal to the delta us of reaction. And conveniently we are given that in the problem right over here. So from the problem, it's that the delta S of the reaction is negative 1 54 jewels per kelvin. Now, solving for the second component of this equation is that the delta US of the surroundings And what that is included for that is going to be our negative delta H. Of the reaction divided by R. T. For temperature. Again, conveniently we are given that here and here. All right, so let's go ahead and plug those numerical values in. Of course the delta age of reaction is going to become negative 2 56 killer jewels. We want our units to go ahead and match the delta as of the system. So when we add this together, it'll be easier. So we'll go ahead and convert this into jewels. So on top will have 103 jewels on the bottom will be one killer jewel. Of course you can see that the killer jewels will go ahead and cast out nicely. Now of course this is all going to be divided by R. T. for temperature which is given to us as 3 15 Kelvin. So plugging that into my calculator, the numerical value comes out to be negative 812.69 units being jewels per kelvin. Alright, now that we have both components of the problem, let's go ahead and plug in those american values. So we have delta S. Of the universe equaling two negative 1 54 joules per kelvin plus our negative 8 12.69 joules per kelvin. So putting that into my calculator, I'll get the value of negative 967 jewels per kelvin. So that's one part of my answer or our delta S. Of the universe. Next is to see if the reaction is going to be spontaneous or not. How we can go ahead and solve for that is by our delta G. Delta G is just equal to delta H minus R. T. Multiplied bar delta us. So of course we're given earn have found all these values. So we'll simply go ahead and plug that in. So the delta H. Is going to be 2 56 killer jewels and we're subtracting that by R. T. Which is 3 15 kelvin's multiplied by our delta S. And that's negative 1 54 jewels per kelvin again, we want our units to match. So I'm gonna go ahead and convert our jewels back into kilo jewels. So we have one kg jule on top and 10 to the third jewels on the bottom. As you can see here for units wise, then we have that the jewels will go ahead and cancel and that our kelvin's will go ahead and cancel as well. Alright, so putting my delta G numbers into my calculator, I'll get that numerical value is equal to 305 and units will be just killed jules. Now, what what what can we do with this delta G? We can see that our delta G value is actually positive and whenever we have a positive delta G, this means that the reaction is not spontaneous or non spontaneous. Alright then this will be our second answer for this problem. And again we have, the Delta S of universe is negative, 967 tools per kelvin, and the reaction will not be spontaneous. Alright, thank you all so much for watching
Related Practice
Textbook Question

Without doing any calculations, determine the signs of ΔSsys and ΔSsurr for each chemical reaction. In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction is spontaneous. c. 2 N2(g) + O2(g) → 2 N2O(g) ΔH°rxn = +163.2 kJ

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Textbook Question

Calculate ΔS surr at the indicated temperature for each reaction. d. ΔHrxn ° = +114 kJ; 77 K

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Textbook Question

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) a. ΔH°rxn = +115 kJ; ΔS°rxn = -263 J/K; T = 298 K

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Textbook Question

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) a. ΔH°rxn = -95 kJ; ΔS°rxn = -157 J/K; T = 298 K

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Textbook Question

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) c. ΔH°rxn = +95 kJ; ΔS°rxn = -157 J/K; T = 298 K

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Calculate the free energy change for this reaction at 25 °C. Is the reaction spontaneous? (Assume that all reactants and products are in their standard states.) 2 Ca(s) + O2( g) → 2 CaO(s) ΔH° rxn = -1269.8 kJ; ΔS° rxn = -364.6 J/K

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