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Ch. 18 - Free Energy and Thermodynamics

Chapter 18, Problem 90a

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2( g) a. Show that the reaction is not spontaneous under standard conditions by calculating ΔGrxn ° .

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Hey everyone so asked to prove that using a standard change and gives free energy of the reaction. That the following reaction is not spontaneous at 298 Calvin under standard conditions. So this is already balanced. And we call that the standard change and gives free energy of the reaction. Is used to the change in entropy of the reaction minus the temperature. Time to change in entropy of the reaction. We need to first find delta? S. Other reaction and this is a son comes the number of most. I'm still too S. M. Other products. The sun was the number of moles. I was about to ask em other reactant dr s Poseidon peroxide Is 95.0. Those promote I'm Calvin for dr S. 02. You get 205 138 joe is paramount kelvin. And for delta s designing peroxide It was 15.9 promote how's kelvin other reactions We get to malls of sodium super oxide House 1 15.9 those promote Thanks Calvin minus one mode of Sony and peroxide. 1095. Just promote that was kelvin. Just one more 02. Some 205 138. Those promote how's Calvin but that s of the reaction negative 0.4. Those for Calvin -0.0684. Well jules per Calvin. Now I need to find delta H other reaction and it's equal to the sound. What is the number of malls times? Delta H F. Other products minuses son has a number of malls delta H. F. Other reactive delta H. F. In a tube. Oh champ. Can I get a 50 510 0.9 promote lots of H. F. Oh chip zero. Kill. Just promote and dr H. F. Of N. A. 02 Is negative 26 106. I get 260 0.2. Well just for a month Delta H. The reaction Give us two malls N. A. 0. 2. It was made of 260 tim. Just promote minus one month and a two. I was making 510 0.9 just for a month. Last one more of cartoon zero. We'll just promote for delta H. Of the reaction Getting -9.5. Well jules and we can plug in these values to find delta G. Of the reaction. That's a cheap. Other reaction That is 9.5 kill jules - Alvin, -0.0684. So just from Calvin And we get positive 10.9 Charles dr G. Other reaction, it was a positive but this is not spontaneous. So the reaction is not spontaneous because Delta G. For the reaction, It's 10.9 kg jewels. Thanks for watching my video and I hope it was helpful
Related Practice
Textbook Question

Consider this reaction occurring at 298 K: N2O(g) + NO2(g) ⇌ 3 NO(g) a. Show that the reaction is not spontaneous under standard conditions by calculating ΔG°rxn.

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Textbook Question

Consider this reaction occurring at 298 K: N2O( g) + NO2( g) ⇌ 3 NO( g) b. If a reaction mixture contains only N2O and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous?

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Textbook Question

Consider this reaction occurring at 298 K: N2O(g) + NO2(g) ⇌ 3 NO(g) c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions?

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Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2( g) b. If BaCO3 is placed in an evacuated flask, what is the partial pressure of CO2 when the reaction reaches equilibrium?

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Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g) c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, at what temperature is the partial pressure of carbon dioxide 1.0 atm?

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Textbook Question

Living organisms use energy from the metabolism of food to create an energy-rich molecule called adenosine triphosphate (ATP). The ATP acts as an energy source for a variety of reactions that the living organism must carry out to survive. ATP provides energy through its hydrolysis, which can be symbolized as follows: ATP(aq) + H2O(l) → ADP(aq) + Pi(aq) ΔGrxn ° = -30.5 kJ where ADP represents adenosine diphosphate and Pi represents an inorganic phosphate group (such as HPO42-). a. Calculate the equilibrium constant, K, for the given reaction at 298 K.

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