Consider the decomposition of barium carbonate: BaCO 3 (s) ⇌ BaO(s) + CO 2 (g) Using data from Appendix C, calculate the equilibrium pressure of CO 2 at (a) 298 K.

Hello everyone. So in this video we're trying to solve for the equilibrium pressure of our carbon dioxide as 450 degrees Celsius for this given reaction right over here. So because we're dealing with our delta H values and um its values, we can go ahead and relate that the K. Constant will equal to the equilibrium pressure of carbon dioxide. And the reason why it's just the carbon dioxide term is because we we know that our K. Value is equal to the concentration of our products over the concentration of the reactant and we only deal our input those concentrations if they're in a gaseous phase in this case we only have our carbon dioxide being in a gaseous phase, the serra material and the other agent here is going to be in solid form. So we don't include that. Therefore our K. Constant is just equal to the pressure of our carbon dioxide. So for delta G. Value that is equal to our delta H minus T. Times the delta S. Value. Let's go ahead and solve for our delta H. Value first. So my delta H. Of the reaction is equal to the delta H. Of again my products minus my delta H of the reactant. So given this table we can go ahead and just simply plug in those values. So for my products We have negative 592.0 plus negative 393.5. And then we'll go ahead and subtract that with our Delta H. of products which is just negative 12-0.1. If you put these into the calculator we get the value of 234.6 killer jewels per mole. Now, solving for my delta S again seems like pattern. Here is the values of our products, minus the values of our reactant. So then Delta S. Is equal to the C. For the products we have 54 point four And then we see her. It's just 213.8. Putting that into my calculator, I get 90. I forgot the values for our reactions. Here is subtracted by 97.1. So you put that into the calculator. We get that. This was equal to 1 71. jules, Her mole times cape. All right. Our temperature here were given uh this value in Celsius units. We want to convert this to Calvin's. How we do this is for 15 plus 73.15. And that was equal to 723.15. And our units will now be in Calvin's. So then Mhm. Bring up this equation again, our delta G. Is equal to 34.6. We want to convert our killer jewels into jewels. So we'll multiply this by 1000 to get us into jewels. They were subtracting this with let's see here. 723.15 Kelvin's multiplied by 171.1 Jules Permal Times Kelvin. Once I put all these guys into my calculator, I'll get the final delta G value being 110869 jewels. Now we're not quite done yet. We're solving for R. K. Value. How we relate our delta G to R. K. Value. We're gonna go ahead and use this occasion here, which is delta G. Going to the negative R. T. Not multiplied by the natural log of our cake constant. Will use mathematical manipulation to solve for R. K. Value. And that is K. Is equal to E. To the power of delta G over negative R. Of T. So we have all the numerical components. Let's go ahead and plug those in. So my delta G. Is 110869 jules. Then negative. Let's see. So my art is a constant here. We have 8.314 joules per mole times cape. And then my temperature, we have solved that to be 720, Kelvin's. So putting that into my calculator. I get that like a constant. So again, I'm putting this part into my calculator, I'll get K. is equal to 9.804 times 10 to the -9. And if you're calling the beginning, where he said that the cake value is also equal to the pressure for covered dioxide. And just writing out our final answer then with the correct six figs. Is that P or the pressure of the carbon dioxide is equal to 9.80 times 10 to the negative nine unit being a t. M. So then this answer right over here is going to be the final answer for this problem. Thank you all so much for watching.

Ch.19 - Chemical Thermodynamics

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Brown 15th Edition

Ch.19 - Chemical Thermodynamics

Problem 81a

Chapter 19, Problem 81a

Consider the decomposition of barium carbonate: BaCO₃(s) ⇌ BaO(s) + CO₂(g) Using data from Appendix C, calculate the equilibrium pressure of CO₂ at (a) 298 K.

Verified step by step guidance

Identify the reaction: BaCO_3(s) \rightleftharpoons BaO(s) + CO_2(g).

Use the standard Gibbs free energy change (\Delta G^\circ) for the reaction to find the equilibrium constant (K) at 298 K. \Delta G^\circ = \sum \Delta G^\circ_{\text{products}} - \sum \Delta G^\circ_{\text{reactants}}.

Calculate \Delta G^\circ using the values from Appendix C for BaCO_3, BaO, and CO_2.

Relate \Delta G^\circ to the equilibrium constant using the equation \Delta G^\circ = -RT \ln K, where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin.

Solve for K, which is equal to the equilibrium pressure of CO_2, since the reaction involves only one gaseous product.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In the context of the decomposition of barium carbonate, the system reaches equilibrium when the amount of BaCO3 decomposing into BaO and CO2 is balanced by the amount of CO2 reacting back to form BaCO3.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) for a gaseous reaction is defined in terms of the partial pressures of the products and reactants at equilibrium. For the decomposition of BaCO3, Kp can be expressed as the ratio of the partial pressure of CO2 to the concentration of BaCO3 and BaO, allowing for the calculation of the equilibrium pressure of CO2 at a given temperature.

Ideal Gas Law

The Ideal Gas Law (PV=nRT) relates the pressure, volume, temperature, and number of moles of a gas. In this scenario, it can be used to determine the pressure of CO2 at equilibrium by rearranging the equation to solve for pressure (P) when the number of moles and temperature are known, providing a practical approach to finding the equilibrium pressure in the reaction.