Consider this reaction occurring at 298 K:
BaCO3(s) ∆ BaO(s) + CO...

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?

Accepted Answer

Hi everyone here we have a question asking us to calculate the partial pressure of carbon dioxide. When the decomposition of calcium carbonate in an evacuated flask reaches equilibrium at 298 kelvin hint, Use the free energy of the reaction calcium carbonate solid forms calcium oxide solid plus carbon dioxide gasses at equilibrium. Q will equal K. So are Step one is to calculate our free energy for calcium carbonate That is negative 1,129. For calcium oxide. That is negative 603. for carbon dioxide, That is -394.4. And our change in free energy of our reaction equals the change in free energy of our product minus the change in free energy of our reactant. So in this case, our change in free energy for our reaction is going to equal one mole of calcium oxide Times negative. 603 0. plus one mole of carbon dioxide Times -394. mine is one mole of calcium carbonate Times negative. 1129. And that equals 131.4 killed jules. Our next step is to calculate R. K. So our free energy of our reaction equals negative our times temperature times the natural log of K. So K equals E. To the power of negative Free energy divided by R. T. So E. To the negative 31.4 kg jewels, Times one, Times 10 to the 3rd rules Over one kill a jewel divided by 8. jules per mole times kelvin Times 298 Kelvin Equals 9.26, 5 times 10 to the -24. And our last step, our pressure of carbon dioxide equals K Equals 9.26, 5 times 10 to the - atmosphere equals 9.27 times 10 to the -24 atmosphere. And that's just rounding up. And that is our final answer. Thank you for watching. Bye.

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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