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)
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?

Accepted Answer

Hey everyone today we're being asked if raising or lowering the temperature of the reaction from strontium carbonate to form strontium oxide and carbon dioxide would make the reaction more spontaneous. And if it does increase the spontaneity, what the temperature would be when the partial pressure of carbon dioxide is one atmosphere. So to do this we first need to take a look at what the delta H. Or standard entropy and the delta S. Or standard entropy of the reaction will be. And we can do this using the standard entropy and entropy of formation. Remember the standard entropy and or entropy of formation. Our experimental values that are calculated for different compounds and can be found in tables either online or in a textbook. So going ahead, taking a look first at the delta H. We can use the uh equation that the standard entropy will be equal to the standard entropy of formation of the product minus the standard entropy of formation of the reactant. So we can write that out as So for the sake of brevity I will just add them as we go along. But for one mole of strong team oxide Which has a standard entropy of formation of - killer joules per mole Plus the other product which is one mole of carbon dioxide c. 0. 2 Which has a standard entropy of formation of -393 93.5 kg jewels, Permal minus the one mole of strontium carbonate, one mole of S. R. C. 03 which has a centered entropy of formation of negative 1220.1 kg joules per mole which gives us a final value Of positive 234. killer whales. We can use the same method for the entropy as well. Using the same formula. Even the standard entropy of the form of the reaction will be the standard entropy of formation of the product minus the standard entropy of formation of the reactant reactant. Yeah, so we can go ahead and use a similar setup. One mole of strontium oxide which has a standard entropy of 54.4 killed jewels per mole Kelvin 54. jules Permal Calvin plus one more Carbon dioxide which has a standard entropy of formation of 213. jewels for Mole Kelvin minus one mole of strontium oxide. Or sorry, strontium carbonate Which has a standard entropy of formation of or sorry, 97.1 my bad 97.1 jules per mole kelvin which gives us a final value of 171.1 Jewels Per Kelvin. However, to find out the Delta G, which we're going to do very shortly, we also need to convert jewels to kill a jewels because Delta G is calculated in killer jules. This is done very easily with a transcription or a conversion factor. Remember that one killer jewel is equal to 10 To the 3rd jewels. So these cancel out leaving us with 0. Kill it jules per kelvin. So looking at this we can see that both the standard entropy and standard entropy are both positive. What this means is when we put this into the formula that delta G. The standard delta G. Of the reaction is equal to delta H. Plus or sorry, delta H. The reaction minus T. The temperature times the standard entropy of the reaction. If both Delta S. H. And delta S are positive, then the only way for the reaction to become spontaneous would be if temperature was very high because think about it, assume the temperature is just an arbitrary value. Say one and we have two positives. Well a positive minus a positive. Like let's assume temperature is just one kelvin for now even though that's absurd. But if we just take 234 kila jewels as an entropy minus the 0. Killer Jules well will have a positive value which means the reaction will not be spontaneous. So we need very high temperature. Very high temperature is needed which means the temperature needs to be raised in order to make the reaction spontaneous. So since this is the case, let's go ahead and find out what the temperature would need to be. If carbon had a partial pressure or carbon dioxide, sorry had a partial pressure of one atmosphere. So if it had a partial pressure of one atmosphere, then that means that we are at standard temperature and pressure. Well, not standard temperature, but we're at standard pressure. Because one atmosphere is standard conditions, since we'd be at standard conditions, if we want the Delta G to be spontaneous, if you want the reaction to be spontaneous, we can actually equate this equation 20. So assume delta G is zero As such, we'll get that. zero is equal to 2 34.6 Killer Jewels minus T. Which we still have yet to find times The 0.1711 killer jewels per Kelvin. And solving for T. We get that tea or temperature is equal to 1,371 Kelvin. Which means that if we want the reaction to be spontaneous and at uh 1 80 M, when carbon dioxide has a partial pressure of 1 80 M, we will need A temperature of 1,371 Kelvin or higher. In order for the reaction to be spontaneous. I hope this helps. And I look forward to seeing you all in the next one

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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