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Problem 90

Consider this reaction occurring at $298 \mathrm{K} :$
$$\mathrm{BaCO}_{3}(s) \Longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)$$
$$\begin{array}{l}{\text { a. Show that the reaction is not spontaneous under standard }} \\ {\text { conditions by calculating } \Delta G_{\mathrm{rxn} \text { . }}^{\circ} .} \\ {\text { b. If } \mathrm{BaCO}_{3} \text { is placed in an evacuated flask, what is the partial }} \\ {\text { pressure of } \mathrm{CO}_{2} \text { when the reaction reaches equilibrium? }} \\ {\text { c. Can the reaction be made more spontaneous by an increase }} \\ {\text { or decrease in temperature? If so, at what temperature is the }} \\ {\text { partial pressure of carbon dioxide } 1.0 \text { atm? }}\end{array}$$

PART $\mathrm{A} : \Delta G^{\circ}=219.7 \mathrm{kJ}$
PART $\mathrm{B} : P_{C O_{2}}=3.1 \times 10^{-39} \mathrm{atm}$
PART C: spontaneous at higher temperatures and $T=1562 \mathrm{K}$

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## Video Transcript

So this is a multi step problem. And the first part of this problem is wanting you to show the dope G. This reaction is non spontaneous. So the reaction is barium carbonate B A c A three as a solid going to bury him outside, which is also solid and also making CIA to gas. And so you can calculate this using the formula listed below. Where don't g ve reaction is, um, the Gibbs for energy of your product. Amount of skip gives for energy of your reactant ce accounting for the mole's every products and reactions. For this problem, all the moles are one, so you can just ignore the ends. And so now we just plug in numbers that were confined in our appendix of our textbook, which I've included up here. And so we have Dr Jeannot equals your products Master act. It's too. Let's add up the products. So for barium oxide, you have minus about 20.3 close Negative. 3 94 0.4 Modest. You're acting. It's a minus negative 11334 point for for your barium carbonate. And so this comes out to be to 19.7 killed Jules Promote. And so that's the first part of this problem. And since those numbers positive, this reaction is not spontaneous. Next part of this problem is trying to figure out what the partial pressure would be at equilibrium under nonstandard state commissions. So the question we used for this part is don't G equals, don't g? Not so the standard state gives free of the reaction plus Artie Island of Cuba, where Q is equal to partial pressure of CEO, too, and atmospheres and so at equilibrium. Doctor Jeez, equal to zero because everything's at a standstill. This number we nose to 19 killed Jules Promote plus R, which is 8.314 jewels per Kelvin Moore. It's a constant given to you times the temperature. So the room temperatures to 98 kelvin times natural log You and so one thing we need to keep track of is that this is an killer. Jules, this is n jul somebody to do unit conversion. So 2 19 terms 1000 negative to 19 because you subtract to get on the left side, divided by 24 777.5 70. And so that equals natural log a cue, so I do to get the key with raised both sides to e together a natural log. So now Q equals three point Oh, no times 10 to the minus 39 atmospheres says part. Be Part C is wanting to know whether or not you want to go to higher or lower temperature to make distraction more spontaneous. So when you think what's happening, Q will always be less than one one or less than one, which will give this make this value negative overall. So in this study is positive, so you won't this value to be larger than this one to make it more spontaneous, So we want temperature to increase. So if reports see temperature increases, more spontaneous became. The last part is, question is when you know what temperature is. The reaction at one atmosphere of CO two so we can sell for that Dr G equals Delta T Night plus R T. Ellen of Q. In this case, accused equal the one. So there's actually goes all the way because the natural order of 10 and so the Delta G because Dr Jeannot, that means it is that one atmosphere understanding state conditions. Call 25 degrees Celsius or 2 98 degrees she and the Eight Hilton, and that's the last part for Part C.