Consider the following three reactions:
(i) Ti1s2 + 2 Cl21g2¡TiCl...

Question

Consider the following three reactions:
(i) Ti1s2 + 2 Cl21g2¡TiCl41g2
(ii) C2H61g2 + 7 Cl21g2 ¡2 CCl41g2 + 6 HCl1g2
(iii) BaO1s2 + CO21g2¡BaCO31s2
(c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

Accepted Answer

Hello everyone. So in this video we're trying to calculate for the change in free energy of this reaction right over here and were given a table with the delta H and delta S values. So first of all we want to recognize that if we're trying to calculate the change in free energy, we're gonna go ahead and actually use the Gibbs free energy equation. If we recall that equation is that delta G. Is equal to the delta H minus temperature, multiplied by our delta S. So my delta H. Is equal to the delta H. Of my products, minus my delta age of my reactant. So in my table and the given reaction that's going to be negative 4 71.5 minus negative 2.1 plus two. So put that into my calculator, I'll get the value for delta H. To be negative 469.4 kg jewels. As for my delta S. That's going to be of again the products minus of the reactant since and again using this table. And my given reaction here, we can go ahead and fill in the numerical values. That's going to be 3 65.8 minus 44.1 plus two times our 2 to 3.1. And the reason why we have this to multiplication is because we have two moles here. Alright, so plugging this into my calculator, I'll get the numerical value of negative 1 to 4. units being jewels per k kelvin's. Alright, let's go ahead and scroll down to give us a little bit more space. So again we're going to go ahead and use this Gibbs free energy equation and one last thing we need to go ahead and do calculations for it before we plug it into this equation here is for my T. Value which is going to be my temperature. So we're going to convert my degree Celsius into kelvin's. So we have 25 degrees Celsius. We're gonna go ahead and add this with 700 or 273.15 to give us 98.15 kelvin's. So we're just converting this 25 degrees Celsius into kelvin's here. Alright now we can go ahead and plug in our values. So again we have delta H. That's what we're solving for. We're plugging in our delta H. Value that we solve for negative four or negative 469.4 kg jewels. Want to convert this actually to be in jewels. Every 1000 joules is equal to one. Kill a jewel. And we're going to subtract this by while T. Is a value 298.15 Calvin's. We're multiplying this by our delta s. That we are said to be negative 124. jewels per kelvin so we can see that are killed jules will cancel. And then over here we have the calvins counseling. So basically we're subtracting of the same units and that's exactly what we want and that's why we converted this kill jewels into jewels. So then we can say that the delta G. Then is going to be equal to negative 432280 jewels. We want to actually convert this into killer jewels again. So for every one, killer jewel will have 1000 jewels. As you can see, Jules unit will cancel out. Giving us the Delta G. Is equal to negative 432.28 kill jules. So we can conclude that the relationship of the regions is we have, well, we have negative values of delta G. We have negative values of delta H. As well as negative delta S. So as T increases as T increases, delta G will be more positive. So our final answer, our reform answer is that the change in free energy of the reaction increases as temperature or temp for short increases. All right. And this right here is going to be my final answer for this problem. Thank you all so much for watching

Consider the following three reactions: (i) Ti1s2 + 2 Cl21g2¡TiCl41g2 (ii) C2H61g2 + 7 Cl21g2 ¡2 CCl41g2 + 6 HCl1g2 (iii) BaO1s2 + CO21g2¡BaCO31s2 (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

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