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Question 20 Not yet answeredMarked out of 1.0 Flag questionYou are given the following data: Hzig) ~2Hig) AH" = 872.8 kJlmol Hzig) Clz(g) -2HClig) AHo = -7.24 ...

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Question 20 Not yet answeredMarked out of 1.0 Flag questionYou are given the following data: Hzig) ~2Hig) AH" = 872.8 kJlmol Hzig) Clz(g) -2HClig) AHo = -7.24 x104 Jmol Clz(g) ~2Clig) AHo = 582 kJlmol Calculate 4Ho for the reaction Hig) + Clig) ~HClig)Select one: 645.2 kJlmolb. -727.4 kJmol-763.6 kJmold. -350.7 kJlmol

Question 20 Not yet answered Marked out of 1.0 Flag question You are given the following data: Hzig) ~2Hig) AH" = 872.8 kJlmol Hzig) Clz(g) -2HClig) AHo = -7.24 x104 Jmol Clz(g) ~2Clig) AHo = 582 kJlmol Calculate 4Ho for the reaction Hig) + Clig) ~HClig) Select one: 645.2 kJlmol b. -727.4 kJmol -763.6 kJmol d. -350.7 kJlmol



Answers

At $298 \mathrm{K}, \Delta_{\mathrm{f}} G^{\circ}[\mathrm{CO}(\mathrm{g})]=-137.2 \mathrm{kJ} \mathrm{mol}^{-1}$ and $K=$ $6.5 \times 10^{11}$ for the reaction $\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons$ $\mathrm{COCl}_{2}(\mathrm{g}) . \quad$ Use these data to determine $\Delta_{\mathrm{f}} G^{\circ}\left[\mathrm{COCl}_{2}(\mathrm{g})\right],$ and compare your result with the value in Appendix D.

To calculate the standard delta H. For a given reaction, we simply need to look up the standard delta H. Is the formation of the products multiplied by the coefficient minus the standard delta formation of the reactions. So the standard delta H. For this reaction is going to be Delta formation of N 20 plus two times the delta H. A. Formation of H +20. Liquid minus the delta formation of ammonium nitrate. Giving us 1 23 95 killer jewels. For this reaction it will be the delta H. Of formation of F. F. E. 203 solid. Which is this value Plus four times the delta formation of so two minus two times the delta H. Formation of FPs too, And 11/2 that of oxygen which is zero. Giving us -1656 4 is the delta standard for this reaction. Then going to appendix K. We'll do the same thing. Will take the delta H. A. Formation of S. I. C. Solid found in the back of the book, in appendix K. And then two times the delta formation of carbon monoxide and then minus the delta of formation of a. C. 02 And three times the delta formation of carbon graphite. Giving us Delta H. for this reaction of 624.6 killer jewels

All right, So for part A we'll have uh the progress of the reaction on the X axis and then the energy on the uh Y axis. And we'll start with some arbitrary amount. We'll go up always for the activation energy and then there's a negative sign. So the reaction is exact karmic. So the products have to be lower in energy and then this is not to scale. But um let's do it in a different color. The difference between the reactant and the peak here is the activation energy which in this case is 75 kilometers per mole. And then the delta H. Is the difference between the reactions and the products And in this case -145 killer jewels. Herbal. Yeah. Alright. For part B. It's also an exit thermic reaction. So it would have the same general shape. But we would just have to change the values. So label this be Okay, we'll just change these. So the activation energy is told to be 65 kg per mole. And delta H is negative 70. Alright. For part C. Um It's end of term because delta is positive. The products have to be higher up than the reactant. Since yeah, progress here. Which can also be reaction coordinate. So we'll start with some arbitrary amount. Then we'll have our activation energy and then the product will be up here. Yeah. Okay, So this is going to be 70. None of these are to scale 70 km from all for the activation energy. And then Yeah, okay, I'm sorry, 85 And then 70 kg per mole. Okay, This is feeling, uh, for the Delta H we're done.

In this problem, I can write DDX in S three minus plus B i three plus changes into Beihai multiplication food to end to the power minus to change it into two and not plus 40 minus multiplication three. Also, I can write the value of any. The call to pool B minus. The value of Dell G note is equal to minus tool multiplication 96500 multiplication 965000.55, which is equal to minus 637 L. A. Jewel, but more according to the option. Option A is correct of some age, correct and said for this problem.

From to calculate delta G. Standard. For the process we first need to calculate delta H. Standard and delta S. Standard for the process. Given the values in the table, The reaction or processes to a plus B goes to three C. So Delta H standard is going to be equal to three times delta h. of formation of C -2 times delta formation of a and one times delta formation of B. Giving us negative 1 43 8 killer jewels. Delta standard will be very similar. It'll be three times the standard S. Of C -2 times the standard s. of a. And one times the standard s of B. All these values are provided in the table. Given us negative 296 Jewels Per Kelvin. So delta G then will be equal to delta H minus T. Delta S. Remember to convert the delta asked from the jewels per kelvin to kill jules per kelvin. So we get so that with the killer killer jewel units here we can calculate kill julie units on Delta G standard and we get negative 956 killed jules. The answer then, is a


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